The primary source of scattered radiation is the

A patient.

B tabletop.

C x-ray tube.

D grid.

The Correct Answer is: A
The scatterer between the target and the image recorder is the patient. After the radiation has scattered once, it has been significantly attenuated. The intensity of scattered radiation 1 m from the patient is approximately 0.1% of the intensity of the primary beam. (Bushong, p 552)

x-ray photon beam attenuation is influenced by

1. tissue type.

2. subject thickness.

3. photon quality.

A 1 only B 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
Attenuation (decreased intensity through scattering or absorption) of the x-ray beam is a result of its original energy and its interactions with different types and thicknesses of tissue. The greater the original energy/quality (the higher the kilovoltage) of the incident beam, the less the attenuation. The greater the effective atomic number of the tissues (tissue type determines absorbing properties), the greater the beam attenuation. The greater the volume of tissue (subject density and thickness), the greater the beam attenuation. (Bushong, p 185)

If a duration of 0.05 second was selected for a particular exposure, what milliamperage would be necessary to produce 30 mAs?

A 900

B 600

C 500

D 300

he Correct Answer is: B
The formula for mAs is mA × s = mAs. Substituting known values:

(Selman, 9th ed., p. 214)

Recommended method(s) of minimizing motion unsharpness include

1. suspended respiration
2. short exposure time
3. patient instruction

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D
The shortest possible exposure time should be used to minimize motion unsharpness. Motion causes unsharpness that destroys detail. Careful and accurate patient instruction is essential for minimizing voluntary motion. Suspended respiration eliminates respiratory motion. Using the shortest possible exposure time is essential to decreasing involuntary motion. Immobilization also can be useful in eliminating motion unsharpness. (Selman, 9th ed., p. 210)

If 300 mA has been selected for a particular exposure, what exposure time should be selected to produce 18 mAs?

A 40 ms B 60 ms C 400 ms D 600 ms

he Correct Answer is: B
The exposure factor that regulates radiographic density is milliampere-seconds (mAs). The equation used to determine mAs is mA × s = mAs. Substituting known factors:

(Selman, 9th ed., p. 214)

The radiographic accessory used to measure the thickness of body parts in order to determine optimum selection of exposure factors is th

e A fulcrum.

B caliper.

C densitometer.

D ruler.

he Correct Answer is: B
Radiographic technique charts are highly recommended for use with every x-ray unit. A technique chart identifies the standardized factors that should be used with that particular x-ray unit, for various examinations/positions, of anatomic parts of different sizes. To be used effectively, these technique charts require that the anatomic part in question be measured correctly with a caliper.

A fulcrum is of importance in tomography, a densitometer is used in sensitometry and QA. (Ballinger & Frank, vol 3, p 237)

Why is a very short exposure time essential in chest radiography?

A To avoid excessive focal-spot blur

B To maintain short-scale contrast

C To minimize involuntary motion

D To minimize patient discomfort

The Correct Answer is: C
Radiographers usually are able to stop voluntary motion using suspended respiration, careful instruction, and immobilization. However, involuntary motion also must be considered. To have a “stop action” effect on the heart when radiographing the chest, it is essential to use a short exposure time. (Fauber, pp. 87–88)

Which of the following can affect histogram appearance?

1. Centering accuracy
2. Positioning accuracy
3. Processing algorithm accuracy

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: D
The computed radiography (CR) laser scanner recognizes the various tissue-density values and constructs a representative grayscale histogram. A histogram is a graphic representation showing the distribution of pixel values. Histogram analysis and use of the appropriate LUT together function to produce predictable image quality in CR. Histogram appearance can be affected by a number of things. Degree of accuracy in positioning and centering can have a significant effect on histogram appearance (as well as patient dose). Change is effected in average exposure level and exposure latitude; these changes will be reflected in the images informational numbers (i.e., S number and exposure index). Other factors affecting histogram appearance, and therefore these informational numbers, include selection of the correct processing algorithm (e.g., chest vs. femur vs. cervical spine) and changes in scatter, SID, OID, and collimation. Figure 7–21 illustrates the effect of incorrect collimation on histogram appearance—in short, anything that affects scatter and/or dose. (Carlton and Adler, 4th ed., pp. 361–

For which of the following examinations might the use of a grid not be necessary in an adult patient?

A Hip

B Knee

C Abdomen

D Lumbar spine

The Correct Answer is: B
The abdomen is a thick structure that contains many structures of similar density, and thus it requires increased exposure and a grid to absorb scattered radiation. The lumbar spine and hip are also dense structures requiring increased exposure and use of a grid. The knee, however, is frequently small enough to be radiographed without a grid. The general rule is that structures measuring more than 10 cm should be radiographed with a grid. (Bontrager and Lampignano, 6th ed.

Tree-like branching black marks on a radiograph are usually the result of

A bending the film acutely

B improper development

C improper film storage

D static electricity

The Correct Answer is: D
X-ray film is sensitive and requires proper handling and storage. Several kinds of artifacts can be produced by careless handling during production of the radiographic image. Tree-like, branching black marks on a radiograph usually are due to static electrical discharge. Problems with static electricity are especially prevalent during cold, dry weather and can be produced simply by removing a sweater in the darkroom. (Selman, 9th ed., p. 197)

Involuntary motion can be caused by 1. peristalsis. 2. severe pain. 3. heart muscle contraction.

A 1 only

B 2 only

C 1 and 2 only

D 1, 2, and 3

he Correct Answer is: D
Patients who are able to cooperate are usually able to control voluntary motion. However, certain body functions and responses create involuntary motion that is not controllable by the patient. Severe pain, muscle spasm, and chills all cause involuntary movements. Peristaltic activity of the intestinal tract and motion caused by contraction of the heart muscle are other sources of involuntary motion. (Ballinger & Frank, vol 1, p 12)

Which of the following absorbers has the highest attenuation coefficient?

A Bone

B Muscle

C Fat

D Air

he Correct Answer is: A
The radiographic subject, the patient, is composed of many different tissue types that have varying tissue densities, resulting in varying degrees of photon attenuation and absorption. The atomic number (Z) of the tissues under investigation is directly related to its attenuation coefficient. This differential absorption contributes to the various shades of gray (scale of radiographic contrast) on the finished x-ray image. Air has an effective Z number of 7.78, fat is about 6.46, water is 7.51, muscle is 7.64, and bone is 12.31. (Carlton and Adler, 3rd ed., p. 249)

The variation in photon distribution between the anode and cathode ends of the x-ray tube is known as

A the line focus principle.

B the anode heel effect.

C the inverse square law.

D Bohr's theory.

The Correct Answer is: B
Because the focal spot (track) of an x-ray tube is along the anode's beveled edge, photons produced at the target are able to diverge toward the cathode end of the tube, but are absorbed by the "heel" of the anode at the opposite anode end of the tube. This results in a greater number of x-ray photons distributed toward the cathode end and is known as the anode heel effect. The line focus principle is a geometric principle illustrating that the effective focal spot is always smaller than the actual focal spot. The inverse square law of radiation deals with the relationship between distance and radiation intensity. Bohr's theory refers to an atom's resemblance to the solar system. (Selman, p 253)

If 300 mA has been selected for a particular exposure, what exposure time would be required to produce 6 mAs?

A 5 ms

B 10 ms

C 15 ms

D 20 ms

he Correct Answer is: D
Milliampere-seconds (mAs) is the exposure factor that regulates radiographic density. The equation used to determine mAs is mA × s = mAs. Substituting the known factors:

(Fauber, p. 55)

the direction of electron travel in the x-ray tube is

A filament to cathode

B cathode to anode

C anode to focus

D anode to cathode

The Correct Answer is: B
The x-ray tube is a diode tube; that is, it has two electrodes—a negative and a positive. The cathode assembly is the negative terminal of the x-ray tube, and the anode is the positive terminal. Electrons are released by the cathode filament (thermionic emission) as it is heated to incandescence. When kilovoltage is applied, the electrons are driven across to the anode's focal spot. Upon sudden deceleration of electrons at the anode surface, x-rays are produced. Hence, electrons travel from cathode to anode within the x-ray tube. (Bushong, 9th ed., pp. 122-125)

The reduction in x-ray photon intensity as the photon passes through material is termed

A absorption

B scattering

C attenuation

D divergence

he Correct Answer is: C
Absorption occurs when an x-ray photon interacts with matter and disappears, as in the photoelectric effect. Scattering occurs when there is partial transfer of energy to matter, as in the Compton effect. The reduction in the intensity of an x-ray beam as it passes through matter is called attenuation. (Bushong, 8th ed., p. 185)

the radiographic accessory used to measure the thickness of body parts in order to determine optimal selection of exposure factors is the

A fulcrum

B caliper

C densitometer

D ruler

he Correct Answer is: B
Radiographic technique charts are highly recommended for use with every x-ray unit. A technique chart identifies the standardized factors that should be used with that particular x-ray unit for various examinations/positions of anatomic parts of different sizes. To be used effectively, these technique charts require that the anatomic part in question be measured correctly with a caliper. A fulcrum is of importance in tomography; a densitometer is used in sensitometry and QA. (Bushong, 8th ed., p. 308)

Which of the following matrix sizes is most likely to produce the best image resolution?

A 128 × 128

B 512 × 512

C 1,024 × 1,024

D 2,048 × 2,048

The Correct Answer is: D
The matrix is the number of pixels in the xy direction. The larger the matrix size, the better is the image resolution. Typical image matrix sizes used in radiography are

A digital image is formed by a matrix of pixels in rows and columns. A matrix having 512 pixels in each row and column is a 512 × 512 matrix. The term field of view is used to describe how much of the patient (e.g., 150-mm diameter) is included in the matrix. The matrix or field of view can be changed without affecting the other, but changes in either will change pixel size. As in traditional radiography, spatial resolution is measured in line pairs per millimeter (lp/mm). As matrix size is increased, there are more and smaller pixels in the matrix and, therefore, improved spatial resolution. Fewer and larger pixels result in a poor-resolution “pixelly” image, that is, one in which you can actually see the individual pixel boxes. (Fosbinder and Kelsey, p. 286)

How are mAs and radiographic density related in the process of image formation?

A mAs and radiographic density are inversely proportional

B mAs and radiographic density are directly proportional

C mAs and radiographic density are related to image unsharpness

D mAs and radiographic density are unrelated

The Correct Answer is: B
Radiographic density is described as the overall degree of blackening of a radiograph or a part of it. The milliampere-seconds value regulates the number of x-ray photons produced at the target and thus regulates radiographic density. If it is desired to double the radiographic density, one simply doubles the milliampere-seconds; therefore, milliampere-seconds and radiographic density are directly proportional. (Selman, 9th ed., p. 214)

Which of the following chemicals is used in the production of radiographic film emulsion?

A Sodium sulfite

B Potassium bromide

C Silver halide

D Chrome alum

The Correct Answer is: C
Film emulsion consists of silver halide crystals suspended in gelatin. Sodium sulfite is a film-processing preservative, and potassium bromide is a developer restrainer. Potassium and chrome alum are emulsion hardeners used in fixer solution. (Selman, 9th ed., p. 174)

A grid is usually employed 1. when radiographing a large or dense body part. 2. when using high kilovoltage. 3. when less patient dose is required.

A 1 only B 3 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: C
Significant scattered radiation is produced when radiographing large or dense body parts and when using high kilovoltage. A radiographic grid is made of alternating lead strips and interspace material; it is placed between the patient and the IR to absorb energetic scatter emerging from the patient. Although a grid prevents much scattered radiation fog from reaching the radiograph, its use does necessitate a significant increase in patient exposure. (Shephard, pp 244–245)

The line-focus principle expresses the relationship between

A the actual and the effective focal spot

B exposure given the IR and resulting density

C SID used and resulting density

D grid ratio and lines per inch

The Correct Answer is: A
The line-focus principle is a geometric principle illustrating that the actual focal spot is larger than the effective (projected) focal spot. The actual focal spot (target) is larger, to accommodate heat over a larger area, and is angled so as to project a smaller focal spot, thus maintaining recorded detail by reducing blur. The relationship between the exposure given the IR and the resulting density is expressed in the reciprocity law; the relationship between the SID and resulting density is expressed by the inverse-square law. Grid ratio and lines per inch are unrelated to the line-focus principle. (Selman, p. 138; Shephard, pp. 218–219).

Distortion can be caused by

1. tube angle
2. the position of the organ or structure within the body
3. the radiographic positioning of the part

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
Distortion is caused by improper alignment of the tube, body part, and IR. Anatomic structures within the body are rarely parallel to the IR in a simple recumbent position. In an attempt to overcome this distortion, we position the part to be parallel with the IR or angle the central ray to “open up” the part. Examples of this technique are obliquing the pelvis to place the ilium parallel to the IR or angling the central ray cephalad to “open up” the sigmoid colon. (Shephard, pp. 228–234)

Which of the following pathologic conditions would require an increase in exposure factors?

A Pneumoperitoneum

B Obstructed bowel

C Renal colic

D Ascites

he Correct Answer is: D
Because pneumoperitoneum is an abnormal accumulation of air or gas in the peritoneal cavity, it would require a decrease in exposure factors. Obstructed bowel usually involves distended, air- or gas-filled bowel loops, again requiring a decrease in exposure factors. With ascites, there is an abnormal accumulation of fluid in the abdominal cavity, necessitating an increase in exposure factors. Renal colic is the pain associated with the passage of renal calculi; no change from the normal exposure factors is usually required. (Carlton & Adler, p 25

If 300 mA has been selected for a particular exposure, what exposure time would be required to produce 60 mAs?

A 1/60 second

B 1/30 second

C 1/10 second

D 1/5 second

The Correct Answer is: D
The mAs is the exposure factor that regulates radiographic density. The equation used to determine mAs is mA × s = mAs. Substituting the known factors,

300x = 60

x = 0.2 (1/5) second

(Fauber, p 55)

Which of the following devices is used to overcome severe variation in patient anatomy or tissue density, providing more uniform radiographic density?

A Compensating filter

B Grid

C Collimator

D Intensifying screen

he Correct Answer is: A
A compensating filter is used when the part to be radiographed is of uneven thickness or density (in the chest, mediastinum vs. lungs). The filter (made of aluminum or lead acrylic) is constructed in such a way that it will absorb much of the primary radiation that would expose the low-tissue-density area while allowing the primary radiation to pass unaffected to the high-tissue-density area. A collimator is used to decrease the production of scattered radiation by limiting the volume of tissue irradiated. The grid functions to trap scattered radiation before it reaches the IR, thus reducing scattered radiation fog. (Selman, 9th ed., pp. 254–255)

Compression of the breast during mammographic imaging improves the technical quality of the image because

1. geometric blurring is decreased
2. less scattered radiation is produced
3. patient motion is reduced

A 1 only B 3 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: D
Compression of the breast tissue during mammographic imaging improves the technical quality of the image for several reasons. Compression brings breast structures into closer contact with the IR, thus reducing geometric blur and improving detail. As the breast tissue is compressed and essentially becomes thinner, less scattered radiation is produced. Compression serves as excellent immobilization as well. (Peart, p. 49)

Which of the following pathologic conditions would require an increase in exposure factors?

A Pneumoperitoneum

B Obstructed bowel

C Renal colic

D Ascites

The Correct Answer is: D
Because pneumoperitoneum is an abnormal accumulation of air or gas in the peritoneal cavity, it would require a decrease in exposure factors. Obstructed bowel usually involves distended, air- or gas-filled bowel loops, again requiring a decrease in exposure factors. With ascites, there is an abnormal accumulation of fluid in the abdominal cavity, necessitating an increase in exposure factors. Renal colic is the pain associated with the passage of renal calculi; no change from the normal exposure factors is usually required. (Carlton and Adler, 4th ed., p. 248)

Which of the following will produce the greatest distortion?

A AP projection of the skull

B PA projection of the skull

C 37° AP axial of the skull

D 20° PA axial of the skull

The Correct Answer is: C
Distortion is the result of misalignment of the x-ray tube, the anatomic part, and the IR. If these three parts are not parallel with one another, shape distortion occurs. The greater the misalignment, the greater the distortion. In the example cited, the image made with the greatest tube angle will produce the greatest distortion. Distortion is often introduced intentionally to visualize some structure to better advantage. The 37° (caudad) AP axial projection of the skull, for example, projects the facial bones inferiorly so that the occipital bone can be visualized to better advantage. (Shephard, pp 231–234)

A grid usually is employed in which of the following circumstances?

1. When radiographing a large or dense body part
2. When using high kilovoltage
3. When a lower patient dose is required

A 1 only B 3 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: C
Significant scattered radiation is generated within the part when imaging large or dense body parts and when using high kilovoltage. A radiographic grid is made of alternating lead strips and interspace material; it is placed between the patient and the IR to absorb energetic scatter emerging from the patient. Although a grid prevents much of the scattered radiation from reaching the radiograph, its use does necessitate a significant increase in patient exposure. (Bushong, 8th ed., p. 248)

Of the following groups of technical factors, which will produce the greatest radiographic density?

A 10 mAs, 74 kV, 44-in. SID

B 10 mAs, 74 kV, 36-in. SID

C 5 mAs, 85 kV, 48-in. SID

D 5 mAs, 85 kV, 40-in. SID

The Correct Answer is: B
If (A) and (B) are reduced to 5 mAs for consistency, the kilovoltage will increase to 85 kV in both cases, thereby balancing radiographic densities. Thus, the greatest density is determined by the shortest SID (greatest exposure rate). (Shephard, pp. 306–307)

In which of the following examinations would a cassette front with very low absorption properties be especially desirable?

A Extremity radiography

B Abdominal radiography

C Mammography

D Angiography

The Correct Answer is: C
Because mammographic techniques operate at very low kilovoltage levels, the cassette front material becomes especially important. The use of soft, low-energy x-ray photons is the underlying principle of mammography; any attenuation of the beam would be most undesirable. Special plastics that resist impact and heat softening, such as polystyrene and polycarbonate, are used frequently as cassette front material. (Shephard, p. 49)

Which of the following groups of exposure factors would be most appropriate to control involuntary motion?

A 400 mA, 0.03 second

B 200 mA, 0.06 second

C 600 mA, 0.02 second

D 100 mA, 0.12 second

he Correct Answer is: C
Control of motion, both voluntary and involuntary, is an important part of radiography. Patients are unable to control certain types of motion, such as heart action, peristalsis, and muscle spasm. In these circumstances, it is essential to use the shortest possible exposure time in order to have a “stop action” effect. (Carlton and Adler, 4th ed., p. 451)

The exposure factors of 400 mA, 17 ms, and 82 kV produce a milliampere-seconds value of

A 2.35 B 6.8 C 23.5 D 68

The Correct Answer is: B
To calculate milliampere-seconds, multiply milliamperage times exposure time. In this case, 400 mA × 0.017 second (17 ms) = 6.8 mAs. Careful attention to proper decimal placement will help to avoid basic math errors. (Shephard, p. 170)

graphic diagram of signal values representing various densities within the part being imaged is called a

A processing algorithm

B DICOM

C histogram

D window

he Correct Answer is: C
A histogram is a graph usually having several peaks and valleys representing the pixel values/absorbing properties of the various tissues, and so on that make up the imaged part. These various attenuators include such things as bone, muscle, air, contrast agents, foreign bodies, and pathology. The various pixel values, then, represent image contrast. If the histogram has a rather flat “tail,” this represents underexposed areas at the periphery of the image, which can skew the overall histogram analysis. The radiographer selects the particular processing algorithm on the computer/control panel that corresponds to the anatomic part and projection being performed. DICOM (Digital Imaging and Communications in Medicine) refers to the standard for communication between PACS and HIS/RIS systems. Windowing refers to the radiographer's postprocessing adjustment of contrast and density (at the workstation). (Shephard, pp. 341, 345)

Exposure factors of 90 kVp and 4 mAs are used for a particular nongrid exposure. What should be the new mAs if an 8:1 grid is added?

A 8

B 12

C 16

D 20

The Correct Answer is: C
To change nongrid to grid exposure or to adjust exposure when changing from one grid ratio to another, it is necessary to recall the factor for each grid ratio:

No grid = 1 × the original mAs

5:1 grid = 2 × the original mAs

6:1 grid = 3 × the original mAs

8:1 grid = 4 × the original mAs

12:1 (or 10:1) grid = 5 × the original mAs

16:1 grid = 6 × the original mAs

Therefore, to change from nongrid to an 8:1 grid, multiply the original mAs by a factor of 4. A new mAs of 16 is required. (Saia, p 328)

The major function of filtration is to reduce

A image noise.

B scattered radiation.

C operator dose.

D patient dose.

The Correct Answer is: D
X-rays produced at the target make up a heterogeneous primary beam. There are many "soft," low-energy photons that, if not removed, would contribute only to greater patient (skin) dose. They do not have enough energy to penetrate the patient and expose the film; they just penetrate a small thickness of the patient's tissue and are absorbed. These photons are removed by aluminum filters. (Fauber, pp 32–33)

In a PA projection of the chest being used for cardiac evaluation, the heart measures 15.2 cm between its widest points. If the magnification factor is known to be 1.3, what is the actual diameter of the heart?

A 9.7 cm

B 11.7 cm

C 19.7 cm

D 20.3 c

The Correct Answer is: B
The formula for magnification factor is MF = image size/object size. In the stated problem, the anatomic measurement is 15.2 cm, and the magnification factor is known to be 1.3. Substituting the known factors in the appropriate equation,

x = 11.69 cm (actual anatomic size)

(Fauber, pp 90–92)

If 85 kV and 20 mAs were used for a particular abdominal exposure with single-phase equipment, what mAs would be required to produce a similar radiograph with 3-phase, 12-pulse equipment.

