Exposure Chapter29 Distortion test bank

1. Distortion is a misrepresentation of

a.size only.

b.shape only.

c.size and shape.

d.detail.

size and shape

2. Size distortion in radiography can be ____ only.

a. foreshortening

b. magnification

c. elongation

d. minimization

b. Magnification

3. Size distortion is controlled by

a. SID.

b. OID.

c. radiographic distances.

d. all of the above.

all of the above

4. As size distortion decreases, the resolution of recorded detail

a. increases

b. decreases.

c. is not affected.

d. is altered due to digital reprocessing.

increases

5. The ____ the SID, the ____ the degree of magnification.

a. greater; larger

b. greater; smaller

c. lesser; smaller

d. SID has no effect on magnification.

Greater; Smaller

6. Examinations of body parts with a large inherent OID warrant a ____ whenever possible.

a.small SID

b.large SID

c.small SOD

d.large focal spot

large SID

7. Which examination does not exhibit an inherently large OID?

a.AP lumbar spine

b.lateral cervical spine

c.lateral chest

d.AP facial bones

AP lumbar spine

8. As OID increases and SID remains constant, entrance skin exposure (ESE)

a. remains unchanged.

b. decreases.

c. increases.

d. none of the above.

increases.

9. Large patients receive a greater exposure than small patients because their

a. SOD is decreased.

b. OID is increased.

c. entrance skin surface is closer to the source.

d. all of the above.

All of the above

10. The magnification factor for a radiographic procedure is calculated at 1.25. It may be assumed that there is a ____ magnification of the object size.

a. 25 percent

b. 125 percent

c. 0.25 percent

d. 1.25 percent

b. 125%

1.25*100 = 125%

11. Which of the following occurs when the tube or the image receptor are not properly aligned?

a. elongation

b. magnification

c. foreshortening

d. minification

elongation

12. Which one of the following occurs when the part is improperly aligned?

a. elongation

b. magnification

c. foreshortening

d. minification

Foreshortening

13. Proper alignment is achieved when the central ray is ____ to the part and ____ to the image receptor.

a. perpendicular; parallel

b. parallel; perpendicular

c. perpendicular; perpendicular

d. parallel; parallel

c. perpendicular; perpendicular

14. When the position of the patient is reversed, the direction of the tube angle must be ____ to maintain the relationship.

a. maintained

b. decreased

c. increased

d. reversed

d. reversed

15. Because the x-ray beam is divergent,

a. minification is impossible.

b. minification is collimated off from the edges of the beam.

c. minification is greater with a smaller target angle.

d. minification is greater with a larger target angle.

a. minification is impossible.

16. X-ray tube angulations inherently

a. increase spatial resolution.

b. change the SID.

c. introduce magnification to some degree.

d. both b and c.

c. magnification to some degree.

17. In order to achieve a true SID of 40" with an x-ray tube angle of 30 degrees, the overhead scale should read

a. 23 inches.

b. 95 cm.

c. 34.6 inches.

d. both b and c.

34.6 inches

18. Performing a routine chest radiograph in the anterioposterior (AP) projection will

a. decrease heart size on the image.

b. demonstrate higher resolution of the anterior ribs.

c. produce improved resolution of the thoracic spine.

d. decrease breast exposure.

produce improved resolution of the thoracic spine.

19. A renal calculus that measures 0.2 mm in size

a. cannot be visualized with an effective focal spot of 0.5 mm focal spot due to penumbral overlap.

b. is best seen with magnification and a large focal spot.

c. most likely will be visualized with a small radiographic tube angle using a large target angle.

d. all of the above.

cannot be visualized with an effective focal spot of 0.5 mm focal spot due to penumbral overlap.

20. Opening up the intervertebral joints of the cervical spine would best be accomplished with

a. the patient in a PA projection at 72" SID.

b. the patient in an AP projection with a 40" SID.

c. a perpendicular central ray at 72" SID.

d. the patient in a PA projection with a 40" SID.

the patient in an AP projection with a 40" SID.

21. With digital image receptors (CR/DR)

a. the geometric factors of distortion do not apply.

b. EI#s are inaccurate due to elongation distortion.

c. computer postprocessing corrects for shape distortion.

d. none of the above.

none of the above

22. A radiographic magnification factor(MF) of −.15 indicates

a.the object size and image size are nearly equal.

b.nothing, since negative MF is not possible.

c.the image size is smaller than the object size.

d.a change in OID is needed.

nothing, since negative MF is not possible.

23. Enlargement of a digital radiographic image on the review monitor, is an example of

a. purposeful geometric magnification.

b. electronic magnification.

c. shape distortion.

d. digital histogram distortion.

electronic magnification.

24. Positioning patients is intended to eliminate anatomical superimposition, from a radiologist perspective. To optimize image resolution it is best to

a. keep the central ray and receptor perpendicular to the patient's anatomy, and rotate the part.

b. minimize central angulation and rotate the part only slightly.

c. use the small focal spot and extend the degree of part rotation.

d. lower the SID and increase the OID with part rotation.

keep the central ray and receptor perpendicular to the patient's anatomy, and rotate the part.

25. Shape distortion is calculated

a. as the ratio of SID and SOD.

b. using the tube angle and OID.

c. automatically in the computer postprocessing.

d. none of the above.

none of the above

MF = SID/SOD (MF=magnification factor)-not shape distortion

26. ratio between image size and MF

object size

object size = image size/MF

27. an image in which the object is actually shorter

elongation

The image appears longer than the object.

28. best seen with the patient supine

kidneys

The kidneys lie retroperitoneally (behind the peritoneum) in the abdomen, either side of the vertebral column.

29. MF

MF = SID ÷ SOD

30. SID ÷ MF

SOD

31. inherently magnified in AP projection

patella

There is the large OID between patella and IR.

32. Calculate the magnification factor when the SID is 40 in. and the SOD is 25 in.

40/25 = 1.6

33. Calculate the magnification factor when the SID is 72 in. and the SOD is 25 in.

2.9

34. Calculate the magnification factor when the SID is 40 in. and the OID is 5 in.

SOD; 40 - 5 = 35

40/35 = 1.14

35. Calculate the magnification factor when the SID is 72 in. and the OID is 2 in.

1.0

36. If a projected image measures 3 in. and the magnification factor is 1.1, what is the size of the actual object?

o = I/MF

o = 3/1.1

object size = 2.7 inches

37. If an object is measured as 4 in. in diameter on the image and 2 in. from the film, what is the actual size of the object if the SID is 40 in.?

Image size = 4 inches

OID = 2 inches

SID = 40 inches

SOD = 40 - 2 = 38inches

MF = 40/38 = 1.05

Object size = Image size/MF

O = 4/1.05 = 3.8 inches

38. If an object is measured as 4 in. in diameter on the image and 4 in. from the film, what is the actual size of the object if the SID is 40 in.?

3.6 inches

39. If an object measures 2 cm and the image measures 4 cm, what is the percent magnification of the object?

100%

Object size = Image size/MF

MF = 2

2*100=200%

Object is 100% and image is 200%

so the image is magnified from the object 200%-100%=100%

40. If an object measures 2 cm and the image measures 3 cm, what would be the percent magnification of the object?

50%

MF = 3/2 = 1.5

The object is 100% and the image is 150%

The image is magnified 150% - 100% = 50%