a. 2.43 in
b. 3.24 in
c. 3.63 in
d. 6.53 in
b. 3.24 in
In order to determine object size, the image size is divided by the magnification factor; in this case 4.6 in is divided by 1.42.
REF: p.165
a. off-centering the central ray.
b. angling the central ray.
c. increasing the SID.
d. A and B only.
d. A and B only.
Shape distortion results from misalignment or angulation of the tube, part, or image receptor.
REF: p.165
a. adding SID and OID.
b. adding the MF to the OID.
c. subtracting the OID from SID.
d. subtracting the SID from the MF.
c. subtracting the OID from SID.
The source-to-object distance (SOD) can be calculated by subtracting the OID from the SID.
REF: p.163
a. Increased recorded detail
b. Decreased recorded detail
c. No effect on recorded detail
a. Increased recorded detail
Increasing SID increases recorded detail by reducing the amount of geometric unsharpness and decreasing size distortion.
REF: p.173
a. 2 to 3 cm
b. 4 to 5 cm
c. 6 to 7 cm
d. 8 to 9 cm
b. 4 to 5 cm
For every increase of 4 to 5 cm of part thickness, the mAs should be doubled. It should be halved when the part is 4 to 5 cm thinner.
REF: p.172
a. 1 mAs
b. 2 mAs
c. 6 mAs
d. 12 mAs
c. 6 mAs
According to the mAs/distance compensation formula, increasing the distance from 40 to 50 in will require an increase of mAs from 4 to 6 in order to maintain exposure to the IR.
REF: p.160
a. increases
b. decreases
c. has no effect on
a. increases
Increased collimation (a smaller field size) reduces the amount of scatter produced by the tissue and therefore increases the image contrast, because less scatter radiation reaches the image receptor.
REF: p.169
a. Use 40 ms exposure time.
b. Use 60 ms exposure time.
c. Use 200 ms exposure time.
d. Use at least 80 kVp.
a. Use 40 ms exposure time.
When imaging an infant for a chest radiograph, a short exposure time is critical in order to stop patient motion.
REF: p.169
a. Asthenic
b. Sthenic
c. Hypersthenic
d. Hyposthenic
c. Hypersthenic
The hypersthenic patient has a large, stocky build and will require the highest exposure factors.
REF: p.170
a. quality.
b. quantity.
c. divergence.
d. restriction.
c. divergence
Due to the divergence of the x-ray beam, a geometric relationship exists among the source of x-rays, the object, and the image receptor.
REF: p.158
a. one fourth
b. one half
c. two times
d. four times
c. two times
Only after the part has been adequately penetrated, doubling mAs results in twice the exposure to the IR.
REF: p.172
a. 200 mA at 0.06 s
b. 400 mA at 0.03 s
c. 600 mA at 20 ms
d. 800 mA at 0.15 s
d. 800 mA at 0.15 s
800 mA at 0.15 s results in 120 mAs.
REF: p.149
a. mA.
b. exposure time.
c. mAs.
d. all of the above.
d. all of the above.
mAs, consisting of mA and exposure time, control film-screen density.
REF: p.151
a. decreased; lower
b. decreased; higher
c. increased; lower
d. increased; higher
c. increased; lower
With film-screen imaging, increasing the kVp results in an image with increased density and lower (more gray) contrast.
REF: p.152
a. very low; increased
b. appropriate; increased
c. excessive; decreased
d. unacceptable; decreased
b. appropriate; increased
With digital imaging, if too low kVp is set, the image will have appropriate brightness (due to the computer making the adjustment) and increased quantum mottle.
REF: p.152
a. appropriate; increased
b. appropriate; appropriate
c. excessive; increased
d. excessive; appropriate
a. appropriate; increased
With digital imaging, using twice as much mAs as needed will result in an image with appropriate brightness (due to the computer making the adjustment) and increased patient exposure.
REF: p.150
a. 1.5 mAs
b. 3 mAs
c. 12 mAs
d. 24 mAs
c. 12 mAs
As a rule of thumb, changing from 40 in SID to 56 in SID requires twice as much mAs.
REF: p.161
a. Recorded detail
b. Image contrast
c. Amount of radiation reaching the IR
d. Magnification
b. Image contrast
Image contrast is not affected by SID changes.
REF: p.161
a. inside the x-ray tube.
b. between the tube and patient.
c. between the patient and the IR.
d. beneath the IR.
c. between the patient and the IR.
In order to absorb the scattered radiation exiting the patient, the grid is located between the patient and IR.
REF: p.167
a. lower; higher
b. lower; lower
c. higher; higher
d. higher; lower
a. lower; higher
With film-screen imaging, decreasing collimation (using a larger field size) results in lower image contrast (due to increased scatter) and higher patient exposure.
REF: p.169
a. less; lower
b. less; higher
c. more; lower
d. more; higher
b. less; higher
Increasing tube filtration results in an x-ray beam with less quantity (due to increased attenuation by the filter) and higher energy (because the lower-energy photons were absorbed).
REF: p.170
a. less; lower
b. more; lower
c. less; higher
d. more; higher
c. less; higher
As compared to a patient with a 24 cm thick abdomen, imaging a patient with a 20 cm thick abdomen will result in less scatter being produced and therefore higher image contrast.
REF: p.172
1 T/F With film-screen imaging, for a given exposure technique, increasing part thickness decreases radiographic density.
True
A thicker part absorbs more radiation, resulting in less exit radiation and reduced density on a film-screen image.
REF: p.172
2 T/F Kilovoltage does not affect the quantity of exit radiation until adequate penetration is reached.
True
The area of interest must be adequately penetrated before kilovoltage can be adjusted to affect the quantity of exit radiation.
REF: p.152