A 40

B 25

C 20

D 10

The Correct Answer is: D
Single-phase radiographic equipment is much less efficient than three-phase equipment because it has a 100% voltage ripple. With three-phase equipment, voltage never drops to zero, and x-ray intensity is significantly greater. To produce similar density, only two thirds of the original mAs would be used for three-phase, six-pulse equipment (2/3 × 20 = 13 mAs). With 3-phase, 12-pulse equipment, the original mAs would be cut in half; thus, 10 mAs should be used. (Saia, pp 329, 330)

Which of the following is (are) classified as rare earth phosphors?

1. Lanthanum oxybromide
2. Gadolinium oxysulfide
3. Cesium iodide

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
Rare earth phosphors have a greater conversion efficiency than do other phosphors. Lanthanum oxybromide is a blue-emitting rare earth phosphor, and gadolinium oxysulfide is a green-emitting rare earth phosphor. Cesium iodide is the phosphor used on the input screen of image intensifiers; it is not a rare earth phosphor. (Shephard, p. 68)

Which of the following adult radiographic examinations usually require(s) use of a grid?

1. Ribs
2. Vertebrae
3. Shoulder

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D
Generally speaking, anatomic parts measuring in excess of 10 cm require a grid. (The major exception to this rule can be the chest). The larger the part, the more scattered radiation is generated. To avoid degradation of the image as a result of scattered radiation fog, a grid is used to absorb scatter. Parts generally requiring the use of a grid include the skull, spine, ribs, pelvis, shoulder, and femur. (Shephard, p. 245)

Which of the following conditions will require an increase in x-ray photon energy/penetration?

A Fibrosarcoma

B Osteomalacia

C Paralytic ileus

D Ascites

he Correct Answer is: D
The ability of x-ray photons to penetrate a body part has a great deal to do with the composition of that part (e.g., bone vs. soft tissue vs. air) and the presence of any pathologic condition. Pathologic conditions can alter the normal nature of the anatomic part. Some conditions, such as osteomalacia, fibrosarcoma, and paralytic ileus (obstruction), result in a decrease in body tissue density. When body tissue density decreases, x-rays will penetrate the tissues more readily; that is, there is more x-ray penetrability. In conditions such as ascites, where body tissue density increases as a result of the accumulation of fluid, x-rays will not readily penetrate the body tissues; that is, there is less x-ray penetrability. (Carlton and Adler, 4th ed., p. 250)

Because of the anode heel effect, the intensity of the x-ray beam is greatest along the

A path of the central ray

B anode end of the beam

C cathode end of the beam

D transverse axis of the IR

The Correct Answer is: C
Because the anode's focal track is beveled (angled, facing the cathode), x-ray photons can freely diverge toward the cathode end of the x-ray tube. However, the “heel” of the focal track prevents x-ray photons from diverging toward the anode end of the tube. This results in varying intensity from anode to cathode, with fewer photons at the anode end and more photons at the cathode end. The anode heel effect is most noticeable when using large IRs, short SIDs, and steep target angles. (Carlton and Adler, 4th ed., p. 407)

Angulation of the central ray may be required

1. to avoid superimposition of overlying structures.

2. to avoid foreshortening or self-superimposition.

3. to project through certain articulations.

A 1 only B 2 only C 1 and 3 only D 1, 2, and 3

he Correct Answer is: D
If structures are overlying or underlying the area to be demonstrated (eg, the medial femoral condyle obscuring the joint space in the lateral knee projection), central ray angulation is employed (eg, 5° cephalad angulation to see the joint space in the lateral knee). If structures would be foreshortened or self-superimposed (eg, the scaphoid in a PA wrist), central ray angulation may be employed to place the structure more closely parallel with the IR. Another example is the oblique cervical spine, where cephalad or caudad angulation is required to "open" the intervertebral foramina. (Frank, Long, and Smith, vol. 1, 11th ed., p. 307)

Which of the following groups of exposure factors will produce the most radiographic density?

A 100 mA, 50 ms

B 200 mA, 40 ms

C 400 mA, 70 ms

D 600 mA, 30 ms

he Correct Answer is: C
Milliampere-seconds (mAs) is the exposure factor that is used to regulate radiographic density. Using the equation milliamperage × time = mAs, determine each mAs: (A) = 5 mAs, (B) = 8 mAs, (C) = 28 mAs, (D) = 18 mAs. Group C will produce the most radiographic density. (Selman, 9th ed., p. 214)

Shape distortion is influenced by the relationship between the

1. x-ray tube and the part to be imaged
2. part to be imaged and the IR
3. IR and the x-ray tube

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D
Shape distortion is caused by misalignment of the x-ray tube, the part to be radiographed, and the IR/film. An object can be falsely imaged (foreshortened or elongated) by incorrect placement of the tube, the body part, or the IR. Only one of the three need be misaligned for distortion to occur. (Shephard, pp. 231–234)

Which of the following would be appropriate cassette front material(s)?

1. Tungsten
2. Magnesium
3. Bakelite

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: C
The cassette is used to support the intensifying screens and x-ray film. It should be strong and should provide good screen–film contact. The cassette front should be made of a sturdy material with a low atomic number because attenuation of the remnant beam is undesirable. Bakelite (the forerunner of today's plastics) and magnesium (the lightest structural metal) are the materials used most commonly for cassette fronts. The high atomic number of tungsten makes it inappropriate as a cassette front material. (Shephard, p. 41)

Cassette-front material can be made of which of the following?

1. Carbon fiber
2. Magnesium
3. Lead

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: B
The cassette–IR front material must not attenuate the remnant beam yet must be sturdy enough to withstand daily use. Bakelite has long been used as the material for tabletops and IR fronts, but now it has been replaced largely by magnesium and carbon fiber. Lead would not be a suitable material because it would absorb the remnant beam, and no image would be formed. (Shephard, p. 41)

An increase in kilovoltage will serve to

A produce a longer scale of contrast

B produce a shorter scale of contrast

C decrease the radiographic density

D decrease the production of scattered radiation

he Correct Answer is: A
An increase in kilovoltage increases the overall average energy of the x-ray photons produced at the target, thus giving them greater penetrability. (This can increase the incidence of Compton interaction and, therefore, the production of scattered radiation.) Greater penetration of all tissues serves to lengthen the scale of contrast. However, excessive scattered radiation reaching the IR will cause a fog and carries no useful information. (Selman, 9th ed., pp. 127–128)

Which of the following examinations might require the use of 70 kV?

1. AP abdomen
2. Chest radiograph
3. Barium-filled stomach

A 1 only B 2 only C 1 and 2 only D 2 and 3 only

The Correct Answer is: A
It is appropriate to perform an AP abdomen radiograph with lower kilovoltage because it has such low subject contrast. Abdominal tissue densities are so similar that it takes high- or short-scale contrast (using low kilovoltage) to emphasize the little difference there is between tissues. However, high-kilovoltage factors are used frequently to even out densities in anatomic parts having high tissue contrast (e.g., the chest). However, since high kilovoltage produces added scattered radiation, it generally must be used with a grid. Barium-filled structures frequently are radiographed using 120 kV or more to penetrate the barium—to see through to posterior structures (Carlton and Adler, 4th ed., pp. 423–424)

n which of the following examinations should 70 kV not be exceeded?

A Upper GI (UGI)

B Barium enema (BE)

C Intravenous urogram (IVU)

D Chest

The Correct Answer is: C
The iodine-based contrast material used in IVU gives optimal opacification at 60 to 70 kV. Use of higher kilovoltage will negate the effect of the contrast medium; a lower contrast will be produced, and poor visualization of the renal collecting system will result. GI and BE examinations employ high-kilovoltage exposure factors (about 120 kV) to penetrate through the barium. In chest radiography, high-kilovoltage technical factors are preferred for maximum visualization of pulmonary vascular markings made visible with long-scale contrast. (Saia, 4th ed., p. 347)

An exposure was made using 600 mA and 18 ms. If the mA is changed to 400, which of the following exposure times would most closely approximate the original radiographic density?

A 16 ms

B 0.16 second

C 27 ms

D 0.27 second

he Correct Answer is: C
Since 18 ms is equal to 0.018 s, and since mA × time = mAs, the original mAs was 10.8. Now it is only necessary to determine what exposure time must be used with 400 mA to provide the same 10.8 mAs (and thus the same radiographic density). Because mA × time = mAs,

400x = 10.8

x = 0.027 second (27 milliseconds)

(Selman, p 214)

Phosphors classified as rare earth include

1. lanthanum oxybromide.
2. gadolinium oxysulfide.
3. cesium iodide.

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
Rare earth phosphors have a greater conversion efficiency than do other phosphors. Lanthanum oxybromide is a blue-emitting phosphor, and gadolinium oxysulfide is a green-emitting phosphor. Cesium iodide is the phosphor used on the input screen of image intensifiers; it is not a rare earth phosphor. (Shephard, p. 66)

Which of the following contribute to the radiographic contrast present in the finished radiograph?

1. Tissue density
2. Pathology
3. Muscle development

A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
The radiographic subject (the patient) is composed of many different tissue types of varying densities, resulting in varying degrees of photon attenuation and absorption. This differential absorption contributes to the various shades of gray (scale of radiographic contrast) on the finished radiograph. Normal tissue density may be significantly altered in the presence of pathology. For example, destructive bone disease can cause a dramatic decrease in tissue density. Abnormal accumulation of fluid (as in ascites) will cause a significant increase in tissue density. Muscle atrophy or highly developed muscles similarly will decrease or increase tissue density. (Shephard, p. 203)

A technique chart should be prepared for each AEC x-ray unit and should contain which of the following information for each type of examination?

1. Photocell(s) used
2. Optimum kilovoltage
3. Backup time

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: D
The AEC automatically adjusts the exposure required for body parts that have different thicknesses and densities. Proper functioning of the phototimer depends on accurate positioning by the radiographer. The correct photocell(s) must be selected, and the anatomic part of interest must completely cover the photocell to achieve the desired density. If collimation is inadequate and a field size larger than the part is used, excessive scattered radiation from the body or tabletop can cause the AEC to terminate the exposure prematurely, resulting in an underexposed radiograph. Backup time always should be selected on the manual timer to prevent patient overexposure and to protect the x-ray tube from excessive heat production should the AEC malfunction. Selection of the optimal kilovoltage for the part being radiographed is essential—no practical amount of milliampere-seconds can make up for inadequate penetration (kilovoltage), and excessive kilovoltage can cause the AEC to terminate the exposure prematurely. A technique chart, therefore, is strongly recommended for use with AEC; it should indicate the optimal kilovoltage for the part, the photocells that should be selected, and the backup time that should be set. (Carlton and Adler, 4th ed., p. 540)

The factors that control recorded detail include 1. Focal spot size 2. Type of rectification 3. SID

A 1 and 2 only

B 1 and 3 only

C 2 and 3 only

D 1, 2, and 3

The Correct Answer is: B
Focal spot size affects recorded detail by its effect on focal spot blur: The larger the focal spot size, the greater the blur produced. Recorded detail is significantly affected by distance changes because of their effect on magnification. As SID increases, magnification decreases and recorded detail increases. The method of rectification has no controlling effect on recorded detail. Single-phase rectified units produce intermittent radiation at fluctuating voltage, whereas three-phase units produce almost constant potential. Single phase equipment exposures could require longer exposures, possibly resulting in motion unsharpness, though that equipment is seldom used today. (Bushong 10th ed p251)

What should be done to correct for magnification when using air-gap technique?

A Decrease OID

B Increase OID Decrease

SID D Increase SID

he Correct Answer is: D
OID is used to effect an increase in contrast in the absence of a grid, usually in chest radiography. If a 6-in. air gap (OID) is introduced between the part and the IR, much of the scattered radiation emitted from the body will not reach the IR; thus, the OID acts as a low-ratio grid and increases image contrast. However, the 6-in. OID air gap will make a very noticeable increase in magnification. To correct for this, the SID must be increased. Generally speaking, the SID needs to be increased 7 in. for every 1 in. of OID. With a 6-in. OID, the SID usually is increased from 6 to 10 ft (120 in.). (Shephard, pp. 263, 264)

If 84 kV and 8 mAs were used for a particular abdominal exposure with single-phase equipment, what milliampere-seconds value would be required to produce a similar radiograph with three-phase, 12-pulse equipment?

A 24 mAs B 16 mAs C 8 mAs D 4 mAs

he Correct Answer is: D
Single-phase radiographic equipment is much less efficient than three-phase equipment because it has a 100% voltage ripple. With three-phase equipment, voltage never drops to zero, and x-ray intensity is significantly greater. To produce similar density, only two-thirds of the original milliampere-seconds would be used for three-phase, six-pulse equipment ( 2 / ₃ × 8 = 5.3 mAs). With three-phase, 12-pulse equipment, the original milliampere-seconds would be cut in half ( 1 / ₂ × 8 = 4 mAs). (Bushong, 8th ed., p. 124)

OID is related to recorded detail in which of the following ways?

A Radiographic detail is directly related to OID.

B Radiographic detail is inversely related to OID.

C As OID increases, so does radiographic detail.

D OID is unrelated to radiographic detail.

he Correct Answer is: B
As the distance from the object to the IR (OID) increases, so does magnification distortion, thereby decreasing recorded detail. Some magnification is inevitable in radiography because it is not possible to place anatomic structures directly on the IR. However, our understanding of how to minimize magnification distortion is an important part of our everyday work. (Fauber, pp. 90–91)

An increase in the kilovoltage applied to the x-ray tube increases the

1. percentage of high-energy photons produced.
2. beam intensity.
3. patient absorption.

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: B
As the kilovoltage is increased, a greater number of electrons are driven across to the anode with greater force. Therefore, as energy conversion takes place at the anode, more high-energy photons are produced. However, because they are higher-energy photons, there will be less patient absorption. (Fauber, 2nd ed., p. 58)

Several types of exposure timers may be found on x-ray equipment. Which of the following types of timers functions to accurately duplicate radiographic densities?

A Synchronous

B Impulse

C Electronic

D Phototimer

The Correct Answer is: D
The synchronous timer is an older type of x-ray timer that does not permit very precise, short exposures. The impulse timer permits a shorter, more precise exposure, and the electronic timer may be used for exposures as short as 0.001 s. The photo timer, however, automatically terminates the exposure when the proper density has been recorded on the IR. The important advantage of the photo timer, then, is that it can accurately duplicate radiographic densities. It is, therefore, very useful for providing accurate comparison in follow-up examinations and for decreasing patient dose by decreasing the number of “retakes” required because of improper exposure. Remember that proper functioning of the photo timer depends on accurate positioning (and centering) by the radiographer. (Bushong, 8th ed., pp. 115–117)

What type of x-ray imaging uses an area beam and a photostimulable phosphor as the IR?

A Traditional radiography

B Computed radiography

C Digital radiography

D Cineradiography

he Correct Answer is: B
Traditional radiography uses an area x-ray beam, but the IR is film emulsion sandwiched between intensifying screens in a cassette. Computed radiography (CR) also uses an area x-ray beam, but the IR is a photostimulable phosphor such as europium-activated barium fluorohalide coated on an image plate. Digital radiography (DR) can use an area x-ray beam detected by a direct-capture solid-state device—there are no traditional cassette-like devices. DR can also use a fan-shaped x-ray beam. The fan-shaped beam is “read” by a linear array of detectors. (Bushong, 8th ed., pp. 401, 403

Shape distortion is influenced by the relationship between the

1. x-ray tube and the part to be imaged.
2. body part to be imaged and the IR.
3. IR and the x-ray tube.

A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: D
Shape distortion is caused by misalignment of the x-ray tube, the body part to be radiographed, and the IR. An object can be falsely imaged (foreshortened or elongated) as a result of incorrect placement of the tube, the part, or the IR. Only one of the three need be misaligned for distortion to occur. (Selman, 9th ed., pp. 225–226)

An exposure was made using 8 mAs and 60 kV. If the kilovoltage was changed to 70 to obtain longer-scale contrast, what new milliampere-seconds value is required to maintain density?

A 2

B 4

C 16

D 32

The Correct Answer is: B
According to the 15% rule, if the kilovoltage is increased by 15%, radiographic density will be doubled. Therefore, to compensate for this change and to maintain radiographic density, the milliampere-seconds value should be reduced to 4 mAs. (Shephard, p. 178)

Which of the following is likely to contribute to the radiographic contrast present on the finished radiograph?

1. Atomic number of tissues radiographed
2. Any pathologic processes
3. Degree of muscle development

A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
The radiographic subject, the patient, is composed of many different tissue types that have varying densities, resulting in varying degrees of photon attenuation and absorption. The atomic number of the tissues under investigation is directly related to their attenuation coefficient. This differential absorption contributes to the various shades of gray (scale of radiographic contrast) on the finished radiograph. Normal tissue density may be altered significantly in the presence of pathologic processes. For example, destructive bone disease can cause a dramatic decrease in tissue density (and subsequent increase in radiographic density). Abnormal accumulation of fluid (as in ascites) will cause a significant increase in tissue density. Muscle atrophy or highly developed muscles similarly will decrease or increase tissue density. (Bushong, 8th ed., pp. 290, 298–299

Exposure factors of 90 kV and 3 mAs are used for a particular nongrid exposure. What should be the new milliampere-seconds (mAs) value if a 12:1 grid is added?

A 86

B 9

C 12

D 15

The Correct Answer is: D
To change nongrid to grid exposure or to adjust exposure when changing from one grid ratio to another, it is necessary to recall the factor for each grid ratio:

Therefore, to change from nongrid to a 12:1 grid, multiply the original milliampere-seconds value by a factor of 5. A new milliampere-seconds value of 15 is required. (Shephard, pp. 247–248)

Conditions contributing to poor radiographic film archival quality include

1. fixer retention
2. insufficient developer replenishment
3. poor storage conditions

A 1 only B 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
The archival quality of a film refers to its ability to retain its image for a long period of time. Many states have laws governing how long a patient's medical records, including films, must be retained. Very importantly, they must be retained in their original condition. Archival quality is poor if radiographic films begin to show evidence of stain after being stored for only a short time. Probably the most common cause of stain, and hence of poor archival quality, is retained fixer within the emulsion. Fixer may be retained as a result of poor washing or because there was insufficient hardener (underreplenishment) in the developer, thus permitting fixer to be retained by the swollen emulsion. A test for quantity of retained fixer in film emulsion is often included as part of a QA program. Stain may also be caused by poor storage conditions. Storage in a hot, humid place will cause even the smallest amount of retained fixer to react with silver, causing stain. (Shephard, pp. 110, 135, 137)

Characteristics of high-ratio focused grids, compared with lower-ratio grids, include which of the following?

1. They allow more positioning latitude.
2. They are more efficient in collecting SR.
3. They absorb more of the useful beam.

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: C
Two of a grid's physical characteristics that determine its degree of efficiency in the removal of scattered radiation are grid ratio (the height of the lead strips compared with the distance between them) and the number of lead strips per inch. As the lead strips are made taller or the distance between them decreases, scattered radiation is more likely to be trapped before reaching the IR. A 12:1 ratio grid will absorb more scattered radiation than an 8:1 ratio grid. An undesirable but unavoidable characteristic of grids is that they do absorb some primary/useful photons as well as scattered photons. The higher the ratio grid, the more scatter radiation the grid will clean up, but more useful photons will be absorbed as well. The higher the primary to scattered photon transmission ratio, the more desirable is the grid. Higher-ratio grids restrict positioning latitude more severely—grid centering must be more accurate (than with lower-ratio grids) to avoid grid cutoff. (Shephard, pp. 245–246)

Underexposure of a radiograph can be caused by all the following except insufficient

A milliamperage (mA)

B exposure time

C Kilovoltage

D SID

he Correct Answer is: D
Insufficient milliamperage and/or exposure time will result in lack of radiographic density. Insufficient kilovoltage will result in underpenetration and excessive contrast. Insufficient SID, however, will result in increased exposure rate and radiographic overexposure. (Selman, 9th ed., pp. 214–215)

With all other factors constant, as digital image matrix size increases,

1. pixel size decreases. 2. resolution increases. 3. pixel size increases.

A 1 only B 2 only C 1 and 2 only D 2 and 3 only

The Correct Answer is: C
A digital image is formed by a matrix of pixels (picture elements) in rows and columns. A matrix that has 512 pixels in each row and column is a 512 × 512 matrix. The term field of view is used to describe how much of the patient (eg, 150-mm diameter) is included in the matrix. The matrix and the field of view can be changed independently, without one affecting the other, but changes in either will change pixel size. As in traditional radiography, spatial resolution is measured in line pairs per mm (lp/mm). As matrix size is increased, there are more and smaller pixels in the matrix, and therefore improved resolution. Fewer and larger pixels result in a poor resolution, "pixelly" image, that is, one in which you can actually see the individual pixel boxes. (Fosbinder & Kelsey, p 286; Shephard, p 336)

Radiographic contrast is the result of

A transmitted electrons

B differential absorption

C absorbed photons

D milliampere-seconds selection

The Correct Answer is: B
Differential absorption refers to the x-ray absorption characteristics of neighboring anatomic structures—determined by the atomic number of the tissue being examined. The radiographic representation of these various density structures is referred to as radiographic contrast; it may be enhanced with high-contrast technical factors, especially using low kilovoltage levels. At low kilovoltage levels, the photoelectric effect predominates. If photons are absorbed, there will be no contrast. The technical factor milliampere-seconds is used to regulate image density. (Bushong, pp. 181–184)

Exposure factors of 80 kVp and 8 mAs are used for a particular nongrid exposure. What should be the new milliampere-seconds value if an 8:1 grid is added?

