Comprehensive Image Production and Evaluation: Selection of Technical Factors

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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 answer is C.
EXPLANATION: 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.

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 answer is B.
EXPLANATION: 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.

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

The answer is C.
EXPLANATION: 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.

The interaction between x-ray photons and matter illustrated in the figure below is most likely to occur
1. in structures having a high atomic number.
2. during radiographic examination of the abdomen.
3. using high kV and low mAs exposure factors.

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

The answer is C.
EXPLANATION: Diagnostic x-ray photons interact with tissue in a number of ways, but mostly they are involved in the production of Compton scatter or in the photoelectric effect. Compton scatter is pictured; it occurs when a relatively high-energy (kV) photon uses some of its energy to eject an outer-shell electron. In doing so, the photon is deviated in direction and becomes a scattered photon. Compton scatter causes objectionable scattered radiation fog in large structures such as the abdomen and poses a radiation hazard to personnel during procedures such as fluoroscopy. In the photoelectric effect, a relatively low-energy x-ray photon uses all its energy to eject an inner-shell electron, leaving a "hole" in the K shell. An L-shell electron then drops down to fill the K vacancy, and in so doing emits a characteristic ray whose energy is equal to the difference between the binding energies of the K and L shells. The photoelectric effect occurs with high-atomic-number absorbers such as bone and positive contrast media, and is responsible for the production of radiographic contrast. It is helpful for the production of the radiographic image, but it contributes to the dose received by the patient (because it involves complete absorption of the incident photon).

Screen-film imaging is one example of a (n)

A. analog system.
B. digital system.
C. electromagnetic system.
D. direct-action radiation system.

The answer is A.
EXPLANATION: Screen-film imaging consists of an exposure method of converting x-ray energy to light energy, then converting light energy to electrochemical energy in the development process. Processing changes the invisible electrochemical image to a visible/manifest radiographic image. This process ends with analog data. Digital imaging is an electronic imaging method that allows data capture and manipulation in an electron pattern. The resulting image can be turned into an analog image after going through several energy changes (electron to light to film or TV screen). The direct action of x-rays has very little influence on a radiographic image produced with intensifying screens (fluorescent light is responsible for the majority of film exposure).

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 answer is B.
EXPLANATION: 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.

A grid is usually 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 answer is C.
EXPLANATION: 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.

The continued emission of light by a phosphor after the activating source has ceased is termed

A. fluorescence.
B. phosphorescence.
C. image intensification.
D. quantum mottle.

The answer is B.
EXPLANATION: Fluorescence occurs when an intensifying screen absorbs x-ray photon energy, emits light, and then ceases to emit light as soon as the energizing source ceases. Phosphorescence occurs when an intensifying screen absorbs x-ray photon energy, emits light, and continues to emit light for a short time after the energizing source ceases. Quantum mottle is the freckle-like appearance on some radiographs made using a very fast imaging system. The brightness of a fluoroscopic image is amplified through image intensification.

Which of the following is (are) method(s) that would enable the radiographer to reduce the exposure time required for a particular radiograph?
1. Use higher mA.
2. Use higher kVp.
3. Use faster film-screen combination.

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

The answer is D.
EXPLANATION: If it is desired to reduce the exposure time for a particular radiograph, as it might be when radiographing those who are unable to cooperate fully, the milliamperage must be increased sufficiently to maintain the original mAs, and thus radiographic density. A higher kilovoltage could be useful because it would allow further reduction of the mAs (exposure time) according to the 15% rule. Use of a higher-speed film-screen combination also helps reduce mAs (exposure time) through more efficient conversion of photon energy to fluorescent light energy.

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.

The answer is A.
EXPLANATION: 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.

The absorption of useful radiation by a grid is called

A. grid selectivity.
B. contrast improvement factor.
C. grid cutoff.
D. latitude.

The answer is C.
EXPLANATION: Grids are used in radiography to absorb scattered radiation before it reaches the IR, thus improving radiographic contrast. Contrast obtained with a grid compared to contrast without a grid is termed contrast improvement factor. The greater the percentage of scattered radiation absorbed compared to absorbed primary radiation, the greater 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.

Which of the following factors influence(s) the production of scattered radiation?
1. Kilovoltage level
2. Tissue density
3. Size of field

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

The answer is D.
EXPLANATION: As photon energy (kVp) increases, so does the production of scattered radiation. The greater the density of the irradiated tissues, the greater the production of scattered radiation. As the size of the irradiated field increases, there is an increase in the volume of tissue irradiated, and the percentage of scatter again increases. Beam restriction is the single most important way to limit the amount of scattered radiation produced.