A 16 mAs

B 24 mAs

C 32 mAs

D 40 mAs

The Correct Answer is: C
To change nongrid to grid exposure, or to adjust exposure when changing from one grid ratio to another, remember the factor for each grid ratio:

Therefore, to change from nongrid exposure to an 8:1 grid, multiply the original milliampere-seconds value by a factor of 4. Thus, a new setting of 32 mAs is required. (Shephard, p. 248

Recorded detail can be improved by decreasing

1. the SID
2. the OID
3. patient/part motion

A 1 only B 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C
Motion, voluntary or involuntary, is most detrimental to good recorded detail. Even if all other factors are adjusted to maximize detail, if motion occurs during exposure, detail is lost. The most important ways to reduce the possibility of motion are using the shortest possible exposure time, careful patient instruction (for suspended respiration), and adequate immobilization when necessary. Minimizing magnification through the use of increased SID and decreased OID functions to improve recorded detail. (Carlton and Adler, 4th ed., p. 451)

All the following affect the exposure rate of the primary beam except

A milliamperage B kilovoltage C distance D field size

The Correct Answer is: D
Exposure rate is regulated by milliamperage. Distance significantly affects the exposure rate according to the inverse-square law of radiation. Kilovoltage also has an effect on exposure rate because an increase in kilovoltage will increase the number of high-energy photons produced at the target. The size of the x-ray field determines the volume of tissue irradiated, and hence the amount of scattered radiation generated, but is unrelated to the exposure rate. (Selman, 9th ed., p. 117)

Characteristics of low ratio focused grids include the following:

1. they have a greater focal range

2. they are less efficient in collecting SR

3. they can be used inverted

A 1 only

B 1 and 2 only

C 2 and 3 only

D 1, 2, and 3

he Correct Answer is: B
Grid ratio compares the height of the lead strip to the distance between the lead strips. Focused grids have their lead strips angled so as to parallel the divergent x-ray beam. The higher the grid ratio, the greater the grid's efficiency in absorbing scattered radiation before it reaches the image receptor—but the more critical the centering and distance specifications. Although higher ratio focused grids absorb more SR they have a narrower focal range (focusing distance) and grid/tube centering becomes much more critical. Focused grids must not be accidentally inverted—to do so would cause the lead strips to be placed exactly in the path of the lead strips (grid cutoff), everywhere but in the center of the grid. (Ballinger & Frank, vol 3, p 235)

The best way to control voluntary motion is

A immobilization of the part.

B careful explanation of the procedure.

C short exposure time.

D physical restraint.

The Correct Answer is: B
Patients who are able to cooperate are usually able to control voluntary motion if they are provided with an adequate explanation of the procedure. Once patients understand what is needed, most will cooperate to the best of their ability (by suspending respiration and holding still for the exposure). Certain body functions and responses, such as heart action, peristalsis, pain, and muscle spasm, cause involuntary motion that is uncontrollable by the patient. The best and only way to control involuntary motion is by always selecting the shortest possible exposure time. Involuntary motion may also be minimized by careful explanation, immobilization, and (as a last resort and only in certain cases) restraint. (Ballinger & Frank, vol 1, pp 12-13)

In electronic imaging, as digital image matrix size increases

1. pixel size decreases
2. resolution decreases
3. pixel depth decreases

A 1 only B 2 only C 1 and 2 only D 2 and 3 only

The Correct Answer is: A
Pixel depth is directly related to shades of gray—called dynamic range—and is measured in bits. The greater the number of bits, the more shades of gray. For example, a 1-bit (2 1) pixel will demonstrate 2 shades of gray, whereas a 6-bit (2 6) pixel can display 64 shades and a 7-bit (2 7) pixel 128 shades. However, pixel depth is unrelated to resolution.

A digital image is formed by a matrix of pixels (picture elements) in rows and columns. A matrix that has 512 pixels in each row and column is a 512 × 512 matrix. The term field of view is used to describe how much of the patient (e.g., 150-mm diameter) is included in the matrix. The matrix and the field of view can be changed independently without one affecting the other, but changes in either will change pixel size. As in traditional radiography, spatial resolution is measured in line pairs per millimeter (lp/mm). As matrix size is increased (e.g., from 512 × 512 to 1,024 × 1,024) there are more and smaller pixels in the matrix and, therefore, improved resolution. Fewer and larger pixels result in poor resolution, a “pixelly” image, that is, one in which you can actually see the individual pixel boxes. (Fosbinder and Kelsey, p. 286; Shephard, p. 336)

Compared to a low ratio grid, a high ratio grid will

1. absorb more primary radiation.

2. absorb more scattered radiation.

3. allow more centering latitude.

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
Grid ratio is defined as the height of the lead strips to the width of the interspace material (see the figure below). The higher the lead strips (or the smaller the distance between the strips), the greater the grid ratio and the greater the percentage of scattered radiation absorbed. However, a grid does absorb some primary radiation as well. The higher the lead strips, the more critical the need for accurate centering, as the lead strips will more readily trap photons whose direction do not parallel them. (Shephard, pp. 245, 255)

All the following have an impact on radiographic contrast except

A photon energy

B grid ratio

C OID

D focal-spot size

he Correct Answer is: D
As photon energy increases, more penetration and greater production of scattered radiation occur, producing a longer scale of contrast. As grid ratio increases, more scattered radiation is absorbed, producing a shorter scale of contrast. As OID increases, the distance between the part and the IR acts as a grid, and consequently, less scattered radiation reaches the IR, producing a shorter scale of contrast. Focal-spot size is related only to recorded detail. (Shephard, p. 203)

A particular radiograph was produced using 6 mAs and 110 kVp with an 8:1 ratio grid. The radiograph is to be repeated using a 16:1 ratio grid. What should be the new mAs?

A 3

B 6

C 9

D 12

The Correct Answer is: C
To change nongrid exposures to grid exposures, or to adjust exposure when changing from one grid ratio to another, you must remember the factor for each grid ratio:

No grid = 1 × the original mAs

5:1 grid = 2 × the original mAs

6:1 grid = 3 × the original mAs

8:1 grid = 4 × the original mAs

12:1 grid = 5 × the original mAs

16:1 grid = 6 × the original mAs

To adjust exposure factors, you simply compare the old with the new:

x = 9 mAs using 16:1 grid.

(Shepard, pp. 247–248)

Which of the following terms/units is used to express the resolution of a diagnostic image?

A Line pairs per millimeter (lp/mm)

B Speed

C Latitude

D Kiloelectronvolts (keV)

The Correct Answer is: A
Resolution describes how closely fine details may be associated and still be recognized as separate details before seeming to blend into each other and appear “as one.” The degree of resolution transferred to the IR is a function of the resolving power of each of the system components and can be expressed in line pairs per millimeter (lp/mm). It can be measured using a resolution test pattern; a variety of resolution test tools are available. The star pattern generally is used for focal-spot-size evaluation, whereas the parallel-line type is used for evaluating intensifying screens. Resolution can also be expressed in terms of line-spread function (LSP) or modulation transfer function (MTF). LSP is measured using a 10-×m x-ray beam; MTF measures the amount of information lost between the object and the IR. (Carlton and Adler, 4th ed., p. 334

Which of the following is (are) characteristic(s) of a 16:1 grid?

1. It absorbs a high percentage of scattered radiation.

2. It has little positioning latitude.

3. It is used with high-kVp exposures.

A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
High-kilovoltage exposures produce large amounts of scattered radiation, and therefore high-ratio grids are used in an effort to trap more of this scattered radiation. However, accurate centering and positioning become more critical to avoid grid cutoff. (Selman, p 243)

Which of the following can affect radiographic contrast?

1. Processing 2. Pathology 3. OID

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

he Correct Answer is: D
All three factors can affect radiographic contrast. The type of chemistry used in the automatic processor and especially the temperature of the solution can have a big impact on the resulting image contrast. As temperature increases, contrast decreases. Since pathology can alter the degree of attenuation of the x-ray beam, it can affect contrast. The type of pathology will determine how contrast is affected. An additive pathology such as Paget's disease will increase contrast, while a destructive disease such as osteoporosis will decrease contrast. OID can affect contrast when it is used as an air gap. If a 6-inch air gap (OID) is introduced between the part and the IR, much of the scattered radiation emitted from the body will not reach the IR; the air gap thus acts as a grid and increases image contrast. (Carlton & Adler, pp 397–398)

Exposure factors of 400 mA, 20 ms, 68 kVp, 400-speed system, at 40-in. SID were used to produce a satisfactory radiographic image.

A change to 4 mAs can be best compensated for by which of the following?

A Increasing the SID to 60 in.

B Decreasing the SID to 20 in.

C Decreasing the speed to 200

D Increasing the kilovoltage to 78 kVp

he Correct Answer is: D
The milliampere-seconds value used was 8 (400 mA × 0.02 s); the new milliampere-seconds value is half that—4 mAs—which will result in half the original density. An increase in SID, as suggested by choice (A), would further decrease density. The decrease in milliampere-seconds could be compensated for by a decrease in SID—but according to the density-maintenance formula, that distance should be 18 in.. Cutting the system speed in half would also decrease the density by another half. Increasing the kilovoltage by 15%, as suggested by choice (D), will best simulate the original density. (Shephard, pp. 178–181)

To be suitable for use in intensifying screens, a phosphor should have which of the following characteristics?

1. High conversion efficiency
2. High x-ray absorption
3. High atomic number

A 1 only B 3 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: D
Intensifying-screen phosphors that have a high atomic number are more likely to absorb a high percentage of the incident x-ray photons and convert x-ray photon energy to fluorescent light energy. How efficiently the phosphors detect and interact with the x-ray photons is termed quantum detection efficiency. How effectively the phosphors make this energy conversion is termed conversion efficiency. (Shephard, p. 65)

Which of the following is (are) characteristic(s) of a 5:1 grid?

1. It allows some positioning latitude.
2. It is used with high-kilovoltage exposures.
3. It absorbs a high percentage of scattered radiation.

A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: A
Low-ratio grids, such as 5:1, 6:1, and 8:1, are used with moderate-kilovolt techniques and are not recommended for use beyond 85 kV. They are not able to clean up the amount of scatter produced at high kilovoltages, but their low ratio permits more positioning latitude than high-ratio grids. High-kilovoltage exposures produce large amounts of scattered radiation, and therefore, high-ratio grids are used in an effort to trap more of this scattered radiation. However, accurate centering and positioning become more critical to avoid grid cutoff. (Selman, 9th ed., p. 243)

Greater latitude is available to the radiographer in which of the following circumstances?

1. Using high-kV technical factors
2. Using a low-ratio grid
3. Using low-kV technical factors

A 1 only B 1 and 2 only C 2 and 3 only D 3 only

The Correct Answer is: B
In the low-kilovoltage ranges, a difference of just a few kilovolts makes a very noticeable radiographic difference, therefore offering little margin for error/latitude. High-kilovolt technical factors offer much greater margin for error; in the high-kV ranges, an error of a few kV makes little/no difference in the resulting image. Lower-ratio grids offer more tube-centering latitude than high-ratio grids. (Saia, 4th ed., p. 360)

Which of the following is most likely to result from the introduction of a grid to a particular radiographic examination?

A Increased patient dose and increased contrast

B Decreased patient dose and decreased contrast

C Increased patient dose and decreased contrast

D Decreased patient dose and increased contrast

The Correct Answer is: A
A grid is a device interposed between the patient and image receptor that absorbs a large percentage of scattered radiation before it reaches the image receptor. It is constructed of alternating strips of lead foil and radiolucent filler material. X-ray photons traveling in the same direction as the primary beam pass between the lead strips. X-ray photons, having undergone interactions within the body and deviated in various directions, are absorbed by the lead strips; this is referred to as "clean-up" of scattered radiation and results in shorter scale (i.e., more or increased) contrast.

If a grid is introduced, there will be significantly fewer photons reaching the image receptor, hence a very significant decrease in radiographic density. To maintain adequate density then, the addition of a grid must be accompanied by an appropriately substantial increase in mAs, hence, increased patient dose. (Bushong, 8th ed, p 248)

If a radiograph exposed using a 12:1 ratio grid exhibits a loss of density at its lateral edges, it is probably because the

A SID was too great

B grid failed to move during the exposure

C x-ray tube was angled in the direction of the lead strips

D central ray was off-center

he Correct Answer is: A
If the SID is above or below the recommended focusing distance, the primary beam will not coincide with the angled lead strips at the lateral edges. Consequently, there will be absorption of the primary beam, termed grid cutoff. If the grid failed to move during the exposure, there would be grid lines throughout. Central ray angulation in the direction of the lead strips is appropriate and will not cause grid cutoff. If the central ray were off-center, there would be uniform loss of density. (Carlton and Adler, 4th ed., p. 260)

Factor(s) that can be used to regulate radiographic density is (are)

1. milliamperage.

2. exposure time.

3. kilovoltage.

A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: D
Factors that regulate the number of x-ray photons produced at the target can be used to control radiographic density, namely milliamperage and exposure time (mAs). Radiographic density is directly proportional to mAs; if the mAs is cut in half, the radiographic density will decrease by one-half. Although kilovoltage is used primarily to regulate radiographic contrast, it may also be used to regulate radiographic density in variable-kVp techniques, according to the 15% rule. (Selman, pp 213–214)

In an AP abdomen radiograph taken at 105-cm SID during an IVU series, one renal shadow measures 9 cm in width. If the OID is 18 cm, what is the actual width of the kidney?

A 5 cm

B 7.5 cm

C 11 cm

D 18 cm

The Correct Answer is: B
As OID increases, magnification increases. Viscera and structures within the body will be varying distances from the IR depending on their location within the body and the position used for the exposure. The size of a particular structure or image can be calculated using the following formula:

Changes in milliampere-seconds can affect all the following except

A quantity of x-ray photons produced

B exposure rate

C optical density

D recorded detail

The Correct Answer is: D
Milliampere-seconds (mAs) are the product of milliamperes (mA) and exposure time (seconds). Any combinations of milliamperes and time that will produce a given milliampere-seconds value (i.e., a particular quantity of x-ray photons) will produce identical optical density. This is known as the reciprocity law. Density is a quantitative factor because it describes the amount of image blackening. The milliampere-seconds value is also a quantitative factor because it regulates x-ray-beam intensity, exposure rate, quantity, or number of x-ray photons produced (the milliampere-seconds value is the single most important technical factor associated with image density and is the factor of choice for regulating radiographic/optical density). The milliampere-seconds value is directly proportional to the intensity (i.e., exposure rate, number, and quantity) of x-ray photons produced and the resulting radiographic density. If the milliampere-seconds value is doubled, twice the exposure rate and twice the density occur. If the milliampere-seconds value is cut in half, the exposure rate and resulting density are cut in half. The milliampere-seconds value has no effect on recorded detail. (Shephard, p. 170)

Which of the following examinations might require the use of 120 kVp?

1. AP abdomen

2. Chest radiograph

3. Barium-filled stomach

A 1 only B 2 only C 1 and 2 only D 2 and 3 only

he Correct Answer is: D
High-kilovoltage factors are frequently used to even out densities in anatomic parts with high tissue contrast (eg, the chest). However, as high kilovoltage produces added scattered radiation, it generally must be used with a grid. It would be inappropriate to perform an AP abdomen with high kilovoltage because it has such low subject contrast. Barium-filled structures are frequently radiographed using 120 kV or more to penetrate the barium—to see through to structures behind. (Carlton & Adler, p 468)

An exposure was made using 300 mA and 50 ms. If the exposure time is changed to 22 ms, which of the following milliamperage selections would most closely approximate the original radiographic density?

A 300 mA

B 400 mA

C 600 mA

D 700 mA

he Correct Answer is: D
Since 50 ms is equal to 0.050 s, and since mA × time mAs, the original milliampere-seconds value was 15 mAs. Now, it is only necessary to determine what milliamperage must be used with 22 ms to provide the same 15 mAs (and thus the same radiographic density). Because mA × time = mAs,

(Selman, 9th ed., p. 214)

The attenuation of x-ray photons is not influenced by

1. pathology
2. effective atomic number
3. photon quantity

A 1 only B 3 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: B
Attenuation (decreased intensity through scattering or absorption) of the x-ray beam is a result of its original energy and its interactions with different types and thicknesses of tissue. The greater the original energy/quality (the higher the kilovoltage) of the incident beam, the less is the attenuation. The greater the effective atomic number of the tissues (tissue type and pathology determine absorbing properties), the greater is the beam attenuation. The greater the volume of tissue (subject density and thickness), the greater is the beam attenuation. (Bushong, 8th ed., p. 185)

An automatic exposure control (AEC) device can operate on which of the following principles?

1. A photomultiplier tube charged by a fluorescent screen
2. A parallel-plate ionization chamber charged by x-ray photons
3. Motion of magnetic fields inducing current in a conductor

A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

he Correct Answer is: C
A phototimer is one type of automatic exposure control (AEC) that actually measures light. As x-ray photons penetrate and emerge from a part, a fluorescent screen beneath the cassette glows, and the fluorescent light charges a photomultiplier tube. Once a predetermined charge has been reached, the exposure terminates automatically. A parallel-plate ionization chamber is another type of AEC. A radiolucent chamber is beneath the patient (between the patient and the IR). As photons emerge from the patient, they enter the chamber and ionize the air within it. Once a predetermined charge has been reached, the exposure is terminated automatically. Motion of magnetic fields inducing a current in a conductor refers to the principle of mutual induction. (Fauber, 2nd ed., pp. 232–233

types of moving grid mechanisms include

1. oscillating. 2. reciprocating. 3. synchronous.

A 1 only

B 1 and 2 only

C 1 and 3 only

D 2 and 3 only

The Correct Answer is: B
Grids are devices constructed of alternating strips of lead foil and radiolucent interspacing material. They are placed between the patient and the IR, and they function to remove scattered radiation from the remnant beam before it forms the latent image. Stationary grids will efficiently remove scattered radiation from the remnant beam; however, their lead strips will be imaged on the radiograph. If the grid is made to move (usually in a direction perpendicular to the lead strips) during the exposure, the lead strips will be effectively blurred. The motion of a moving grid, or Potter-Bucky diaphragm, may be reciprocating (equal strokes back and forth), oscillating (almost circular direction), or catapult (rapid forward motion and slow return). Synchronous refers to a type of x-ray timer. (Bushong, p 256)

Which of the following factors impact(s) recorded detail?

1. Focal-spot size
2. Subject motion
3. SOD

A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
Focal-spot size affects recorded detail by its effect on focal-spot blur: The larger the focal-spot size, the greater is the blur produced. Recorded detail is affected significantly by distance changes because of their effect on magnification. As SID increases and as OID decreases, magnification decreases and recorded detail increases. SOD is determined by subtracting OID from SID. (Shephard, p. 215)

A 3-inch object to be radiographed at a 36-inch SID lies 4 inches from the image recorder. What will be the image width?

A 2.6 inches

B 3.3 inches

C 26 inches

D 33 inches

he Correct Answer is: B
Magnification is part of every radiographic image. Anatomic parts within the body are at various distances from the image recorder and therefore have various degrees of magnification. The formula used to determine the amount of image magnification is:

Substituting known values:

x = 3.37 inches image width

(Bushong, p 284)

If the center photocell were selected for a lateral projection of the lumbar spine that was positioned with the spinous processes instead of the vertebral bodies centered to the grid, how would the resulting radiograph look?

A The image would be underexposed.

B The image would be overexposed.

C The image would be correctly exposed.

D An exposure could not be made.

The Correct Answer is: A
If the photocell were centered more posteriorly to a thinner and less dense structure, then the exposure received would be correct for that less-dense structure. The spinous processes would be well visualized, but the denser vertebral bodies and surrounding structures (pedicles and lamina) would be underexposed. Accurate selection of photocells and precise positioning are critical with the use of automatic exposure devices. (Carlton and Adler, 4th ed., p. 540)

X-ray photon energy is inversely related to

1. photon wavelength
2. applied milliamperes (mA)
3. applied kilovoltage (kV)

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

he Correct Answer is: A
As kilovoltage is increased, more high-energy photons are produced, and the overall energy of the primary beam is increased. Photon energy is inversely related to wavelength; that is, as photon energy increases, wavelength decreases. An increase in milliamperage serves to increase the number of photons produced at the target but is unrelated to their energy. (Selman, 9th ed., p. 118)

Combinations of milliamperage and exposure time that produce a particular milliampere-seconds value will produce identical radiographic density. This statement is an expression of the

A inverse-square law

B line-focus principle

C reciprocity law D

D log E curve

The Correct Answer is: C
A number of milliamperage and exposure time settings can produce the same milliampere-seconds value. Each of the following milliamperage and time combinations produces 10 mAs: 100 mA and 0.1 s, 200 mA and 0.05 s, 300 mA and, and 400 mA and 0.025 s. These milliamperage and exposure-time combinations should produce identical radiographic density. This is known as the reciprocity law. The radiographer can make good use of the reciprocity law when manipulating exposure factors to decrease exposure time and decrease motion unsharpness. (Selman, 9th ed., p. 214)

The exposure factors used for a particular nongrid x-ray image were 300 mA, 4 ms, and 90 kV. Another image, using an 8:1 grid, is requested. Which of the following groups of factors is most appropriate?

A 400 mA, 3 ms, 110 kV

B 400 mA, 12 ms, 90 kV

C 300 mA, 8 ms, 100 kV

D 200 mA, 240 ms, 90 kV

The Correct Answer is: B
The addition of a grid will help to clean up the scattered radiation produced by higher kilovoltage, but the grid requires an adjustment of milliampere-seconds. According to the grid conversion factors listed here, the addition of an 8:1 grid requires that the original milliampere-seconds be multiplied by a factor of 4:

The original milliampere-seconds value is 1.2. The ideal adjustment, therefore, requires a 4.8 mAs at 90 kV. Although 2.4 mAs with 100 kV (choice C), or 1.2 mAs with 110 kV (choice A), also might seem workable, an increase in kilovoltage would further compromise contrast, nullifying the effect of the grid. Additionally, kilovoltage exceeding 100 should not be used with an 8:1 grid. (Shephard, pp. 247–24

A compensating filter is used to

A absorb the harmful photons that contribute only to patient dose

B even out widely differing tissue densities

C eliminate much of the scattered radiation

D improve fluoroscopy

The Correct Answer is: B
A compensating filter is used to make up for widely differing tissue densities. For example, it is difficult to obtain a satisfactory image of the mediastinum and lungs simultaneously without the use of a compensating filter to “even out” the densities. With this device, the chest is radiographed using mediastinal factors, and a trough-shaped filter (thicker laterally) is used to absorb excess photons that would overexpose the lungs. The middle portion of the filter lets the photons pass to the mediastinum almost unimpeded. Filters that absorb the photons contributing to skin dose are inherent and added filters. Compensating filtration is unrelated to elimination of scattered radiation or fluoroscopy. (Selman, 9th ed., p. 254)

An x-ray image of the ankle was made at 40-SID, 200 mA, 50 ms, 70 kV, 0.6 mm focal spot, and minimal OID. Which of the following modifications would result in the greatest increase in magnification?

A 1.2 mm focal spot

B 36-in. SID

C 44-in. SID

D 4-in. OID

The Correct Answer is: D
All the factor changes affect recorded detail, but focal spot size does not affect magnification. An increase in SID would decrease magnification. Although a decrease in SID will increase magnification, it does not have as significant an effect as an increase in OID. In general, it requires an increase of 7 in. SID to compensate for every inch of OID. (Carlton and Adler, 4th ed., pp. 458–460)

Which of the following may be used to reduce the effect of scattered radiation on a finished radiograph?

1. Grids
2. Collimators
3. Compression bands

A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
Collimators restrict the size of the irradiated field, thereby limiting the volume of irradiated tissue, and hence less scattered radiation is produced. Once radiation has scattered and emerged from the body, it can be trapped by the grid's lead strips. Grids effectively remove much of the scattered radiation in the remnant beam before it reaches the IR. Compression can be applied to reduce the effect of excessive fatty tissue (e.g., in the abdomen), in effect reducing the thickness of the part to be radiographed. (Selman, 9th ed., pp. 234, 247, 289)

Exposure factors of 110 kVp and 12 mAs are used with an 8:1 grid for a particular exposure. What should be the new mAs if a 12:1 grid is substituted?

A 3

B 9

C 15

D 18

The Correct Answer is: C
To change nongrid to grid exposure, or to adjust exposure when changing from one grid ratio to another, recall the factor for each grid ratio:

The grid conversion formula is

Substituting known quantities:

Compared with a low-ratio grid, a high-ratio grid will

1. allow more centering latitude
2. absorb more scattered radiation
3. absorb more primary radiation

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C
Grid ratio is defined as the height of the lead strips to the width of the interspace material (Figure 4–32). The higher the lead strips (or the smaller the distance between the strips), the higher the grid ratio, and the greater the percentage of scattered radiation absorbed. However, a grid does absorb some primary/useful radiation as well. The higher the lead strips, the more critical is the need for accurate centering because the lead strips will more readily trap photons whose direction does not parallel them. (Shephard, pp. 245, 255)

How is SID related to exposure rate and radiographic density?

A As SID increases, exposure rate increases and radiographic density increases.

B As SID increases, exposure rate increases and radiographic density decreases

. C As SID increases, exposure rate decreases and radiographic density increases.

D As SID increases, exposure rate decreases and radiographic density decreases.

he Correct Answer is: D
According to the inverse-square law of radiation, the intensity or exposure rate of radiation is inversely proportional to the square of the distance from its source. Thus, as distance from the source of radiation increases, exposure rate decreases. Because exposure rate and radiographic density are directly proportional, if the exposure rate of a beam directed to an IR is decreased, the resulting radiographic density would be decreased proportionately. (Selman, 9th ed., p. 117)

A focal-spot size of 0.3 mm or smaller is essential for

A small-bone radiography

B magnification radiography

C tomography

D fluoroscopy

The Correct Answer is: B
A fractional focal spot of 0.3 mm or smaller is essential for reproducing fine detail without focal-spot blurring in magnification radiography. As the object image is magnified, so will be any associated blur unless a fractional focal spot is used. Use of a fractional focal spot on a routine basis is unnecessary; it is not advised because it causes unnecessary wear on the x-ray tube and offers little radiographic advantage. (Shephard, p. 21

The functions of automatic beam limitation devices include

1. reducing the production of scattered radiation
2. increasing the absorption of scattered radiation
3. changing the quality of the x-ray beam

A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: A
Beam restrictors function to limit the size of the irradiated field. In so doing, they limit the volume of tissue irradiated (thereby decreasing the percentage of scattered radiation generated in the part) and help to reduce patient dose. Beam restrictors do not affect the quality (energy) of the x-ray beam—that is, the function of kilovoltage and filtration. Beam restrictors do not absorb scattered radiation—that is a function of grids. (Shephard, p. 27)

Which of the following affect(s) both the quantity and the quality of the primary beam?

1. Half-value layer (HVL)
2. Kilovoltage (kV)
3. Milliamperage (mA)

A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

he Correct Answer is: C
Kilovoltage and the HVL affect both the quantity and the quality of the primary beam. The principal qualitative factor for the primary beam is kilovoltage, but an increase in kilovoltage will also create an increase in the number of photons produced at the target. HVL is defined as the amount of material necessary to decrease the intensity of the beam to one-half its original value, thereby effecting a change in both beam quality and quantity. The milliampere-seconds value is adjusted to regulate the number of x-ray photons produced at the target. X-ray-beam quality is unaffected by changes in milliampere-seconds. (Carlton and Adler, 4th ed., p. 181)

Which of the following is most likely to produce a high-quality image?

A Small image matrix

B High signal-to-noise ratio (SNR)

C Large pixel size

D Low resolution

he Correct Answer is: B
SNR can refer to home television images, magnetic resonance images (MRIs), ultrasound images, x-ray images, and so on. Noise interferes with visualization of image details, for example, scattered radiation fog, graininess from quantum mottle, and so on. The actual signal can be from x-rays, sound waves, and so on. The signal is desirable, the noise is not, therefore, a higher SNR produces a higher-quality image. Low SNR severely impairs contrast resolution. (Bushong, 8th ed., p. 412)

An overall image density arising from factors other than the light or radiation used to expose the image is called

A fog.

B log-relative exposure.

C optical density.

D artifact.

The Correct Answer is: A
This is the definition of fog. Anything other than intensifying screen light or primary x-radiation is undesirable in terms of image exposure. Log-relative exposure is the amount of exposure required to produce a given density as measured on the sensitometric graph. Optical density is normal radiographic density. An artifact is anything foreign to the image; the term could include fog, but it also covers many physical interferences. (Selman, 9th ed., pp. 196–197)

Which of the following conditions would require an increase in exposure factors?

1. Congestive heart failure

2. Pleural effusion

3. Emphysema

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: B
Emphysema is abnormal distention of the alveoli (or tissue spaces) with air. The presence of abnormal amounts of air makes it necessary to decrease from normal exposure factors. Congestive heart failure and pleural effusion involve abnormal amounts of fluid in the chest and thus require an increase in exposure factors. (Carlton & Adler, p 258)

Low-kilovoltage exposure factors usually are indicated for radiographic examinations using

1. water-soluble, iodinated media
2. a negative contrast agent
3. barium sulfate

A 1 only B 1 and 2 only C 3 only D 1 and 3 only

he Correct Answer is: B
Positive contrast medium is radiopaque; negative contrast material is radioparent. Barium sulfate (radiopaque, positive contrast material) is used most frequently for examinations of the intestinal tract, and high-kilovoltage exposure factors are used to penetrate (to see through and behind) the barium. Water-based iodinated contrast media (Conray, Amipaque) are also positive contrast agents. However, the K-edge binding energy of iodine prohibits the use of much greater than 70 kV with these materials. Higher kilovoltage values will obviate the effect of the contrast agent. Air is an example of a negative contrast agent, and high-kilovoltage factors are clearly not indicated. (Shephard, pp. 200–201)

Which of the following groups of exposure factors will produce the longest scale of contrast?

A 200 mA, 0.25 second, 70 kVp, 12:1 grid

B 500 mA, 0.10 second, 90 kVp, 8:1 grid

C 400 mA, 0.125 second, 80 kVp, 12:1 grid

D 300 mA, 0.16 second, 70 kVp, 8:1 grid

he Correct Answer is: B
Of the given factors, kilovoltage and grid ratio will have a significant effect on the scale of radiographic contrast. The mAs values are almost identical. Because an increased kilovoltage and low-ratio grid combination would allow the greatest amount of scattered radiation to reach the IR, thereby producing more gray tones, B is the best answer. Group D also uses a low-ratio grid, but the kilovoltage is too low to produce as many gray tones as B. (Shepard, p. 308)

Which of the following is most likely to occur as a result of using a 30-in. SID with a 14 × 17 in. IR to radiograph a fairly homogeneous structure?

A Production of quantum mottle

B Density variation between opposite ends of the IR

C Production of scatter radiation fog

D Excessively short-scale contrast

Which of the following is most likely to occur as a result of using a 30-in. SID with a 14 × 17 in. IR to radiograph a fairly homogeneous structure?

A Production of quantum mottle

B Density variation between opposite ends of the IR

C Production of scatter radiation fog

D Excessively short-scale contrast

The Correct Answer is: A
The focal-spot size selected will determine the amount of focal-spot, or geometric, blur produced in the image. Different screen speeds will create differing degrees of fluorescent light diffusion, affecting recorded detail. OID is responsible for image magnification and hence recorded detail. The milliamperage is unrelated to recorded detail; it affects only the quantity of x-ray photons produced and thus the radiographic density. (Selman, 9th ed., pp. 206–210)

Types of shape distortion include

1. magnification
2. elongation
3. foreshortening

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C
Size distortion (magnification) is inversely proportional to SID and directly proportional to OID. Increasing the SID and decreasing the OID decreases size distortion. Aligning the tube, part, and IR so that they are parallel reduces shape distortion. There are two types of shape distortion—elongation and foreshortening. Angulation of the part with relation to the IR results in foreshortening of the object. Tube angulation causes elongation of the object. (Shephard, pp. 228, 231–234)

Of the following groups of exposure factors, which will produce the most radiographic density?

A 400 mA, 30 ms, 72-in. SID

B 200 mA, 30 ms, 36-in. SID

C 200 mA, 60 ms, 36-in. SID

D 400 mA, 60 ms, 72-in. SID

he Correct Answer is: C
The formula mA × s = mAs is used to determine each milliampere-second setting (remember to first change milliseconds to seconds). The greatest radiographic density will be produced by the combination of highest milliampere-seconds value and shortest SID. The groups in choices (B) and (D) should produce identical radiographic density, according to the inverse-square law, because group (D) includes twice the distance and 4 times the milliampere-seconds value of group (B). The group in (A) has twice the distance of the group in (B) but only twice the milliampere-seconds; therefore, it has the least density. The group in (C) has the same distance as the group in (B) and twice the milliampere-seconds, making group in (C) the group of technical factors that will produce the greatest radiographic density. (Selman, 9th ed., p. 214)

If a particular grid has lead strips 0.40 mm thick, 4.0 mm high, and 0.25 mm apart, what is its grid ratio?

A 8:1 B 10:1 C 12:1 D 16:1

he Correct Answer is: D
Grid ratio is defined as the ratio between the height of the lead strips and the width of the distance between them (i.e., their height divided by the distance between them). If the height of the lead strips is 4.0 mm and the lead strips are 0.25 mm apart, the grid ratio must be 16:1 (4.0 divided by 0.25). The thickness of the lead strip is unrelated to grid ratio. (Selman, 9th ed., p. 236)

Of the following groups of exposure factors, which will produce the shortest scale of radiographic contrast?

A 500 mA, 0.040 second, 70 kV

B 100 mA, 0.100 second, 80 kV

C 200 mA, 0.025 second, 92 kV

D 700 mA, 0.014 second, 80 kV

The Correct Answer is: A
The single most important factor regulating radiographic contrast is kilovoltage. The lower the kilovoltage, the shorter is the scale of contrast. All the milliampere-seconds values in this problem have been adjusted for kilovoltage changes to maintain density, but just a glance at each of the kilovoltages is often a good indicator of which will produce the longest scale or shortest scale contrast. (Shephard, pp. 306, 308)

If 32 mAs and 50-speed screens were used to produce a particular radiographic density, what new mAs value would be required to produce the same density if the screen speed were changed to 400?

A 4 mAs B 40 mAs C 175 mAs D 256 mAs

The Correct Answer is: A
With all other factors remaining the same, as intensifying-screen speed increases, radiographic density increases. Radiographic density is directly proportional to intensifying-screen speed; that is, if screen speed doubles, density doubles. The formula to determine how milliampere-seconds should be corrected with screen-speed changes is

Screen speed and milliampere-seconds conversion factors are as follows:

Substituting known quantities:

Thus, x = 4 mAs with 400-speed screens. (Shephard, p. 68

Which of the following groups of exposure factors would be most appropriate for a sthenic adult IVU?

A 300 mA, 0.02 s, 72 kVp

B 300 mA, 0.01 s, 82 kVp

C 150 mA, 0.01 s, 94 kVp

D 100 mA, 0.03 s, 82 kVp

The Correct Answer is: A
IVU requires the use of iodinated contrast media. Low kilovoltage (about 70 kVp) is usually used to enhance the photoelectric effect and, in turn, to better visualize the renal collecting system. High kilovoltage will produce excessive scattered radiation and obviate the effect of the contrast agent. A higher milliamperage with a short exposure time generally is preferable. (Fauber, 2nd ed., p. 264)

The fact that x-ray intensity across the primary beam can vary as much as 45% describes the

A line-focus principle.

B transformer law.

C anode heel effect.

D inverse-square law.

he Correct Answer is: C
A beveled focal track extends around the periphery of the anode disk; when a small angle is used, the beveled edge allows for a smaller effective focal spot and better detail. The disadvantage, however, is that photons are noticeably absorbed by the “heel” of the anode, resulting in a smaller percentage of x-ray photons at the anode end of the x-ray beam and a concentration of x-ray photons at the cathode end of the beam. This is known as the anode heel effect and can cause a primary beam variation of up to 45%. The anode heel effect becomes more pronounced as the SID decreases, as IR size increases, and as target angle decreases. (Bushong, 8th ed., pp. 138–140)

A satisfactory radiograph of the abdomen was made at a 42-inch SID using 300 mA, 0.06-second exposure, and 80 kVp. If the distance is changed to 38 inches, what new exposure time would be required?

A 0.02 second

B 0.05 second

C 0.12 second

D 0.15 second

The Correct Answer is: B
According to the inverse square law of radiation, as the distance between the radiation source and the IR decreases, the exposure rate increases. Therefore, a decrease in technical factors is indicated. The density maintenance formula is used to determine new mAs values when changing distance:

Then, to determine the new exposure time (mA × s = mAs),

Which of the following will influence recorded detail?

1. Screen speed
2. Screen–film contact
3. Focal spot

A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
The faster the imaging system, the greater is the sacrifice of image clarity (recorded detail). As intensifying-screen speed increases, recorded detail decreases. Perfect screen–film contact is essential for good detail. Any areas of poor contact result in considerable blurriness in the radiographic image. Focal-spot blur is related to focal-spot size; smaller focal spots produce less blur and thus better recorded detail. (Selman, 9th ed., pp. 206–210)

The advantage(s) of high-kilovoltage chest radiography is (are) that

1. exposure latitude is increased
2. it produces long-scale contrast
3. it reduces patient dose

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
The chest is composed of tissues with widely differing densities (bone and air). In an effort to “even out” these tissue densities and better visualize pulmonary vascular markings, high kilovoltage generally is used. This produces more uniform penetration and results in a longer scale of contrast with visualization of the pulmonary vascular markings as well as bone (which is better penetrated) and air densities. The increased kilovoltage also affords the advantage of greater exposure latitude (an error of a few kilovolts will make little, if any, difference). The fact that the kilovoltage is increased means that the milliampere-seconds value is reduced accordingly, and thus patient dose is reduced as well. A grid usually is used whenever high kilovoltage is required. (Carlton and Adler, 4th ed., pp. 423–424)

A particular radiograph was produced using 12 mAs and 85 kV with a 16:1 ratio grid. The radiograph is to be repeated using an 8:1 ratio grid. What should be the new milliampere-seconds value?

A 3

B 6

C 8

D 10

he Correct Answer is: C
To change nongrid exposures to grid exposures, or to adjust exposure when changing from one grid ratio to another, you must remember the factor for each grid ratio:

To adjust exposure factors, you simply compare the old with the new:

A satisfactory radiograph of the abdomen was made at a 38-in. SID using 400 mA, 60-ms exposure, and 80 kV. If the distance is changed to 42 in., what new exposure time would be required?

A 25 ms

B 50 ms

C 73 ms

D 93 ms

The Correct Answer is: C
According to the inverse-square law of radiation, as the distance between the radiation source and the IR decreases, the exposure rate increases. Therefore, a decrease in technical factors is indicated. The density-maintenance formula is used to determine new milliampere-seconds values when changing distance:

Thus, x = 29.31 mAs at 42-in. SID. Then, to determine the new exposure time (mA × s = mAs),

Thus, x = 0.073 second (73 ms) at 400 mA. (Selman, 9th ed., p. 214

A satisfactory radiograph was made without a grid using a 72-in. SID and 8 mAs. If the distance is changed to 40 in. and an 12:1 ratio grid is added, what should be the new milliampere-seconds value?

A 9.5 mAs

B 12 mAs

C 21 mAs

D 26 mAs

The Correct Answer is: B
According to the inverse-square law of radiation, as the distance between the radiation source and the IR decreases, the exposure rate increases. Therefore, a decrease in technical factors is first indicated to compensate for the distance change. The following formula (density-maintenance formula) is used to determine new milliampere-seconds values when changing distance:

Substituting known values:

Thus, x = 2.47 mAs at 40-in. SID. To then compensate for adding a 12:1 grid, you must multiply the 2.47 mAs by a factor of 5. Thus, 12 mAs is required to produce an image density similar to the original radiograph. The following are the factors used for milliampere-seconds conversion from nongrid to grid:

A technique chart should include which of the following information?

1. Recommended SID
2. Grid ratio
3. Screen–film combination

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D
Technique charts are exposure factor guides that help technologists produce radiographs with consistent density and contrast. They suggest a group of exposure factors to be used at a particular SID with a particular grid ratio, screen–film combination, focal-spot size, and CR angulation. Technique charts do not take into account the nature of the part (disease and atrophy). (Shephard, p. 298)

Which of the following factors is/are related to grid efficiency?

1. Grid ratio
2. Number of lead strips per inch
3. Amount of scatter transmitted through the grid

A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

he Correct Answer is: D
Grid ratio is defined as the ratio of the height of the lead strips to the width of the interspace material; the higher the lead strips, the more scattered radiation they will trap and the greater is the grid's efficiency. The greater the number of lead strips per inch, the thinner and less visible they will be on the finished radiograph. The function of a grid is to absorb scattered radiation in order to improve radiographic contrast. The selectivity of a grid is determined by the amount of primary radiation transmitted through the grid divided by the amount of scattered radiation transmitted through the grid. (Selman, 9th ed., pp. 236–237)

As grid ratio is decreased,

A the scale of contrast becomes longer

B the scale of contrast becomes shorter

C radiographic density decreases

D radiographic distortion decreases

he Correct Answer is: A
Because lead content decreases when grid ratio decreases, a smaller amount of scattered radiation is trapped before reaching the IR. More grays, therefore, are recorded, and a longer scale of contrast results. Radiographic density would increase with a decrease in grid ratio. Grid ratio is unrelated to distortion. (Carlton and Adler, 4th ed., p. 432)

What is the single most important factor controlling size distortion?

A Tube, part, IR alignment

B IR dimensions

C SID

D OID

he Correct Answer is: D
Shape distortion (foreshortening, elongation) is caused by improper alignment of the tube, part, and image receptor. Size distortion, or magnification, is caused by too great an object–image distance or too short a source–image distance. OID is the primary factor influencing magnification, followed by SID. (Bushong, 8th ed, p 284)

A radiograph exposed using a 12:1 ratio grid may exhibit a loss of density at its lateral edges because the

A SID was too great.

B grid failed to move during the exposure.

C x-ray tube was angled in the direction of the lead strips.

D CR was off-center.

The Correct Answer is: A
If the SID is above or below the recommended focusing distance, the primary beam at the lateral edges will not coincide with the angled lead strips. Consequently, there will be absorption of the primary beam, termed grid cutoff. If the grid failed to move during the exposure, there would be grid lines throughout. CR angulation in the direction of the lead strips is appropriate and will not cause grid cutoff. If the CR were off-center, there would be uniform loss of density. (Selman, 9th ed., p. 240)

If a radiograph were made of an average-size knee using automatic exposure control (AEC) and all three photocells were selected, the resulting radiograph would demonstrate

A excessive density.