The use of which of the following is (are) essential in magnification radiography?
1. High-ratio grid
2. Fractional focal spot
3. Direct exposure film

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

The answer is B.
EXPLANATION: Magnification radiography is used to enlarge details to make them more perceptible. Hairline fractures, minute blood vessels, and microcalcifications 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. A direct-exposure technique would not be likely to be used because of the excessive exposure required.

Which of the following can impact the visibility of the anode heel effect?
1. SID
2. Image recorder size
3. Screen speed

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

The answer is B.
EXPLANATION: 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 considerably toward the cathode end of the x-ray tube but are absorbed by the heel of the anode at the opposite end of the tube. This results in a greater number of x-ray photons distributed toward the cathode end, which is known as the anode heel effect. The effect of this restricting heel is most pronounced when the x-ray photons are required to diverge more, as would be the case with short SID, large-size IRs, and steeper (smaller) target angles.

Greater latitude is available to the radiographer in which of the following circumstances?
1. Using high-kVp technical factors
2. Using a slow film-screen combination
3. Using a low-ratio grid

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

The answer is D.
EXPLANATION: In the low kilovoltage ranges, a difference of just a few kVp makes a very noticeable radiographic difference. High-kVp technical factors offer much greater margin for error, as do slow film-screen combinations. Lower-ratio grids offer more tube-centering latitude than high-ratio grids.

In radiography of a large abdomen, which of the following is (are) effective way(s) to minimize the amount of scattered radiation reaching the IR?
1. Use of close collimation
2. Use of compression devices
3. Use of a low-ratio grid

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

The answer is B.
EXPLANATION: One way to minimize scattered radiation reaching the IR is to use optimal kilovoltage; excessive kVp increases the production of scattered radiation. Close collimation is also important because the smaller the volume of irradiated material, the less scattered radiation is produced. Using compression bands or the prone position in a large abdomen has the effect of making the abdomen "thinner"; it will therefore generate less scattered radiation. 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.

Radiographic contrast is a result of
1. differential tissue absorption.
2. emulsion characteristics.
3. proper regulation of mAs.

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

The answer is B.
EXPLANATION: 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 mAs.

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

A. mAs by 30%.
B. mAs by 15%.
C. kVp by 15%.
D. kVp by 30%

The answer is A.
EXPLANATION: If a radiograph lacks sufficient blackening, an increase in mAs is required. The mAs regulates the number of x-ray photons produced at the target. An increase or decrease in mAs of at least 30% is necessary to produce a perceptible effect. Increasing the kVp by 15% will have about the same effect as doubling the mAs.

All of the following have an impact on radiographic contrast, except
A. photon energy.B. grid ratio.C. OID.D. focal spot size.

The answer is D.
EXPLANATION: 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.

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 answer is D.
EXPLANATION: Distortion is caused by improper alignment of the tube, body part, and image recorder. 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.

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 answer is D.
EXPLANATION: 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.

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 answer is C.
EXPLANATION: 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.

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

The answer is C.
EXPLANATION: 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.

With a given exposure, as intensifying-screen speed decreases, how is radiographic density affected?

A. Decreases
B. Increases
C. Remains unchanged
D. Is variable

The answer is A.
EXPLANATION: As intensifying-screen speed decreases, less fluorescent light is emitted from the phosphors. If less fluorescent light strikes the film emulsion, a smaller number of silver halide grains are changed to black metallic silver in the developer, and hence there is a decrease in radiographic density. As intensifying-screen speed decreases, so does radiographic density. Intensifying-screen speed and radiographic density are directly related. (Shephard, pp. 67-68)

Why are a single intensifying screen and single emulsion film used for select radiographic examinations?

A. To decrease patient dose
B. To achieve longer-scale contrast
C. For better recorded detail
D. To decrease fiscal expenses

The answer is C.
EXPLANATION: The diffusion of fluorescent light from intensifying screens is responsible for a loss of recorded detail on double-emulsion film. Therefore, by changing the system to include only one intensifying screen and single-emulsion film, as in mammographic systems, light diffusion is reduced and better recorded detail results. Patient dose is somewhat greater than with a two-screen cassette system, but the advantage of significantly improved recorded detail greatly offsets this. (Shephard, p 49)

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 answer is B.
EXPLANATION: 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.

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 answer is D.
EXPLANATION: The faster the imaging system, the greater 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.

A change from 100 speed screens to 200 speed screens would require what change in mAs?