B insufficient density.

C poor detail.

D adequate exposure.

The Correct Answer is: B
Proper functioning of the photo timer depends on accurate positioning by the radiographer. The correct photocell(s) must be selected, and the anatomic part of interest must completely cover the photocell(s) to achieve the desired density. If a photocell is left uncovered, scattered radiation from the part being examined will cause premature termination of exposure and an underexposed radiograph. (Carlton and Adler, 4th ed., p. 102)

Geometric unsharpness is influenced by which of the following?

1. Distance from object to image
2. Distance from source to object
3. Distance from source to image

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D
Geometric unsharpness is affected by all three factors listed. As OID increases, so does magnification. OID is directly related to magnification; i.e. as OID increases, so does magnification. Focal-object distance and SID are inversely related to magnification. As focal-object distance and SID decrease, magnification increases. (Bushong, 10th ed pp. 174-175)

A particular milliampere-seconds value, regardless of the combination of milliamperes and time, will reproduce the same radiographic density. This is a statement of the

A line-focus principle

B inverse-square law

C reciprocity law

D law of conservation of energy

The Correct Answer is:C

The reciprocity law states that a particular milliampere-seconds value, regardless of the milliamperage and exposure time used, will provide identical radiographic density. This holds true with direct exposure techniques, but it does fail somewhat with the use of intensifying screens. However, the fault is so slight as to be unimportant in most radiographic procedures. (Shephard, p. 193)

Which of the following focal-spot sizes should be employed for magnification radiography?

A 0.2 mm

B 0.6 mm

C 1.2 mm

D 2.0 mm

The Correct Answer is: A
Proper use of focal spot size is of paramount importance in magnification radiography. A magnified image that is diagnostic can be obtained only by using a fractional focal spot of 0.3 mm or smaller. The amount of blur or geometric unsharpness produced by focal spots that are larger in size render the radiograph undiagnostic. (Selman, 9th ed., p. 226)

Which of the following is most likely to produce a radiograph with a long scale of contrast?

A Increased photon energy

B Increased screen speed

C Increased mAs

D Increased SID

The Correct Answer is: A
An increase in photon energy accompanies an increase in kilovoltage. Kilovoltage regulates the penetrability of x-ray photons; it regulates their wavelength—the amount of energy with which they are associated. The higher the related energy of an x-ray beam, the greater its penetrability (kilovoltage and photon energy are directly related; kilovoltage and wavelength are inversely related). Adjustments in kilovoltage have a big impact on radiographic contrast: As kilovoltage (photon energy) is increased, the number of grays increases, thereby producing a longer scale of contrast. In general, as screen speed increases, so does contrast (resulting in a shorter scale of contrast). An increase in mAs is frequently accompanied by an appropriate decrease in kilovoltage, which would also shorten the contrast scale. SID and radiographic contrast are unrelated. (Shephard, p 204)

What pixel size has a 1024 × 1024 matrix with a 35-cm FOV?

A 30 mm

B 0.35 mm

C 0.15 mm

D 0.03 mm

The Correct Answer is: B
In digital imaging, pixel size is determined by dividing the FOV by the matrix. In this case, the FOV is 35 cm; since the answer is expressed in millimeters, first change 35 cm to 350 mm. Then 350 divided by 1024 equals 0.35 mm.

The FOV and matrix size are independent of one another, that is, either can be changed and the other will remain unaffected. However, pixel size is affected by changes in either the FOV or matrix size. For example, if the matrix size is increased, pixel size decreases. If FOV increases, pixel size increases. Pixel size is inversely related to resolution. As pixel size increases, resolution decreases. (Fosbinder & Kelsey, p

If 85 kVp, 400 mA, and ⅛ s were used for a particular exposure using single-phase equipment, which of the following milliamperage or time values would be required, all other factors being constant, to produce a similar density using three-phase, 12-pulse equipment?

A 200 mA

B 600 mA

C 0.125 s

D 0.25 s

he Correct Answer is: A
With three-phase equipment, the voltage never drops to zero, and x-ray intensity is significantly greater. When changing from single-phase to three-phase, six-pulse equipment, two-thirds of the original milliampere-seconds are required to produce a radiograph with similar density. (When going from three-phase, six-pulse to single-phase, add one-third more milliampere-seconds.) When changing from single-phase to three-phase, 12-pulse equipment, only one-half of the original milliampere-seconds is required. (Going from three-phase, 12-pulse to single-phase requires twice the milliampere-seconds.) In this instance, we are changing from single-phase to three-phase, 12-pulse equipment; therefore, the new milliampere-seconds value should be half the original 50 mAs, or 25 mAs. The only selection that will provide 25 mAs is (A), 200 mA. (B) will produce 75 mAs (600 mA × ⅛ s = 75 mAs); (C) will produce 50 mAs (400 mA × 0.125 s = 50 mAs); (D) will produce 100 mAs (400 × 0.25 = 100 mAs). (Carlton and Adler, 4th ed., p. 96) Comparison of technical factors required Single phase  Three phase  Three phase 6-pulse 12-pulse x mAs ⅔ x mAs ½ x mAs

How is source-to-image distance (SID) related to exposure rate and radiographic density?

A As SID increases, exposure rate increases and radiographic density increases.

B As SID increases, exposure rate increases and radiographic density decreases.

C As SID increases, exposure rate decreases and radiographic density increases.

D As SID increases, exposure rate decreases and radiographic density decreases.

he Correct Answer is: D
According to the inverse-square law of radiation, the intensity or exposure rate of radiation from its source is inversely proportional to the square of the distance. Thus, as distance from the source of radiation increases, exposure rate decreases. Because exposure rate and radiographic density are directly proportional, if the exposure rate of a beam directed to the IR is decreased, the resulting radiographic density would be decreased proportionally. (Selman, 9th ed., p. 117)

In which of the following ways can higher radiographic contrast be obtained in abdominal radiography?

1. By using lower kilovoltage

2. By using a contrast medium

3. By limiting the field size

A 1 only

B 1 and 2 only

C 1 and 3 only

D 1, 2, and 3

The Correct Answer is: D
Higher contrast is shorter scale contrast; it is present in an image that has few shades of gray between white and black. High radiographic contrast is, in part, a result of lower energy photons (lower kVp). High radiographic contrast also results when radiographing anatomic parts that have high subject contrast, such as the chest. The abdomen has low subject contrast, and therefore abdominal radiographs will tend to have very low contrast unless technical factors are selected to increase contrast. To produce high radiographic contrast in abdominal radiography, lower kVp should be used. To better demonstrate high contrast within a viscus, a contrast medium such as barium, iodine, or air can be used. Restricting the size of the field will also function to increase contrast because less scattered radiation will be generated. (Carlton & Adler, p 397)

If 92 kV and 12 mAs were used for a particular abdominal exposure with single-phase equipment, what mAs would be required to produce a similar radiograph with three-phase, six-pulse equipment?

A 36

B 24

C 8

he Correct Answer is: C
Single-phase radiographic equipment is much less efficient than three-phase equipment because it has a 100% voltage ripple. With three-phase equipment, voltage never drops to zero, and x-ray intensity is significantly greater. To produce similar density, only two thirds of the original mAs would be used for three-phase, six-pulse equipment (2/3 × 12 = 8 mAs). With 3-phase, 12-pulse equipment, the original mAs would be cut in half. (Saia, pp 329, 330

Which of the following pathologic conditions are considered additive conditions with respect to selection of exposure factors?

1. Osteoma
2. Bronchiectasis
3. Pneumonia

A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
All these conditions are considered technically additive because they all involve an increase in tissue density. Osteoma, or exostosis, is a (usually benign) bony tumor that can develop on bone. Bronchiectasis is a chronic dilatation of the bronchi with accumulation of fluid. Pneumonia is inflammation of the lung(s) with accumulation of fluid. Additional bony tissues and the pathologic presence of fluid are additive pathologic conditions and require an increase in exposure factors. Destructive conditions such as osteoporosis require a decrease in exposure factors. (Carlton and Adler, 4th ed., p. 249)

An exposure was made of a part using 300 mA and 0.06 second with a 200-speed film–screen combination. An additional radiograph is requested using a 400-speed system to reduce motion unsharpness. Using 400 mA, all other factors remaining constant, what should be the new exposure time?

A 5 ms

B 11 ms

C 22 ms

D 44 ms

The Correct Answer is: C
High-speed imaging systems are valuable for reducing patient exposure and patient motion. However, some detail will be sacrificed, and quantum mottle can cause further image impairment. In general, doubling the film–screen speed doubles the radiographic density, thereby requiring that the milliampere-seconds value be halved to maintain the original radiographic density. Changing from 200 to 400 screens requires halving the milliampere-seconds value to 9 mAs. The new exposure time, using 400 mA, is 400x = 9. Thus, x = 0.0225-s exposure using 400 mA and 400-speed screens (0.0225 = 22.5 ms). (Selman, 9th ed., p. 181)

To produce a just perceptible increase in radiographic density, the radiographer must increase the

A mAs by 30%

B mAs by 15%

C kV by 15%

D kV by 30%

The Correct Answer is: A
If a radiograph lacks sufficient blackening, an increase in milliampere-seconds is required. The milliampere-seconds value regulates the number of x-ray photons produced at the target. An increase or decrease in milliampere-seconds of at least 30% is necessary to produce a perceptible effect. Increasing the kilovoltage by 15% will have about the same effect as doubling the milliampere-seconds. (Shephard, p. 173)

If 40 mAs and a 200-speed screen–film system were used for a particular exposure, what new milliampere-seconds value would be required to produce the same density if the screen–film system were changed to 800 speed?

A 10

B 20

C 80

D 160

The Correct Answer is: A
The screen–film system and radiographic density are directly proportional; that is, if the system speed is doubled, the radiographic density is doubled. In this case, we started at 40 mAs with a 200-speed system. If the system speed is doubled to 400, we should decrease the milliampere-seconds to 20 mAs. If the speed is again doubled to 800, we use half the 20 mAs, or 10 mAs. Or milliampere-seconds conversion factors and the following formula may be used:

Thus, x = 10 mAs with a 800-speed screen–film system. (Shephard, p. 184)

A decrease from 200 to 100 mA will result in a decrease in which of the following?

1. Wavelength
2. Exposure rate
3. Beam intensity

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C
Technical factors can be expressed in terms of milliampere-seconds rather than milliamperes and time. The milliampere-seconds value is a quantitative factor because it regulates x-ray-beam intensity, exposure rate, quantity, or number of x-ray photons produced (the milliampere-seconds value is the single most important technical factor associated with image density and is the factor of choice for regulating radiographic/optical density). The milliampere-seconds value is directly proportional to the intensity (i.e., exposure rate, number, and quantity) of x-ray photons produced and the resulting radiographic density. If the milliampere-seconds value is doubled, twice the exposure rate and twice the density occur. If the milliampere-seconds value is cut in half, the exposure rate and resulting density are cut in half. Kilovoltage is the qualitative exposure factor—it determines beam quality by regulating photon energy (i.e., wavelength). (Carlton and Adler, 5th ed., p. 183)

What is the best way to reduce magnification distortion?

A Use a small focal spot.

B Increase the SID.

C Decrease the OID.

D Use a slow screen-film combination.

The Correct Answer is: C
There are two types of distortion: size and shape. Shape distortion relates to the alignment of the x-ray tube, the part to be radiographed, and the image recorder. There are two kinds of shape distortion: elongation and foreshortening. Size distortion is magnification, and it is related to the OID and the SID. Magnification can be reduced by either increasing the SID or decreasing the OID. However, an increase in SID must be accompanied by an increase in mAs to maintain density. It is therefore preferable, in the interest of exposure, to reduce OID whenever possible. (Fauber, p 90)

With milliamperage adjusted to produce equal exposures, all the following statements are true except

A a single-phase examination done at 10 mAs can be duplicated with three-phase, 12-pulse at 5 mAs.

B There is greater patient dose with three-phase equipment than with single-phase equipment.

C Three-phase equipment can produce comparable radiographs with less heat unit (HU) buildup.

D Three-phase equipment produces lower-contrast radiographs than single-phase equipment.

The Correct Answer is: B
If the same kilovoltage is used with single-phase and three-phase equipment, the three-phase unit will require about 50% fewer milliampere-seconds to produce similar radiographs. Because three-phase equipment has much higher effective voltage than single-phase equipment, the three-phase radiograph will have lower contrast. A lower milliampere-seconds value can be used with three-phase equipment, so heat units are not built up as quickly. When technical factors are adjusted to obtain the same density and contrast, there is no difference in patient dose. (Selman, 9th ed., pp. 162–164)

Exposure rate increases with an increase in 1. mA. 2. kVp. 3. SID.

A 1 only

B 1 and 2 only

C 2 and 3 only

D 1, 2, and 3

The Correct Answer is: B
The quantity of x-ray photons produced at the target is the function of mAs. The quality (wavelength, penetration, energy) of x-ray photons produced at the target is the function of kVp. The kVp also has an effect on exposure rate, because an increase in kVp will increase the number of high-energy x-ray photons produced at the target. Exposure rate decreases with an increase in SID. (Selman, p 117)

Exposures less than the minimum response time of an AEC may be required when

1. using high milliamperage
2. using fast film–screen combinations
3. examining large patients or body parts

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
The minimum response time, or minimum reaction time, is the length of the shortest exposure possible with a particular AEC. If less than the minimum response time is required for a particular exposure, the radiograph will exhibit excessive density. This problem becomes apparent when making exposures that require very short exposure times, such as when using high-milliamperage and fast film–screen combinations. To resolve this problem, the radiographer should decrease the milliamperage rather than the kilovoltage in order to leave contrast unaffected. (Carlton and Adler, 4th ed., p. 102)

The use of which of the following is (are) essential in magnification radiography?

1. High-ratio grid
2. Fractional focal spot
3. Direct exposure technique

A 1 only B 2 only C 1 and 3 only D 1, 2, and 3

he Correct Answer is: B
Magnification radiography is used to enlarge details to a more perceptible degree. Hairline fractures and minute blood vessels are candidates for magnification radiography. The problem of magnification unsharpness is overcome by using a fractional focal spot; larger focal-spot sizes will produce excessive blurring unsharpness. Grids are usually unnecessary in magnification radiography because of the air-gap effect produced by the OID. Direct-exposure technique probably would not be used because of the excessive exposure required. (Selman, 9th ed., pp. 226–228)

In comparison with 60 kV, 80 kV will

1. permit greater exposure latitude
2. produce more scattered radiation
3. produce shorter-scale contrast

A 1 only B 2 only C 1 and 2 only D 2 and 3 only

The Correct Answer is: C
The higher the kilovoltage range, the greater is the exposure latitude (margin of error in exposure). Higher kilovoltage produces more energetic photons, is more penetrating, and produces more grays on the radiographic image, lengthening the scale of contrast. As kilovoltage increases, the percentage of scattered radiation also increases. (Bushong, 9th ed., pp. 256-257)

Which of the following is performed to check the correctness of the developing parameters?

A Densitometry

B A thorough cleaning of rollers

C A warm-up procedure

D Sensitometry

The Correct Answer is: D
Sensitometry is a method of quality control for daily monitoring of an automatic film processor. A densitometer is a device used to read optical density. Crossover rollers should be cleaned daily to prevent the buildup of crystallized solution on the rollers. A warm-up procedure is performed on an x-ray tube for safe operation after prolonged disuse. (Selman, 9th ed., pp. 292–293)

Which of the following will result if developer replenishment is inadequate?

A Images with excessively high contrast

B Images with excessively low contrast

C Images with excessively high density

D Dry, brittle films

The Correct Answer is: B
As films are developed, the developer solution becomes weaker, and oxidation products are produced in the solution. If sufficient replenishment of new developer solution does not take place, the activity of the older solution decreases, and chemical fog is produced. Films lack contrast and have a flat, gray appearance. (Shephard, p. 153)

Which of the following will contribute to the production of longer-scale radiographic contrast?

1. An increase in kV 2. An increase in grid ratio 3. An increase in photon energy

A 1 only

B 1 and 2 only

C 1 and 3 only

D 1, 2, and 3

he Correct Answer is: C
Increased photon energy is caused by an increase in kVp, resulting in more penetration of the part and a longer scale of contrast. Increasing the grid ratio will result in a larger percentage of scattered radiation being absorbed and hence a shorter scale of contrast. (Shephard, pp 203–204)

Which of the following terms is used to describe unsharp edges of tiny radiographic details?

A Diffusion

B Mottle

C Blur

D Umbra

The Correct Answer is: C
Recorded detail is evaluated by how sharply tiny anatomic details are imaged on the radiograph. The area of blurriness that may be associated with small image details is termed geometric blur. The blurriness can be produced by using a large focal spot or by diffused fluorescent light from intensifying screens. The image proper (i.e., without blur) is termed the umbra. Mottle is a grainy appearance caused by fast imaging systems. (Selman, 9th ed., pp. 206–207)

A radiograph made using 300 mA, 0.1 second, and 75 kV exhibits motion unsharpness but otherwise satisfactory technical quality. The radiograph will be repeated using a shorter exposure time. Using 86 kV and 400 mA, what should be the new exposure time?

A 25 ms

B 37 ms

C 50 ms

D 75 ms

The Correct Answer is: B
The milliampere-seconds (mAs) formula is milliamperage × time = mAs. With two of the factors known, the third can be determined. To find the milliampere-seconds value that was used originally, substitute the known values:

We have increased the kilovoltage to 86 kV, an increase of 15%, which has an effect similar to that of doubling the milliampere-seconds. Therefore, only 15 mAs is now required as a result of the kilovoltage increase:

Central ray angulation may be required for

1. magnification of anatomic structures
2. foreshortening or self-superimposition
3. superimposition of overlying structures

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C
If structures are overlying or underlying the area to be demonstrated (e.g., the medial femoral condyle obscuring the joint space in the lateral knee projection), CR angulation is used (e.g., 5-degree cephalad angulation to see the joint space in the lateral knee).

If structures are likely to be foreshortened or self-superimposed (e.g., the scaphoid in a PA wrist), CR angulation may be employed to place the structure more closely parallel with the IR.

Another example is the oblique cervical spine, where cephalad or caudad angulation is required to “open” the intervertebral foramina.

Magnification is controlled by object-to- image-receptor distance (OID) and SID; it is unrelated to CR angulation. (Frank, Long, and Smith, 11th ed., vol. 1, p 307)

Geometric unsharpness is most likely to be greater

A at long SIDs.

B at the anode end of the image.

C with small focal spots.

D at the cathode end of the image.

he Correct Answer is: D
The x-ray tube anode is designed according to the line focus principle, that is, with the focal track beveled (see the figure below). This allows a larger actual focal spot to project a smaller effective focal spot, resulting in improved recorded detail with less blur. However, because of the target angle, penumbral blur varies along the longitudinal tube axis, being greater at the cathode end of the image and less at the anode end of the image. (Bushong, 8th ed., p. 287

A radiograph made using 300 mA, 0.1 second, and 75 kVp exhibits motion unsharpness, but otherwise satisfactory technical quality. The radiograph will be repeated using a shorter exposure time. Using 86 kV and 500 mA, what should be the new exposure time?

A 0.12 second

B 0.06 second

C 0.03 second

D 0.01 second

The Correct Answer is: C
The mAs formula is milliamperage × time = mAs. With two of the factors known, the third can be determined. To find the mAs that was originally used, substitute the known values:

300 × 0.1 = 30

We have increased the kilovoltage to 86, an increase of 15%, which has an effect similar to that of doubling the mAs. Therefore, only 15 mAs is now required as a result of the kV increase:

mA × s = mAs

500 x = 15

x = 0.03-second exposure

(Selman, p 214)

Recorded detail is directly related to

1. SID.
2. tube current.
3. focal-spot size.

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: A
As SID increases, so does recorded detail because magnification is decreased. Therefore, SID is directly related to recorded detail. As focal spot size increases, recorded detail decreases because more penumbra is produced. Focal spot size is thus inversely related to radiographic sharpness or recorded detail. Tube current affects radiographic density and is unrelated to recorded detail. (Fauber, 2nd ed., pp. 79, 81)

Exposure factors of 100 kVp and 6 mAs are used with a 6:1 grid for a particular exposure. What should be the new milliampere-seconds value if a 12:1 grid is substituted?

A 7.5 mAs

B 10 mAs

C 13 mAs

D 18 mAs

The Correct Answer is: B
To change nongrid to grid exposure, or to adjust exposure when changing from one grid ratio to another, recall the factor for each grid ratio:

The grid conversion formula is

Substituting known quantities:

Thus, x = 10 mAs with a 12:1 grid. (Shephard, p. 248)

An exposure was made using 300 mA, 40 ms exposure, and 85 kV. Each of the following changes will decrease the radiographic density by one half except a change to

A 1/50 s exposure

B 72 kV

C 10 mAs

D 150 mA

he Correct Answer is: C
Radiographic density is directly proportional to milliampere-seconds. If exposure time is halved from 40 ms (0.04 or 1 / ₂₅ ) s to 0.02 ( 1 / ₅₀ ) s, radiographic density will be cut in half. Changing to 150 mA also will halve the milliampere-seconds, effectively halving the radiographic density. If the kilovoltage is decreased by 15%, from 85 to 72 kV, radiographic density will be halved according to the 15% rule. To cut the density in half, the milliampere-seconds value must be reduced to 6 mAs (rather than 10 mAs). (Selman, 9th ed., pp. 213–214)

A decrease from 90 to 77 kVp will result in a decrease in which of the following?

1. Wavelength
2. Scale of grays
3. Optical density

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: C
As kilovoltage is increased, more electrons are driven to the anode with greater speed and energy. More high-energy electrons will result in production of more (i.e., increased exposure rate) high-energy (i.e., short-wavelength) x-rays. Thus peak kilovoltage affects both quantity and quality of the x-ray beam. However, although peak kilovoltage and radiographic density are directly related, they are not directly proportional; that is, twice the radiographic density does not result from doubling the kilovoltage. With respect to the effect of peak kilovoltage on image density, there is a convenient rule (15% rule) that can be followed. If it is desired to double the radiographic density yet impossible to adjust the milliampere-seconds, a similar effect can be achieved by increasing the kilovoltage by 15%. Conversely, the density may be cut in half by decreasing the kilovoltage by 15%. (Shephard, pp. 194–197)

Which of the following is (are) associated with subject contrast?

1. Patient thickness
2. Tissue density
3. Kilovoltage

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

he Correct Answer is: D
Radiographic contrast is the sum of film emulsion contrast and subject contrast. Subject contrast has by far the greatest influence on radiographic contrast. Several factors influence subject contrast, each as a result of beam-attenuation differences in the irradiated tissues. As patient thickness and tissue density increase, attenuation increases, and subject contrast is increased. As kilovoltage increases, higher-energy photons are produced, beam attenuation is decreased, and subject contrast decreases. (Carlton and Adler, 4th ed., p. 433)

A 15% increase in kVp accompanied by a 50% decrease in mAs will result in a(n)

A shorter scale of contrast.

B increase in exposure latitude.

C increase in radiographic density.

D decrease in recorded detail.

The Correct Answer is: B
A 15% increase in kVp with a 50% decrease in mAs serves to produce a radiograph similar to the original, but with some obvious differences. The overall blackness (radiographic density) is cut in half because of the decrease in mAs. However, the loss of blackness is compensated for by the addition of grays (therefore, longer-scale contrast) from the increased kVp. The increase in kVp also increases exposure latitude; there is a greater margin for error in higher kVp ranges. Recorded detail is unaffected by changes in kVp. (Fauber, pp 59–60)

Which of the following technical changes would best serve to remedy the effect of very dissimilar tissue densities?

A Use of high-speed screens

B Use of a high-ratio grid

C High-kilovoltage exposure factors

D High milliampere-seconds exposure factors

The Correct Answer is: C
When tissue densities within a part are very dissimilar (e.g., chest x-ray), the radiographic result can be unacceptably high contrast. To “even out” these densities and produce a more appropriate scale of grays, exposure factors using high kilovoltage should be employed. Radiographic contrast generally increases with an increase in screen speed. The higher the grid ratio, the higher is the contrast. Exposure factors using high milliampere-seconds generally result in excessive image density, frequently obliterating much of the gray scale. (Bushong, 8th ed., p. 273; Shephard, p. 200)

Which of the following combinations will result in the most scattered radiation reaching the image receptor?

A Using more mAs and compressing the part

B Using more mAs and a higher ratio grid

C Using fewer mAs and more kVp

D Using more mAs and less kVp

he Correct Answer is: C
As x-ray photons travel through a part, they either pass all the way through to expose the film/image receptor, or they undergo interaction(s) that may result in their being absorbed by the part or deviated in direction. It is those that change direction (scattered radiation) that undermine the image. With respect to the radiographic image, it is responsible for the scattered radiation that reaches the film/image receptor. Scattered radiation adds unwanted, degrading densities to the radiographic image.

The single most important way to reduce the production of scattered radiation is to collimate. Although collimation, use of lower kVp (with appropriately higher mAs), and compression can be used, a large amount of scattered radiation is still generated within the part being radiographed and, because it adds unwanted noninformation-carrying densities, it can have a severely degrading effect on image quality, thus the need for grids. (Bushong, 8th ed, p 236)

The effect described as differential absorption is

1. responsible for radiographic contrast
2. a result of attenuating characteristics of tissue
3. minimized by the use of a high peak kilovoltage

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D
Differential absorption refers to the x-ray absorption characteristics of neighboring anatomic structures. The radiographic representation of these structures is referred to as radiographic contrast; it may be enhanced with high-contrast technical factors, especially using low kilovoltage levels. At low-kilovoltage levels, the photoelectric effect predominates. (Bushong, 8th ed., pp. 181–184)

Decreasing field size from 14 × 17 into 8 × 10 inches will

A decrease radiographic density and decrease the amount of scattered radiation generated within the part

. B decrease radiographic density and increase the amount of scattered radiation generated within the part.

C increase radiographic density and increase the amount of scattered radiation generated within the part.

D increase radiographic density and decrease the amount of scattered radiation generated within the part

The Correct Answer is: A
Limiting the size of the radiographic field serves to limit the amount of scattered radiation produced within the anatomic part. As the amount of scattered radiation generated within the part decreases, so does the resultant density within the radiographic image. Hence, beam restriction is a very effective means of reducing the quantity of non-information-carrying scattered radiation (fog) produced, resulting in a shorter scale of contrast with fewer radiographic densities. (Shephard, p 203)

Disadvantages of using low-kilovoltage technical factors include

1. insufficient penetration
2. increased patient dose
3. diminished latitude

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

he Correct Answer is: D
As the kilovoltage is decreased, x-ray-beam energy (i.e., penetration) is also decreased. Consequently, a shorter scale of contrast is obtained, and at lower kilovoltage levels, there is less exposure latitude (less margin for error in exposure). As kilovoltage is reduced, the milliampere-seconds value must be increased accordingly to maintain adequate density. This increase in milliampere-seconds results in greater patient dose. (Shephard, p. 204)

A decrease in kilovoltage will result in

A a decrease in optical density

B a decrease in contrast

C a decrease in recorded detail

D a decrease in image resolution

The Correct Answer is: A
As kilovoltage is increased, more electrons are driven to the anode with greater speed and energy. More high-energy electrons will result in production of more high-energy x-rays. Thus, kilovoltage affects both quantity and quality (energy) of the x-ray beam. However, although kilovoltage and radiographic density are directly related, they are not directly proportional; that is, twice the radiographic density does not result from doubling the kilovoltage. With respect to the effect of kilovoltage on image density, if it is desired to double the radiographic density yet impossible to adjust the milliampere-seconds, a similar effect can be achieved by increasing the kilovoltage by 15%. Conversely, the density may be cut in half by decreasing the kilovoltage by 15%. Therefore, a decrease in kilovoltage will produce fewer x-ray photons, resulting in decreased density. Additionally, a decrease in kilovoltage will produce fewer shades of gray, that is, a shorter-scale, or higher/increased, contrast. Kilovoltage is unrelated to recorded detail and resolution. (Shephard, pp. 178, 203–204)

Which of the following groups of technical factors will produce the least radiographic density?

A 400 mA, 0.010 second, 94 kV, 100-speed screens

B 500 mA, 0.008 second, 94 kV, 200-speed screens

C 200 mA, 0.040 second, 94 kV, 50-speed screens

D 100 mA, 0.020 second, 80 kV, 200-speed screens

he Correct Answer is: D
Each milliampere-second setting is determined [(A) = 4; (B) = 4; (C) = 8; (D) = 2] and numbered in order of greatest to least density [(C) = 1; (A) and (B) = 2; (D) = 3]. Then, the kilovoltages are reviewed and also numbered in order of greatest to least density [(A), (B), and (C) = 1; (D) = 2]. Next, screen speeds are numbered from greatest density-producing to least density-producing [(D) and (B) = 1; (A) = 2; (C) = 3]. Finally, the numbers assigned to the milliampere-seconds, kilovoltage, and screen speed are added up for each of the four groups [(B) = 4; (A) and (C) = 5; (D) = 6]; the lowest total (B) indicates the group of factors that will produce the greatest radiographic density; the highest total (D) indicates the group of factors that will produce the least radiographic density. This process is illustrated as follows:

(A) 4 mAs (2) + 94 kV (1) + 100 screens (2) = 5

(B) 4 mAs (2) + 94 kV (1) + 200 screens (1) = 4

(C) 8 mAs (1) + 94 kV (1) + 50 screens (3) = 5

(D) 2 mAs (3) + 80 kV (2) + 200 screens (1) = 6

(Shephard, p. 179)

Which of the following quantities of filtration is most likely to be used in mammography?

A 0.5 mm Mo

B 1.5 mm Al

C 1.5 mm Cu

D 2.0 mm Cu

The Correct Answer is: A
Soft tissue radiography requires the use of long-wavelength, low-energy x-ray photons. Very little filtration is used in mammography. Certainly, anything more than 1.0 mm of aluminum would remove the useful soft photons, and the desired high contrast could not be achieved. Dedicated mammographic units usually have molybdenum targets (for the production of soft radiation) and a small amount of molybdenum filtration. (Carlton & Adler, p 581)

Focal-spot blur is greatest

A directly along the course of the central ray

B toward the cathode end of the x-ray beam

C toward the anode end of the x-ray beam

D as the SID is increased

The Correct Answer is: B
Focal-spot blur, or geometric blur, is caused by photons emerging from a large focal spot. The actual focal spot is always larger than the effective (or projected) focal spot, as illustrated by the line-focus principle. In addition, the effective focal-spot size varies along the longitudinal tube axis, being greatest in size at the cathode end of the beam and smallest at the anode end of the beam. Because the projected focal spot is greatest at the cathode end of the x-ray tube, geometric blur is also greatest at the corresponding part (cathode end) of the radiograph. (Bushong, 8th ed., pp. 287–288)

Which of the following groups of exposure factors would be most effective in eliminating prominent pulmonary vascular markings in the RAO position of the sternum?

A 500 mA, 1/30 s, 70 kV

B 200 mA, 0.04 second, 80 kV

C 300 mA, 1/10 s, 80 kV

D 25 mA, 7/10 s, 70 kV

The Correct Answer is: D
In the RAO position, the sternum must be visualized through the thorax and heart. Prominent pulmonary vascular markings can hinder good visualization. A method frequently used to overcome this problem is to use a milliampere-seconds value with a long exposure time. The patient is permitted to breathe normally during the (extended) exposure and by so doing blurs out the prominent vascularities. (Frank, Long, and Smith, 11th ed., vol. 1, p. 470)

Accurate operation of the AEC device depends on

1. the thickness and density of the object
2. positioning of the object with respect to the photocell
3. beam restriction

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
The AEC automatically terminates the exposure when the proper density has been recorded on the film. The important advantage of the phototimer, then, is that it can accurately duplicate radiographic densities. It is very useful in providing accurate comparison in follow-up examinations and in decreasing patient exposure dose by reducing the number of “retakes” needed because of improper exposure. The AEC automatically adjusts the exposure required for body parts with different thicknesses and densities. However, proper functioning of the phototimer depends on accurate positioning by the radiographer. The correct photocell(s) must be selected, and the anatomic part of interest must completely cover the photocell to achieve the desired density. If collimation is inadequate and a field size larger than the part is used, excessive scattered radiation from the body or tabletop can cause the AEC to terminate the exposure prematurely, resulting in an underexposed radiograph. (Carlton and Adler, 4th ed., pp. 540–541)

The quantity of scattered radiation reaching the IR can be reduced through the use of

1. a fast imaging system
2. an air gap
3. a stationary grid

A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C
Scattered radiation adds unwanted degrading densities to the x-ray image. The single most important way to reduce the production of scattered radiation is to collimate. Although collimation, optimal kilovoltage, and compression can be used, a large amount of scattered radiation is still generated within the part being imaged, and because it adds unwanted non–information-carrying densities, it can have a severely degrading effect on image quality. A grid is a device interposed between the patient and IR that functions to absorb a large percentage of scattered radiation before it reaches the IR. It is constructed of alternating strips of lead foil and radiolucent filler material. X-ray photons traveling in the same direction as the primary beam pass between the lead strips. X-ray photons, having undergone interactions within the body and deviated in various directions, are absorbed by the lead strips; this is referred to as cleanup of scattered radiation. An air gap introduced between the object and IR can have an effect similar to that of a grid. As energetic scattered radiation emerges from the body, it continues to travel in its divergent fashion and much of the time will bypass the IR. It is usually necessary to increase the SID to reduce magnification caused by increased OID. Imaging system speed is unrelated to scattered radiation. (Shephard, pp. 244, 263)

A lateral radiograph of the cervical spine was made at 40 in. using 300 mA and 0.03 second exposure. If it is desired to increase the distance to 72 in., what should be the new milliampere (mA) setting, all other factors remaining constant?

A 400 mA

B 800 mA

C 1000 mA

D 1200 mA

The Correct Answer is: C
When exposure rate decreases (as a result of increased SID), an appropriate increase in milliampere-seconds is required to maintain the original radiographic density. Unless exposure is increased, the resulting radiograph will be underexposed. The formula used to determine the new milliampere-seconds value (density-maintenance formula) is substituting known values:

Substituting known values:

Thus, x = 29.16 mAs at 72 in. SID. To determine the required milliamperes (mA × s = mAs),

0.03 x = 29.16

x = 972 mA

(Selman, 9th ed., p. 214)

The squeegee assembly in an automatic processor

1. functions to remove excess solution from films
2. is located near the crossover rollers
3. helps establish the film's rate of travel

A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: C
An exposed radiographic film contains an invisible (latent) image. Only through processing can this image be converted to a permanent, visible (manifest) image. As the film exits the developer section, it passes through the crossover assembly, and before it enters the fixer section, it passes through the squeegee assembly. The squeegee assembly rollers function to remove excess developer solution from the emulsion before the film enters the fixer. This process helps to maintain fixer strength/activity. The rate of travel through the processor is determined by the transport mechanism, that is, the speed of the rollers as established at time of manufacture. (Shephard, p. 143

The appearance of underexposure on an image created using a high-speed film–screen system can be caused by all the following except

A insufficient mAs

B insufficient kV

C insufficient SID

D insufficient development

The Correct Answer is: C
Higher-speed film/screen imaging systems are used often to produce more density with less exposure. Since the milliampere-seconds value is the factor controlling image density, an insufficient amount would result in underexposure. Since kilovoltage has a definite effect on image density, an insufficient amount would result in underexposure. Underdevelopment of the latent film emulsion image also would result in insufficient density. However, insufficient SID (i.e., SID too low) would cause an increase in image density/overexposure. (Shephard p. 179)

If 82 kVp, 300 mA, and 0.05 second were used for a particular exposure using 3-phase, 12-pulse equipment, what mAs would be required, using single-phase equipment, to produce a similar radiograph?

A 7.5

B 20

C 30

D 50

The Correct Answer is: C
With three-phase equipment, the voltage never drops to zero and x-ray intensity is significantly greater. When changing from single-phase to three-phase, six-pulse equipment, two-thirds of the original mAs is required to produce a radiograph with similar density. When changing from single-phase to three-phase, 12-pulse equipment, only one-half of the original mAs is required. In this problem, we are changing from three-phase, 12-pulse to single-phase equipment; therefore, the mAs should be doubled (from 15 to 30 mAs). (Carlton & Adler, p 98)

Which of the following terms is used to express resolution/recorded detail?

A Kiloelectronvolts (keV)

B Modulation transfer function (MTF

C Relative speed

D Latitude

he Correct Answer is: B
Resolution describes how closely fine details may be associated and still be recognized as separate details before seeming to blend into each other and appear as one. The degree of resolution transferred to the image receptor is a function of the resolving power of each of the system components and can be expressed in line pairs per millimeter (lp/mm), line-spread function (LSP), or modulation transfer function (MTF). Line pairs per millimeter can be measured using a resolution test pattern; a number of resolution test tools are available. LSP is measured using a 10-μm x-ray beam; MTF measures the amount of information lost between the object and the IR. (Carlton and Adler, 4th ed., p. 334)

Practice(s) that enable the radiographer to reduce the exposure time required for a particular image include

1. use of a higher milliamperage
2. use of a higher kilovoltage
3. use of a faster film–screen combination

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
If it is desired to reduce the exposure time for a particular radiograph, as it might be when radiographing patients who are unable to cooperate fully, the milliamperage must be increased sufficiently to maintain the original milliampere-seconds value and thus radiographic density. A higher kilovoltage could be useful because it would allow further reduction of the milliampere-seconds (exposure time) according to the 15% rule. Use of a higher-speed film–screen combination also helps to reduce milliampere-seconds (exposure time) through more efficient conversion of photon energy to fluorescent light energy. (Selman, 9th ed., pp. 182, 214)

An increase from 78 to 92 kVp will result in an increase in which of the following?

1. Latitude
2. Scale of grays
3. Optical density

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D
As kilovoltage is increased, more electrons are driven to the anode with greater speed and energy. More high-energy electrons will result in production of more (i.e., increased exposure rate) high-energy (i.e., short-wavelength) x-rays. Thus, kilovoltage affects both quantity and quality of the x-ray beam. However, although kilovoltage and radiographic density are directly related, they are not directly proportional; that is, twice the radiographic density does not result from doubling the kilovoltage. Exposure latitude is related to the technical factors selected. Exposure latitude is the leeway, or margin of error, one has with a given group of exposure factors; the degree of exposure latitude is determined by the kilovoltage. At higher kilovoltages, a difference of a few kilovolts will make little, if any, difference radiographically (recall the 15% rule). At low kilovoltages, however, an error of just a few kilovolts can make a very noticeable difference in the radiographic image. Although the radiographer has little control over film latitude, he or she has much control of exposure latitude. (Shephard, pp. 194–197)

A dry laser printer is generally used when it is necessary to print digital images on film. This laser film is loaded under the following conditions

A daylight

B safelight with red filter

C safelight with amber filter

D total darkness

he Correct Answer is: A
Some physicians or departments might still require some x-ray images to be printed on film; when that is necessary, laser film is used. Laser film is dry imaging film having a blue or clear base. Laser film can be loaded in daylight conditions.

The older, traditional x-ray film was exposed by light and x-rays and required special safelight conditions in a specially designed darkroom. These conditions are not required in digital imaging.

Seerem 2011 pg 181; http://www.premierdigitalxray.com/pdf/Fuji_drypix_4000.pd

Focusing distance is associated with which of the following?

A Computed tomography

B Chest radiography

C Magnification radiography

D Grids

The Correct Answer is: D
Focusing distance is the term used to specify the optimal SID used with a particular focused grid. It is usually expressed as focal range, indicating the minimum and maximum SID workable with that grid. Lesser or greater distances can result in grid cutoff. Although proper distance is important in computed tomography and chest and magnification radiography, focusing distance is unrelated to them. (Selman, 9th ed., pp. 239–240)

An AP radiograph of the femur was made using 300 mA, 0.03 second, 76 kV, 40-in. SID, 1.2-mm focal spot, and a 400-speed film–screen system. With all other factors remaining constant, which of the following exposure times would be required to maintain radiographic density using 87 kV, a 200-speed film–screen system, and the addition of a 12:1 grid?

A 0.15 second

B 0.20 second

C 0.4 second

D 0.6 second

The Correct Answer is: A
If the imaging system speed is cut in half (from 400–200 speed), the result will be half the original density on the radiograph. Therefore, to maintain the original density, the milliampere-seconds value must be doubled from the original 9–18 mAs. A 15% increase in kilovoltage was made, increasing the kilovoltage to 87 kV. Because the kilovoltage change effectively doubles the radiographic density, the milliampere-seconds value now must be cut in half (from 18–9 mAs) to compensate. Grids are used to absorb scattered radiation from the remnant beam before it can contribute to the latent image. Because the grid removes scattered (and some primary/useful photons as well) radiation from the beam, an increase in exposure factors is required. The amount of increase depends on the grid ratio: The higher the grid ratio, the higher is the correction factor. The correction factor for a 12:1 grid is 5; therefore, the milliampere-seconds value (9) is multiplied by 5 to arrive at the new required milliampere-seconds value (45). Using the milliampere-seconds equation mA × s = mAs, it is determined that 0.15 second will be required at 300 mA:

(Fauber, pp. 62, 148)

The exposure factors of 400 mA, 70 ms, and 78 kV were used to produce a particular radiographic density and contrast. A similar radiograph can be produced using 500 mA, 90 kV, and

A 14 ms

B 28 ms

C 56 ms

D 70 ms

The Correct Answer is: B
First, evaluate the change(s): The kilovoltage was increased by 15% (78 + 15% = 90). A 15% increase in kilovoltage will double the radiographic density; therefore, it is necessary to use half the original milliampere-seconds value to maintain the original density. The original milliampere-seconds value was 28 mAs (400 mA × 0.07 second [70 ms] 28 mAs), so we now need 14 mAs, using 500 mA. Because mA × s mAs:

(Fauber, pp. 55, 59–60

Better resolution is obtained wit

h A high SNR.

B low SNR.

C windowing.

D smaller matrix.

The Correct Answer is: A
Spatial resolution increases as SNR (signal-to-noise ratio) increases. A high SNR (e.g., 1000:1) indicates that there is far more signal than noise. A lower SNR (e.g., 200:1) indicates a "noisy" image. Windowing is unrelated to resolution; it allows for contrast and density post-processing manipulation. Image matrix has a great deal to do with resolution. A larger image matrix (1800 × 1800) offers better resolution than a smaller image matrix (700 × 700). Smaller image matrices look "pixelly." (Seeram, p 112)

A satisfactory radiograph was made using a 36-in. SID, 12 mAs, and a 12:1 grid. If the examination will be repeated at a distance of 42 in. and using a 5:1 grid, what should be the new milliampere-seconds value to maintain the original density?