A. mAs should be increased by 15%.
B. mAs should be increased by 30%.
C. mAs should be doubled.
D. mAs should be halved.

The answer is D.
EXPLANATION: As screen speed is increased, exposure factors must be decreased to maintain the original image density. A change from 100 to 200 speed usually requires that the mAs be reduced by one-half. If screen speed were changed from 400 to 200 speed, twice the mAs would be required.

Which of the following is an abnormal intensifying screen action?

A. Fluorescence
B. Luminescence
C. Speed
D. Lag

The answer is D.
EXPLANATION: Luminescence is the production of energy in the form of light. Two types of luminescence are fluorescence and phosphorescence. Fluorescence occurs when an intensifying (radiographic) screen absorbs x-ray photon energy, emits light, and ceases to emit light as soon as the energizing source ceases. Fluoroscopic screens continue to emit light for a short time after the exposure has terminated. This characteristic (phosphorescence) is a desirable quality in fluoroscopic screens. Lag occurs when an intensifying (radiographic) screen continues to fluoresce after the x-ray stimulation has terminated. This characteristic is undesirable and causes excessive density. Screen speed is identified by the amount of light emitted by the phosphors.

Misalignment of the tube-part-IR relationship results in

A. shape distortion.
B. size distortion.
C. magnification.
D. blur.

The answer is A.
EXPLANATION: Shape distortion (foreshortening, elongation) is caused by improper alignment of the tube, part, and IR. Size distortion, or magnification, is caused by too great an OID or too short a SID. Focal spot blur is caused by the use of a large focal spot.

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 image recorder.
3. image recorder and the x-ray tube.

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

The answer is D.
EXPLANATION: Shape distortion is caused by misalignment of the x-ray tube, the part to be radiographed, and the image recorder/film. An object can be falsely imaged (foreshortened or elongated) by incorrect placement of the tube, the body part, or the image recorder. Only one of the three need be misaligned for distortion to occur.

Which of the following affect(s) both the quantity and the quality of the primary beam?
1. Half-value layer (HVL)
2. kVp
3. mA

A. 1 only
B. 2 only
C. 1 and 2 only

The answer is C.
EXPLANATION: Kilovoltage and the HVL affect both the quantity and the quality of the primary beam. The principal qualitative factor for the primary beam is kVp, but an increase in kVp 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 of its original value, thereby effecting a change in both beam quality and quantity. The mAs value is adjusted to regulate the number of x-ray photons produced at the target. X-ray beam quality is unaffected by changes in mAs.

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

The answer is D.
EXPLANATION: 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.

Which of the following have an effect on recorded detail?
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 answer is B.
EXPLANATION: 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 effect on recorded detail. Single-phase rectified units produce "pulsed" radiation, whereas three-phase units produce almost constant potential.

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 answer is C.
EXPLANATION: Radiographers are usually able to stop voluntary motion, using suspended respiration, careful instruction, and immobilization. However, involuntary motion must also be considered. To have a "stop action" effect on the heart when radiographing the chest, it is essential to use a short exposure time.

If the radiographer is unable to achieve a short OID because of the structure of the body part or patient condition, which of the following adjustments can be made to minimize magnification distortion?

A. A smaller focal spot size should be used.
B. A longer SID should be used.
C. Faster intensifying screens should be used.
D. A lower-ratio grid should be used.

The answer is B.
EXPLANATION: An increase in SID will help decrease the effect of excessive OID. For example, in the lateral projection of the cervical spine, there is normally a significant OID that would result in obvious magnification at a 40-inch SID. This effect is decreased by the use of a 72-inch SID. However, especially with larger body parts, increased SID usually requires a significant increase in exposure factors. Focal spot size, screen speed, and grid ratio are unrelated to magnification.

In which of the following examinations should 70 kVp not be exceeded?

A. Upper GI (UGI)
B. Barium enema (BE)
C. Intravenous urogram (IVU)
D. Chest

The answer is C.
EXPLANATION: The iodine-based contrast material used in intravenous (IV) urography gives optimum opacification at 60 to 70 kVp. Use of higher kVp 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-kVp exposure factors (about 120 kVp) to penetrate through the barium. In chest radiography, high-kVp technical factors are preferred for maximum visualization of pulmonary vascular markings, made visible with long-scale contrast.

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 answer is B.
EXPLANATION: Radiographic density is described as the overall degree of blackening of a radiograph or a part of it. The mAs 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 mAs; therefore, mAs and radiographic density are directly proportional.

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 answer is C.
EXPLANATION: 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 IR sizes, short SIDs, and steep target angles

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