A 5.6

B 6.5

C 9.7

D 13

he Correct Answer is: B
According to the density-maintenance formula, if the SID is changed to 48 in., 16.33 mAs is required to maintain the original radiographic density:

Thus, x = 16.33 mAs at 42 in. SID. Then, to compensate for changing from a 12:1 grid to a 5:1 grid, the milliampere-seconds value becomes 6.53 mAs:

Thus, x 6.53 mAs with 5:1 grid at 42 in. SID. Hence, 6.53 mAs is required to produce an image density similar to that of the original radiograph. The following are the factors used for milliampere-seconds conversion from nongrid to grid:

In radiography of a large abdomen, which of the following is (are) effective way(s) to minimize the amount of scattered radiation reaching the image receptor (IR)?

1. Use of close collimation
2. Use of low mAs
3. Use of a low-ratio grid

A 1 only B 1 and 2 only C 1 and 3 only

he Correct Answer is: A
One way to minimize scattered radiation reaching the IR is to use optimal kilovoltage; excessive kilovoltage increases the production of scattered radiation. Close collimation is exceedingly important because the smaller the volume of irradiated material, the less scattered radiation will be produced. The mAs selection has no impact on scattered radiation production or cleanup. Low-ratio grids allow a greater percentage of scattered radiation to reach the IR. Use of a high-ratio grid will clean up a greater amount of scattered radiation before it reaches the IR. Use of a compression band, or the prone position, in a large abdomen has the effect of making the abdomen “thinner”; it will, therefore, generate less scattered radiation. (Shephard, p. 203)

In digital imaging, as the size of the image matrix increases,

1. FOV increases
2. pixel size decreases
3. spatial resolution increases

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: C
The FOV and matrix size are independent of one another; that is, either can be changed, and the other will remain unaffected. However, pixel size is affected by changes in either the FOV or matrix size. For example, if the matrix size is increased, pixel size decreases. If FOV increases, pixel size increases. Pixel size is inversely related to resolution. As pixel size decreases, resolution increases. (Fosbinder and Kelsey, p. 285).

The sensitometric curve may be used to

1. identify automatic processing problems
2. determine film sensitivity
3. illustrate screen speed

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
The sensitometric, or characteristic, curve is used to illustrate the relationship between the exposure, given the film and the resulting film density. It can be used to predict a particular film emulsion's response (i.e., speed and sensitivity) by determining how long it takes to record a particular density. The sensitometric curve is used in sensitometry to monitor automatic processing efficiency and consistency. A film is given a series of predetermined exposures and processed. The resulting densities are plotted, and the resulting curve is compared with a known correct curve. Any deviation between the two may indicate processing difficulties. The sensitometric curve illustrates the effects of exposure and processing on radiographic film emulsion; it is unrelated to film speed. (Shephard, pp. 104–108)

In amorphous selenium flat-panel detectors, the term amorphous refers to a

A crystalline material having typical crystalline structure.

B crystalline material lacking typical crystalline structure.

C toxic crystalline material.

D homogeneous crystalline material.

he Correct Answer is: B
Flat-panel detectors used in DR are often made of an amorphous selenium (a-Se)–coated thin-film transistor (TFT) array. They function to convert the x-ray energy (emerging from the radiographed part) into an electrical signal. The TFT capacitors send the electrical signal to the analog-to-digital converter (ADC) to be changed to a digital signal. Amorphous selenium refers to a crystalline material (selenium) that lacks its crystalline structure. Amorphous selenium or silicon is used to produce the direct-conversion flat-panel detectors used in DR. (Bushong, 8th ed., p. 404)

A lateral radiograph of the lumbar spine was made using 200 mA, 1-second exposure, and 90 kV. If the exposure factors were changed to 200 mA, 0.5 second, and 104 kV, there would be an obvious change in which of the following?

1. Radiographic density
2. Scale of radiographic contrast
3. Distortion

A 1 only B 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
The original milliampere-seconds value (regulating radiographic density) was 200. The original kilovoltage (regulating radiographic contrast) was 90. The milliampere-seconds value was cut in half, to 100, causing a decrease in density. The kilovoltage was increased (by 15%) to compensate for the density loss and thereby increase the scale of contrast. (Shephard, p. 203)

Compared with slow-speed screens, high-speed screens are often used to

1. minimize patient dose
2. keep exposure time to a minimum
3. image fine anatomic details

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
The faster the screen speed, the greater is the quantity of fluorescent light emitted during x-ray exposure. Therefore, high-speed screens require less x-ray exposure to provide adequate radiographic density and can be used when exposure reduction or fast exposure time is essential. However, because they are associated with more diffusion of fluorescent light, they produce less recorded detail and are not used to image structures requiring excellent recorded detail. (Shephard, p. 67)

Which of the following combinations is most likely to be associated with quantum mottle?

A Decreased milliampere-seconds, decreased SID, fast-speed screens

B Increased milliampere-seconds, decreased kilovoltage, slow-speed screens

C Decreased milliampere-seconds, increased kilovoltage, fast-speed screens

D Increased milliampere-seconds, increased SID, fast-speed screens

The Correct Answer is: C
Quantum mottle is a grainy appearance on a finished image that is seen especially in fast-imaging systems. It is very similar in appearance to an enlarged photograph taken with fast film; it has a spotted or freckled appearance. Fast imaging systems (fast film and fast screens, as well as CR/DR systems) with low-milliampere-seconds and high-kilovoltage factors are most likely to be the cause of quantum mottle. (Bushong, 8th ed., p. 273)

Exposure rate will decrease with an increase in

1. SID
2. kilovoltage
3. focal-spot size

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: A
Exposure rate decreases with an increase in SID according to the inverse-square law of radiation. The quantity of x-ray photons produced at the focal spot is the function of milliampere-seconds. The quality (i.e., wavelength, penetration, and energy) of x-ray photons produced at the target is the function of kilovoltage. The kilovoltage also has an effect on exposure rate because an increase in kilovoltage will increase the number of high-energy x-ray photons produced at the anode. (Selman, 9th ed., p. 117)

The microswitch for controlling the amount of replenishment used in an automatic processor is located at the

A receiving bin

B crossover roller

C entrance roller

D replenishment pump

he Correct Answer is: C
The wider dimension of the x-ray film usually is placed on the feed tray so that the film is fed into the processor in that direction. The entrance roller is the first roller of the transport system, located at the end of the feed tray; this is where the microswitch that determines the amount of replenishment is located. The length of the film (the shorter dimension) activates the microswitch, and replenisher is added according to the length of the film; a 10 × 12 in. film will receive less replenisher than will a 14 × 17 in. film. Crossover rollers are located between the different tanks. The receiving bin is where the films exit the processor. The replenishment pump is activated by the microswitch. (Bushong, 8th ed., pp. 212–213)

If a radiograph exhibits insufficient density, this might be attributed to

1. inadequate kVp. 2. inadequate SID. 3. grid cutoff.

A 1 only

B 1 and 2 only

C 1 and 3 only

D 1, 2, and 3

The Correct Answer is: C
As kVp is reduced, the number of high-energy photons produced at the target is reduced; therefore, a decrease in radiographic density occurs. If a grid has been used improperly (off-centered or out of focal range), the lead strips will absorb excessive amounts of primary radiation, resulting in grid cutoff and loss of radiographic density. If the SID is inadequate (too short), an increase in radiographic density will occur. (Selman, pp 214, 240–242)

High-kilovoltage exposure factors are usually required for radiographic examinations using

1. water-soluble, iodinated media. 2. a negative contrast agent. 3. barium sulfate.

A 1 only

B 2 only

C 3 only

D 1 and 3 only

The Correct Answer is: C
Positive-contrast medium is radiopaque; negative-contrast material is radioparent. Barium sulfate (radiopaque, positive-contrast material) is most frequently used for examinations of the intestinal tract, and high-kVp exposure factors are used to penetrate (to see through and behind) the barium. Water-based iodinated contrast media (Conray, Amipaque) are also positive-contrast agents. However, the K-edge binding energy of iodine prohibits the use of much greater than 70 kVp with these materials. Higher kVp values will obviate the effect of the contrast agent. Air is an example of a negative-contrast agent, and high-kVp factors are clearly not indicated. (Shephard, pp 200–201)

Which of the following are methods of limiting the production of scattered radiation?

1. Using moderate ratio grids
2. Using the prone position for abdominal examinations
3. Restricting the field size to the smallest practical size

A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: C
If a fairly large patient is turned prone, the abdominal measurement will be significantly different from the AP measurement as a result of the effect of compression. Thus, the part is essentially “thinner,” and less scattered radiation will be produced. If the patient remains supine and a compression band is applied, a similar effect will be produced. Beam restriction is probably the single most effective means of reducing the production of scattered radiation. Grid ratio affects the cleanup of scattered radiation; it has no effect on the production of scattered radiation. (Shephard, p. 203)

Which of the following has the greatest effect on radiographic density?

A Aluminum filtration

B Kilovoltage

C SID

D Scattered radiation

The Correct Answer is: C
Radiographic density is greatly affected by changes in the SID, as expressed by the inverse-square law of radiation. As distance from the radiation source increases, exposure rate decreases, and radiographic density decreases. Exposure rate is inversely proportional to the square of the SID. Aluminum filtration, kilovoltage, and scattered radiation all have a significant effect on density, but they are not the primary controlling factors. (Selman, 9th ed., p. 214)

How would the introduction of a 6-in. OID affect image contrast?

A Contrast would be increased.

B Contrast would be decreased.

C Contrast would not change.

D The scale of contrast would not change.

The Correct Answer is: A
OID can affect contrast when it is used as an air gap. If a 6-in. air gap (OID) is introduced between the part and IR, much of the scattered radiation emitted from the body will not reach the IR, as shown in Figure 7–20. The OID thus is acting as a low-ratio grid and increasing image contrast. (Shephard, p. 205)

The exposure factors of 300 mA, 0.07 second, and 95 kVp were used to produce a particular radiographic density and contrast. A similar radiograph can be produced using 500 mA, 80 kVp, and

A 0.01 second.

B 0.04 second.

C 0.08 second.

D 0.16 second.

he Correct Answer is: C
First, evaluate the change(s): The kVp was decreased by about 15% [95–15% = 80.7]. A 15% decrease in kVp will cut the radiographic density in half; therefore, it is necessary to use twice the original mAs to maintain the original density. The original mAs was 21, and so we now need 42 mAs, using the 500-mA station. Because mA × s = mAs,

500x = 42

x = 0.084 second

(Fauber, pp 55, 59–60)

If a 4-inch collimated field is changed to a 14-inch collimated field, with no other changes, the radiographic image will possess

A more density/IR exposure.

B less density/IR exposure.

C more detail.

D less detail.

The Correct Answer is: A
More scattered radiation is generated within a part as the kilovoltage is increased, the size of the field is increased, and the thickness and density of tissue increases. As the quantity of scattered radiation increases from any of these sources, more density is added to the radiographic image. (Carlton & Adler, 5th ed p 396)

An exposure was made at a 36-in. SID using 300 mA, a 30-ms exposure, and 80 kV with a 400-speed film–screen combination and an 8:1 grid. It is desired to repeat the radiograph and, in order to improve recorded detail, to use a 40-in. SID and a 200-speed film–screen combination. With all other factors remaining constant, what exposure time will be required to maintain the original radiographic density?

A 0.03 second

B 0.07 second

C 0.14 second

D 0.36 second

he Correct Answer is: B
A review of the problem reveals that three changes are being made: an increase in SID, a change from a 400-speed system to a 200-speed system, and a change in exposure time (to be considered last). Because the original milliampere-seconds value was 9, cutting the speed of the system in half (from 400–200) will require a doubling of the milliampere-seconds value to 18 mAs in order to maintain density. Now we must deal with the distance change. Using the density-maintenance formula (and remembering that 18 is now the old milliampere-seconds value), we find that the required new milliampere-seconds value at 42 in. is 22.

Thus, x = 22.22 mAs at 40-in. SID. Because milliamperage is unchanged, we must determine the exposure time that, when used with 300 mA, will yield 22 mAs.

During film–screen imaging, the manifest image is formed

A on exposure of the film emulsion

B in the developer solution

C in the first half of the fixer process

D in the second half of the fixer process

The Correct Answer is: B
A latent (invisible) image is produced when the film emulsion is exposed by light or x-rays. The invisible silver halide image is composed of exposed silver grains. These are “reduced” to a manifest (visible) black metallic silver image in the developer solution. The fixer solution functions to remove unexposed silver halide crystals from the film. (Shephard, pp. 97–98)

A part whose width is 6 inches will be imaged at 44 inches SID. The part to be imaged lies 9 inches from the IR. What will be the magnification factor?

A 1.25

B 1.86

C 4.9

D 7.3

The Correct Answer is: A
As the object-to-image receptor distance (OID) increases, magnification of that object increases. Depending upon the information provided, we can determine the magnification factor, the percentage magnification, and image width. In the stated scenario, we are looking for image width. The formula used to determine magnification factor is:

MF = SID/SOD

Substituting known factors the equation becomes:

MF = 44/35

MF = 1.257

The "1" in the answer represents the actual object, while the ".257" represents the degree of magnification. The percent magnification can be determined by moving the decimal two places to the right. Thus, the percent magnification is 25.7%. (Shephard, p 230)

The steeper the straight-line portion of a characteristic curve for a particular film, the

1. slower is the film speed
2. higher is the film contrast
3. greater is the exposure latitude

A 1 only B 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: B
The steepness of the characteristic curve is representative of image contrast. The steeper the curve, the greater is the density difference and the higher is the contrast. The speed of the film is determined by the curve's position on the log-relative scale: When comparing two or more characteristic curves, the faster film lies farthest to the left. The faster the film speed, the less is the exposure latitude. (Shephard, p. 105)

Which of the following could be used to improve recorded detail?

1. Slower screen/film combination

2. Smaller focal-spot size

3. Correct photocell selection

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
The factors that affect the recorded detail of traditional screen/film imaging are focal spot size, source-to-image distance (SID), object-to-image distance (OID), film/screen speed, and motion. Recorded detail is improved using small focal-spot size, largest practical SID, shortest possible OID, and slowest practical screen/film combination and avoiding motion of the part being imaged. Other imaging factors such as milliampere-seconds and kilovoltage selection and correct photocell selection influence the visibility of recorded detail by affecting density and contrast. (Shephard, p. 215)

An increase in the kilovoltage applied to the x-ray tube increases the

1. x-ray wavelength
2. exposure rate
3. patient absorption

A 1 only B 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
As the kilovoltage is increased, a greater number of electrons are driven across to the anode with greater force. Therefore, as energy conversion takes place at the anode, more high-energy (short-wavelength) photons are produced. However, because they are higher-energy photons, there will be less patient absorption. (Selman, 9th ed., pp. 117–118)

The term latitude describes

1. an emulsion's ability to record a range of densities
2. the degree of error tolerated with given exposure factors
3. the conversion efficiency of a given intensifying screen

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
The term latitude may refer to either film emulsion latitude or exposure latitude. Exposure latitude refers to the margin of error inherent in a particular group of exposure factors. Selection of high-kilovoltage and low-milliampere-seconds factors will allow greater exposure latitude than low-kilovoltage and high-milliampere-seconds factors. Film emulsion latitude is chemically built into the film emulsion and refers to the emulsion's ability to record a long range of densities from black to white (long-scale contrast). (Bushong, 8th ed., p. 185)

An exposure was made at a 36-in. SID using 12 mAs and 75 kVp with a 400-speed imaging system and an 8:1 grid. A second radiograph is requested with improved recorded detail. Which of the following groups of technical factors will best accomplish this task?

A 15 mAs, 12:1 grid, 75 kVp, 400-speed system, 36-in. SID

B 15 mAs, 12:1 grid, 75 kVp, 400-speed system, 40-in. SID

C 30 mAs, 12:1 grid, 75 kVp, 200-speed system, 40-in. SID

D 12 mAs, 8:1 grid, 86 kVp, 200-speed system, 36-in. SID

The Correct Answer is: C
Look over the choices again, keeping in mind the factors that affect recorded detail. Looking first at SID, the options may be reduced to (B) and (C) because the increase to a 40-in. SID certainly will improve recorded detail. There is one other factor that will affect detail—the speed of the system (intensifying screens). Because a slower system will render better recorded detail, the best answer is (C). The technical factors such as milliampere-seconds, kilovoltage, and grid ratio have no effect on recorded detail. (Shephard, pp. 247, 310)

Recorded detail is directly related to

1. source-image distance (SID). 2. tube current. 3. focal spot size.

A 1 only

B 1 and 2 only

C 2 and 3 only

D 1, 2, and 3

he Correct Answer is: A
As SID increases, so does recorded detail, because magnification is decreased. Therefore, SID is directly related to recorded detail. As focal spot size increases, recorded detail decreases because more penumbra is produced. Focal spot size is thus inversely related to radiographic sharpness or recorded detail. Tube current affects radiographic density and is unrelated to recorded detail. (Fauber, 2nd ed., pp. 79, 81)

A part whose width is 6 inches will be imaged at 44 inches SID. The part to be imaged lies 9 inches from the IR. What will be the projected image width of the part?

A 8 inches

B 10 inches

C 12 inches

D 20 inches

The Correct Answer is: A
As the object-to-image receptor distance (OID) increases, magnification of that object increases. Depending upon the information provided, we can determine the magnification factor, the percentage magnification, and image width. In the stated scenario, we are looking for image width. The formula used to determine image width is:

Substituting known factors the equation becomes:

35x = 254

x = 7.5 inches projected image width

(Shephard, p 230)

A positive contrast agent

1. absorbs x-ray photons
2. results in a dark area on the radiograph
3. is composed of elements having low atomic number

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: A
Radiopaque contrast agents appear white on the finished image because many x-ray photons are absorbed. These are referred to positive contrast agents—composed of dense (i.e., high atomic number) material through which x-rays will not pass easily. Radiolucent contrast agents appear black on the finished image because x-ray photons pass through easily. An example of a radiolucent contrast agent is air. (Shephard, pp. 200–202)

Recorded detail depends on all the following except

1. quantity of filtration.

2. anode angle.

3. intensification factor of screens.

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: A
The factors that affect the recorded detail of traditional screen/film imaging are focal-spot size, source-to-image receptor distance (SID), object-to-image distance (OID), film/screen speed, and motion. Recorded detail is improved using a small focal-spot size, largest practical SID, shortest possible OID, and slowest practical screen film combination and avoiding motion of the part being imaged. The smaller the anode angle, the smaller is the effective focal spot. The greater the intensification factor, the greater is the screen speed. Filtration affects the overall average photon energy, that is, beam quality. Filtration is unrelated to recorded detail. (Shephard, p. 215

The absorption of useful radiation by a grid is called

A grid selectivity.

B grid cleanup.

C grid cutoff.

D latitude.

The Correct Answer is: C
Grids are used in radiography to absorb scattered radiation before it reaches the IR (grid “cleanup”), thus improving radiographic contrast. Contrast obtained with a grid compared with contrast without a grid is termed contrast-improvement factor. The greater the percentage of scattered radiation absorbed compared with absorbed primary radiation, the greater is the “selectivity” of the grid. If a grid absorbs an abnormally large amount of useful radiation as a result of improper centering, tube angle, or tube distance, grid cutoff occurs. (Selman, 9th ed., p. 370)

An exposure was made at 40-in. SID using 5 mAs and 105 kVp with an 8:1 grid. In an effort to improve radiographic contrast, the image is repeated using a 12:1 grid and 90 kVp. Which of the following exposure times will be most appropriate, using 400 mA, to maintain the original density?

A 0.01 s

B 0.03 s

C 0.1 s

D 0.3 s

The Correct Answer is: B
The use of high kilovoltage with a fairly low-ratio grid will be ineffective in ridding the remnant beam of scattered radiation. To improve contrast in this example, it has been decided to decrease the kilovoltage by 15%, thus making it necessary to increase the milliampere-seconds from 5 mAs to 10 mAs. Because an increase in the grid ratio to 12:1 is also desired, another change in milliampere-seconds will be required (remember, 10 mAs is now the old mAs):

Thus, x = 12.5 mAs at 90 kVp. Now determine the exposure time required with 400 mA to produce 12.5 mAs:

An increase in kilovoltage will have which of the following effects?

1. More scattered radiation will be produced.
2. The exposure rate will increase.
3. Radiographic contrast will increase.

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
An increase in kilovoltage (photon energy) will result in a greater number (i.e., exposure rate) of scattered photons (Compton interaction). These scattered photons carry no useful information and contribute to radiation fog, thus decreasing radiographic contrast. (Selman, 9th ed., p. 117)

Greater latitude is available to the radiographer when using

1. high-kilovoltage factors.
2. a slow film–screen combination.
3. a high-ratio grid.

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: B
In the low-kilovoltage ranges, a difference of just a few kilovolts makes a very noticeable radiographic difference; that is, there is little latitude. High-kilovoltage techniques offer a much greater margin for error, as do slower film–screen combinations. Grid ratio is unrelated to exposure latitude, but higher-ratio grids offer less tube-centering latitude (i.e., leeway, margin for error) than low-ratio grids. (Carlton and Adler, 4th ed., p. 185)

Which of the following terms refers to light being reflected from one intensifying screen, through the film, to the opposite emulsion and intensifying screen?

A Reflectance

B Crossover

C Scatter

D Filtration

The Correct Answer is: B
If fluorescent light from one intensifying screen passes through the film to the opposite emulsion and intensifying screen, the associated diffusion creates a type of distortion called crossover. Intensifying screens do need a degree of reflectance to enhance their speed. Scatter and filtration are unrelated to intensifying screens. (Selman, 9th ed., p. 185)

What pixel size has a 2,048 × 2,048 matrix with a 60-cm FOV?

A 0.3 mm

B 0.5 mm

C 0.15 mm

D 0.03 mm

he Correct Answer is: A
In digital imaging, pixel size is determined by dividing the FOV by the matrix. In this case, the FOV is 60 cm; since the answer is expressed in millimeters, first change 60 cm to 600 mm. Then 600 divided by 2,048 equals 0.35 mm:

The FOV and matrix size are independent of one another; that is, either can be changed, and the other will remain unaffected. However, pixel size is affected by changes in either the FOV or matrix size. For example, if the matrix size is increased, pixel size decreases. If FOV increases, pixel size increases. Pixel size is inversely related to resolution. As pixel size increases, resolution decreases. (Fosbinder and Kelsey, p. 285)

Geometric unsharpness is directly influenced by

1. OID
2. SOD
3. SID

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

he Correct Answer is: A
Geometric unsharpness is affected by all three factors listed. As OID increases, so does magnification—therefore, OID is directly related to magnification. As SOD and SID decrease, magnification increases—therefore, SOD and SID are inversely related to magnification. (Carlton and Adler, 4th ed., pp. 444–445)

Methods that help to reduce the production of scattered radiation include using

1. compression
2. beam restriction
3. a grid

A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: A
Limiting the size of the irradiated field is a most effective method of decreasing the production of scattered radiation. The smaller the volume of tissue irradiated, the smaller is the amount of scattered radiation generated; this can be accomplished using compression (prone position instead of supine or a compression band). Use of a grid does not affect the production of scattered radiation but rather removes it once it has been produced. (Carlton and Adler, 4th ed., p. 228)

Using a 48-in. SID, how much OID must be introduced to magnify an object two times?

A 8-in. OID

B 12-in. OID

C 16-in. OID

D 24-in. OID

The Correct Answer is: D
Magnification radiography may be used to delineate a suspected hairline fracture or to enlarge tiny, contrast-filled blood vessels. It also has application in mammography. To magnify an object to twice its actual size, the part must be placed midway between the focal spot and the IR. (Selman, pp. 223–225; Shephard, pp. 229–231)

Which of the following units is (are) used to express resolution?

1. Line-spread function
2. Line pairs per millimeter
3. Line-focus principle

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
Resolution describes how closely fine details may be associated and still be recognized as separate details before seeming to blend into each other and appear “as one.” The degree of resolution transferred to the IR is a function of the resolving power of each of the system components and can be expressed in line pairs per millimeter (lp/mm), line-spread function (LSP), or modulation transfer function (MTF). Lp/mm can be measured using a resolution test pattern; a number of resolution test tools are available. LSP is measured using a 10-μm x-ray beam; MTF measures the amount of information lost between the object and the IR. The effective focal spot is the foreshortened size of the actual focal spot as it is projected down toward the IR, that is, as it would be seen looking up into the emerging x-ray beam. This is called the line-focus principle and is not a unit used to express resolution. (Carlton and Adler, 4th ed., p. 334)

A satisfactory radiograph was made without a grid, using a 72-inch SID and 8 mAs. If the distance is changed to 40 inches and an 8:1 ratio grid is added, what should be the new mAs?

A 10 mAs

B 18 mAs

C 20 mAs

D 32 mAs

The Correct Answer is: A
According to the inverse square law of radiation, as the distance between the radiation source and the IR decreases, the exposure rate increases. Therefore, a decrease in technical factors is first indicated to compensate for the distance change. The following formula (density maintenance formula) is used to determine new mAs values, when changing distance:

Substituting known values,

To then compensate for adding an 8:1 grid, you must multiply the 2.4 mAs by a factor of 4. Thus, 9.6 mAs is required to produce a image density similar to the original radiograph. The following are the factors used for mAs conversion from nongrid to grid:

(Bushong, 8th ed., pp. 69, 252)

No grid= 1 × original mAs 5:1 grid = 2 × original mAs 6:1 grid = 3 × original mAs 8:1 grid = 4 × original mAs 12:1 grid = 5 × original mAs 16:1 grid = 6 × original mAs

Which of the following groups of exposure factors will produce the shortest scale of contrast?

A 200 mA, 0.25 s, 70 kVp, 12:1 grid

B 500 mA, 0.10 s, 90 kVp, 8:1 grid

C 400 mA, 0.125 s, 80 kVp, 12:1 grid

D 300 mA, 0.16 s, 70 kVp, 8:1 grid

he Correct Answer is: A
Of the given factors, kilovoltage and grid ratio will have a significant effect on the scale of radiographic contrast. The milliampere-seconds values are almost identical. Because decreased kilovoltage and high-ratio grid combination would allow the least amount of scattered radiation to reach the IR, thereby producing fewer gray tones, (A) is the best answer. Group (D) also uses low kilovoltage, but the grid ratio is lower, thereby allowing more scatter to reach the IR and producing more gray tones. (Shephard, p. 308)

Which combination of exposure factors most likely will contribute to producing the shortest-scale contrast?

A mAs: 10; kV: 70; Film–screen system: 400; Grid ratio: 5:1; Field size: 14 × 17 in.

B mAs: 12; kV: 90; Film–screen system: 200; Grid ratio: 8:1; Field size: 14 × 17 in.

C mAs: 15; kV: 90; Film–screen system: 200; Grid ratio: 12:1; Field size: 11 × 14 in.

D mAs: 20; kV: 80; Film–screen system: 400; Grid ratio: 10:1; Field size: 8 × 10 i

he Correct Answer is: D
Review the groups of factors. First, because the milliampere-seconds value has no effect on the scale of contrast produced, eliminate milliampere-seconds from consideration by drawing a line through the column. Then check the two entries in each column that are likely to produce shorter-scale contrast. For example, in the kilovoltage column, because lower kilovoltage will produce shorter-scale contrast, place checkmarks next to the 70 and 80 kV. In the film–screen column, the faster screens (400) will produce higher (shorter-scale) contrast than the slower screens; place a checkmark next to each. Because higher-ratio grids permit less scattered radiation to reach the IR, the 10:1 and 12:1 grids will produce a shorter scale of contrast than the lower-ratio grids; check them. As the volume of irradiated tissue decreases, so does the amount of scattered radiation produced, and consequently, the shorter is the scale of radiographic contrast; therefore, check the 11 × 14 and 8 × 10 in. field sizes. An overview shows that the factors in groups (A) and (C) have two checkmarks, whereas the factors in group (D) have four checkmarks, indicating that group (D) will produce the shortest-scale contrast. (Shephard, pp. 306–308)

If a 6-in. OID is introduced during a particular radiographic examination, what change in SID will be necessary to overcome objectionable magnification?

A The SID must be increased by 6 in..

B The SID must be increased by 18 in..

C The SID must be decreased by 6 in..

D The SID must be increased by 42 in.

The Correct Answer is: D
As OID is increased, recorded detail is diminished as a result of magnification distortion. If the OID cannot be minimized, an increase in SID is required to reduce the effect of magnification distortion. However, the relationship between OID and SID is not an equal relationship. In fact, to compensate for every 1 in. of OID, an increase of 7 in. of SID is required. Therefore, an OID of 6 in. requires an SID increase of 42 in.. This is why a chest radiograph with a 6-in. air gap usually is performed at a 10-ft SID. (Saia, 4th ed., p. 290)

A 5-in. object to be radiographed at a 44-in. SID lies 6 in. from the IR. What will be the image width?

A 5.1 in. B 5.7 in. C 6.1 in. D 6.7 in.

he Correct Answer is: B
Magnification is part of every radiographic image. Anatomic parts within the body are at various distances from the IR and, therefore, have various degrees of magnification. The formula used to determine the amount of image magnification is

Substituting known values:

Thus, x = 5.78-in. image width. (Bushong, 8th ed., p. 284)

The term differential absorption is related to

1. beam intensity
2. subject contrast
3. pathology

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: C
The radiographic subject, the patient, is composed of many different tissue types of varying densities (i.e., subject contrast), resulting in varying degrees of photon attenuation and absorption. This differential absorption contributes to the various shades of gray (i.e., scale of radiographic contrast) on the finished image. Normal tissue density may be significantly altered in the presence of pathology. For example, destructive bone disease can cause a dramatic decrease in tissue density. Abnormal accumulation of fluid (as in ascites) will cause a significant increase in tissue density. Muscle atrophy or highly developed muscles similarly will decrease or increase tissue density. (Carlton and Adler, 4th ed., p.

Which of the following is (are) directly related to photon energy?

1. Kilovoltage
2. Milliamperes
3. Wavelength

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

he Correct Answer is: A
Kilovoltage is the qualitative regulating factor; it has a direct effect on photon energy. That is, as kilovoltage is increased, photon energy increases. Photon energy is inversely related to wavelength. That is, as photon energy increases, wavelength decreases. Photon energy is unrelated to milliamperage. (Shephard, pp. 173, 178)

Which of the following methods can be used effectively to decrease differential absorption, providing a longer scale of contrast in the diagnostic range?

1. Using high peak kilovoltage and low milliampere-seconds factors
2. Using compensating filtration
3. Using factors that increase the photoelectric effect

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: B
When differences in absorption characteristics are decreased, body tissues absorb radiation more uniformly, and as a result, more grays are seen on the radiographic image. A longer scale of contrast is produced. High-kilovoltage and low-milliamperage factors achieve this. Compensating filtration is also used to “even out” densities in uneven anatomic parts, such as the thoracic spine. The photoelectric effect is the interaction between x-ray photons and matter that occurs at low-peak kilovoltage levels—levels that tend to produce short-scale contrast. (Shephard, pp. 193, 197, 199)

According to the line-focus principle, an anode with a small angle provides

1. improved recorded detail.
2. improved heat capacity.
3. less heel effect.

A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: A
The line-focus principle illustrates that as the target angle decreases, the effective focal spot decreases (providing improved recorded detail), but the actual area of electron interaction remains much larger (allowing for greater heat capacity). It must be remembered, however, that a steep (small) target angle increases the heel effect, and part coverage may be compromised. (Shephard, p. 21

An anteroposterior (AP) radiograph of the femur was made using 300 mA, 0.03 second, 76 kV, 40-in. SID, 1.2-mm focal spot, and a 400-speed film–screen system. With all other factors remaining constant, which of the following exposure times would be required to maintain radiographic density at a 44-in. SID using 500 mA?

A 12 ms B 22 ms C 30 ms D 36 ms

The Correct Answer is: B
The original milliampere-seconds value was 9 (300 mA × 0.03 second). Using the density-maintenance formula, the new milliampere-seconds value must be determined for the distance change from 40 to 44 in. of the SID:

Thus, x = 10.89 (11) mAs at 44-in. SID. Then, if 500 is the new milliamperage, we must determine what exposure time is required to achieve 8.1 mAs:

Thus, x = 0.022 second (22 ms) at 500 mA and 44-in. SID. (Selman, 9th ed., pp. 214–215

What pixel size has a 2048 × 2048 matrix with an 80-cm FOV?

A 0.04 mm

B 0.08 mm

C 0.2 mm

D 0.4 mm

he Correct Answer is: D
In digital imaging, pixel size is determined by dividing the field of view (FOV) by the matrix. In this case the FOV is 80 cm; since the answer is expressed in mm, first change 80 cm to 800 mm. Then 800 divided by 2048 equals 0.4 mm.

80 cm = 800 mm

The FOV and matrix size are independent of one another, that is, either can be changed and the other will remain unaffected. However, pixel size is affected by changes in either the FOV or matrix size. For example, if the matrix size is increased, pixel size decreases. If FOV increases, pixel size increases. Pixel size is inversely related to resolution. As pixel size increases, resolution decreases. (Fosbinder & Kelsey, p 285)

Terms that refer to size distortion include

1. magnification
2. attenuation
3. elongation

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

he Correct Answer is: A
Distortion is misrepresentation of the actual size or shape of the object being imaged. Size distortion is magnification. Shape distortion is a result of improper alignment of the x-ray tube, the part being radiographed, and the IR; the two types of shape distortion are foreshortening and elongation. The shapes of various structures can be misrepresented radiographically as a result of their position in the body, when the part is out of the central axis of the x-ray beam, or when the CR is angled (Figure 7–19). Parts sometimes are elongated intentionally for better visualization (e.g., sigmoid colon). Some body parts, because of their position in the body, are foreshortened, such as the carpal scaphoid. Attenuation refers to decreasing beam intensity and is unrelated to distortion. (Shephard, pp. 228–231)

An exposure was made using 600 mA, 0.04-s exposure, and 85 kVp. Each of the following changes will serve to decrease the radiographic density by one-half except change to

A 1/50-s exposure

B 72 kVp

C 18 mAs

D 300 mA

he Correct Answer is: C
Radiographic density is directly proportional to milliampere-seconds. If exposure time is halved from 0.04 s to 0.02 s, radiographic density will be cut in half. Changing to 300 mA also will halve the milliampere-seconds, effectively halving the radiographic density. If the kilovoltage is decreased by 15%, from 85 to 72 kVp, radiographic density will be halved according to the 15% rule. To cut the density in half, the milliampere-seconds must be reduced to 12 mAs (rather than to 18 mAs). (Selman, 9th ed., p. 214)

Which one of the following is (are) used to control the production of scattered radiation?

1. Collimators
2. Optimal kV
3. Use of grids

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
As kilovoltage is increased, x-ray photons begin to interact with atoms of tissue via the Compton scattered interaction. Scattered x-ray photons result, which serve only to add unwanted, undiagnostic densities (scattered radiation fog) to the radiologic image. (While Compton scatter reduces patient dose compared with photoelectric interactions, it can pose a significant radiation hazard to personnel during fluoroscopic procedures.) Therefore, the use of optimal kilovoltage is recommended to reduce the production of scattered radiation. Scattered radiation is also a function of the size and content of the irradiated field. The greater the volume and atomic number of the tissue, the greater is the production of scattered radiation. Although there is little that can be done about the atomic number of the structure to be radiographed, every effort can be made to keep the field size restricted to the essential area of interest in an effort to decrease production of scattered radiation. Grids have no effect on the production of scattered radiation, but they are very effective in removing scattered radiation from the beam before it strikes the IR. (Fauber, 2nd ed., pp. 71, 103)

Which of the following is (are) characteristic(s) of a 16:1 grid?

1. It absorbs more useful radiation than an 8:1 grid.
2. It has more centering latitude than an 8:1 grid.
3. It is used with higher-kilovoltage exposures than an 8:1 grid.

A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B
High-kilovoltage exposures produce large amounts of scattered radiation, and high-ratio grids are used often with high-kilovoltage techniques in an effort to absorb more of this scattered radiation. However, as more scattered radiation is absorbed, more primary radiation is absorbed as well. This accounts for the increase in milliampere-seconds required when changing from an 8:1 to a 16:1 grid. In addition, precise centering and positioning become more critical; a small degree of inaccuracy is more likely to cause grid cutoff in a high-ratio grid. (Bushong, 8th ed., pp. 252–255)

A shoulder was imaged using 300 mA, 7 ms, 70 kVp, 40-inch SID, 1.2-mm focal spot, and 100 speed screen/film combination. Which of the following changes, made to compensate for changes in optical density, would result in decreased production/visualization of blur?

1. Use of a 0.6-mm focal spot

2. Use of a 50-inch SID

3. Screens with higher intensification factor

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: B
Blur impairs recorded detail. The factors that affect the recorded detail of traditional screen/film imaging are focal-spot size, source-to-image distance (SID), object-to-image distance (OID), film/screen speed, and motion. Recorded detail is improved using a small focal-spot size, largest practical SID, shortest possible OID, and slowest practical screen/film combination and avoiding motion of the part being imaged. Therefore, reducing the focal spot to 0.6 mm and increasing the SID to 50 inches will result in decreased production/visualization of blur and improved recorded detail. ((Bushong, 10th ed, pp. 139, 177))

Radiographic contrast is a result of

1. differential tissue absorption
2. emulsion characteristics
3. proper regulation of milliampere-seconds

A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

he Correct Answer is: B
Radiographic contrast is defined as the degree of difference between adjacent densities. These density differences represent sometimes very subtle differences in the absorbing properties of adjacent body tissues. The type of film emulsion used also brings with it its own contrast characteristics. Different types of film emulsions have different degrees of contrast “built into” them chemically. The technical factor used to regulate contrast is kilovoltage. Radiographic contrast is unrelated to milliampere-seconds. (Selman, 9th ed., pp. 218–220)

the CR should be directed to the center of the part of greatest interest to avoid

A rotation distortion B magnification C foreshortening D elongation

The Correct Answer is: A
Image details placed away from the path of the CR will be exposed by more divergent rays, resulting in rotation distortion. This is why the CR must be directed to the midpoint of the part of greatest interest. For example, if bilateral hands are requested, they should be examined individually; if imaged simultaneously, the CR will be directed to no anatomic part (between the two hands) and rotation distortion will occur.

Magnification occurs when an OID is introduced, or with a decrease in SID. Foreshortening and elongation are the two types of shape distortion—caused by nonalignment of the x-ray tube, part/subject, and IR.

Geometric unsharpness will be least obvious

1. at long SIDs.
2. with small focal spots.
3. at the anode end of the image.

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: D
The x-ray tube anode is designed according to the line-focus principle, that is, with the focal track beveled (Figure 6–24). This allows a larger actual focal spot to project a smaller effective focal spot, resulting in improved recorded detail with less blur. However, because of the target angle, penumbral blur varies along the longitudinal tube axis, being greater at the cathode end of the image and less at the anode end of the image. Therefore, better detail will be appreciated using small focal spots at the anode end of the x-ray beam and at longer SIDs. (Bushong, 8th ed., p. 287)

Decreasing field size from 14 × 17 in. to 8 × 10 in., with no other changes, wil

l A decrease radiographic density and decrease the amount of scattered radiation generated within the part

B decrease radiographic density and increase the amount of scattered radiation generated within the part

C increase radiographic density and increase the amount of scattered radiation generated within the part

D increase radiographic density and decrease the amount of scattered radiation generated within the part

he Correct Answer is: A
Limiting the size of the radiographic field (irradiated area) serves to limit the amount of scattered radiation produced within the anatomic part. As the amount of scattered radiation produced decreases, so does the resultant density within the radiographic image. Therefore, as field size decreases, scattered radiation production decreases, and overall density decreases. Limiting the size of the radiographic field is a very effective means of reducing the quantity of non–information-carrying scattered radiation (fog) produced, resulting in a shorter scale of contrast with fewer radiographic densities. Limiting the size of the radiographic field is also the most effective means of patient radiation protection. (Shephard, p. 203)

Using a short (25–30 in.) SID with a large (14 × 17 in.) IR is likely to

A increase the scale of contrast

B increase the anode heel effect

C cause malfunction of the AEC

D cause premature termination of the exposure

The Correct Answer is: B
Use of a short SID with a large-size IR (and also with anode angles of 10 degrees or less) causes the anode heel effect to be much more apparent. The x-ray beam needs to diverge more to cover a large-size IR, and it needs to diverge even more for coverage as the SID decreases. The x-ray beam has no problem diverging toward the cathode end of the beam, but as it tries to diverge toward the anode end of the beam, it is eventually stopped by the anode (x-ray photons are absorbed by the anode). This causes a decrease in beam intensity at the anode end of the beam and is characteristic of the anode heel effect. (Carlton and Adler, 4th ed., p. 40

When blue-emitting rare earth screens are matched properly with the correct film emulsion, what type of safelight should be used in the darkroom?

1. Red
2. GBX-2
3. Ultraviolet

A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

he Correct Answer is: B
Most radiographic film (i.e., green- and blue-sensitive) is orthochromatic, that is, sensitive to all colors but red. The GBX-2 is a red filter that is safe with orthochromatic film emulsion. Blue-sensitive film is also sensitive to violet and ultraviolet as well as blue. Green-sensitive film is also sensitive to blue, violet, and ultraviolet, as well as green. (Shephard, p. 90)

An x-ray exposure of a particular part is made and restricted to a 14 × 17 in. field size. The same exposure is repeated, but the x-ray beam is restricted to a 4 × 4 in. field. Compared with the first image, the second image will demonstrate

1. less density
2. more contrast
3. more density

A 1 only B 1 and 2 only C 3 only D 2 and 3 only

The Correct Answer is: B
Less scattered radiation is generated within a part as the kilovoltage is decreased, as the size of the field is decreased, and as the thickness and density of tissue decrease. As the quantity of scattered radiation decreases from any of these sources, the less is the total density of the resulting image. (Carlton and Adler, 4th ed., p. 256)

An increase in kilovoltage with appropriate compensation of milliampere-seconds will result in

1. increased exposure latitude.
2. higher contrast.
3. increased density.

A 1 only B 1 and 2 only C 2 and 3 only D 1 and 3 only

The Correct Answer is: A
As the kilovoltage is increased, more penetration will occur, and a greater range of densities (grays) will be apparent in the image. This is termed long scale or low contrast. In addition, as the kilovoltage and scale of grays increase, the exposure latitude increases; the margin for error in technical factors becomes greater. As the milliampere-seconds value is decreased to compensate for the increased kilovoltage, density should remain the same. (Shephard, p. 204)