Huda US

Question 1

1. The velocity of an ultrasound beam is always:
   (A) Constant for all solids
   (B) Proportional to frequency cubed
   (C) Equal to the velocity of the molecules of the
   medium
   (D) Equal to frequency times wavelength
   (E) 3 X 108 m/second

Answer 1

1-D. The velocity (v) of any wave is always the
product of the frequency ( f ) and wavelength (λ)
(i.e., v = f x λ).

Question 2

2. Sound waves are:
(A) Constant velocity
(B) Low-frequency electromagnetic radiation
(C) Ionizing radiation
(D) Audible at 1 MHz
(E) Longitudinal waves

Answer 2

2-E. In ultrasound, the displacement is along the
direction of travel (electromagnetic waves are
transverse, because the displacement is perpendicular
to the direction of the wave motion).

Question 3

3. A 2 MHz transducer has an approximate
wavelength of:
(A) 0.01 mm
(B) 0.1 mm
(C) 1.0 mm
(D) 10 mm
(E) More than 10 mm

Answer 3

3-C. Because v = f x λ, and the velocity of
sound is 1540 m/second, the wavelength is given
by [(l,540)/(2 X 106)] m. or about 1 mm

Question 4

4. The wavelength of a 3 MHz sound beam is
shortest in:
(A) Air
(B) Castor oil
(C) Fat
(D) Muscle
(E) Bone

Answer 4

4-A. The wavelength (X) is given by v/f, and because
the speed of sound in air (330 m/second) is
much less than in soft tissue (1,540 m/second). the
wavelength is correspondingly shorter.

Question 5

5. Acoustic impedance (Z) is primarily dependent
on tissue:
(A) Density
(B) Attenuation
(C) Atomic number
(D) Temperature
(E) Oxygenation

Answer 5

5-A. Acoustic impedance is dependent on tissue
density and is obtained using the equation Z = ρ x
v, where ρ is the tissue density, and v is the velocity
of sound in the tissue.

Question 6

6. Which of the following has the highest acoustic
impedance?
(A) Bone
(B) Fat
(C) Air
(D) Water
(E) Eye lens

Answer 6

6-A. Acoustic impedance is the product of the
density and velocity of sound, both of which are
the highest for bone.

Question 7

7. The wavelength of a 1 MHz sound beam is not:
(A) The same in all solid media
(B) 0.3 mm in air
(C) 1.5 mm in soft tissue
(D) 4.1 mm in bone
(E) Velocity divided by frequency

Answer 7

7-A. Wavelength generally changes with
medium because frequency will be the same, but
velocity depends on the medium.

Question 8

8. If an ultrasound beam is attenuated by 99%,
the attenuation is:
(A) 1 dB
(B) 3 dB
(C) 10 dB
(D) 20 dB
(E) Greater than 20 dB

Answer 8

8-D. Decibels are 10 x log10 (I0/I), where I0 is

100 and I is 1: this corresponds to 20 decibels.

Question 9

9. The key factor determining the fraction of ultrasound
reflected at a large interface is the:
(A) Depth of the interface
(B) Transducer diameter
(C) Transducer output intensity
(D) Differences in acoustic impedance
(E) Scan mode (A, B. or M)

Answer 9

9-D. The difference in acoustic impedance between
the two tissues (Z1,Z2) determines the fraction
of incident energy reflected.

Question 10

10. What fraction of ultrasound is reflected from
    a liver (Z = 1.55) and soft tissue (Z = 1.65) interface?
    (A) 1/2
    (B) 1/10
    (C) 1/100
    (D) 1/500
    (E) 1/1,000

Answer 10

10-E. The reflected fraction of an ultrasound
beam is given by [(Z1 — Z2)/(Z1 + Z2)]2, which
gives 1/1,000.

Question 11

11. Ultrasound shadowing artifacts are unlikely
behind:
(A) Strong attenuators
(B) Bone
(C) Air
(D) Fluid-filled cysts
(E) Metallic clips

Answer 11

11-D. Shadowing artifacts occur because of a
large loss of transmitted signal intensity caused by
either attenuation or reflection; fluid-filled cysts
transmit ultrasound and result in enhancement of
echoes beyond the cyst.

Question 12

12. Reflections occur from all of the following except:
(A) Smooth surfaces
(B) Kidney interior
(C) Fat-kidney interfaces
(D) Bladder wall
(E) Bladder contents

Answer 12

12-E. There are no reflections from fluids in the

bladder.

Question 13

13. Snell’s law describes the relation between the:
    (A) Angle of incidence and transmission
    (B) Fraunhofer angle and wavelength
    (C) Angle of incidence and angle of reflection
    (D) Focus and transducer curvature
    (E) Fresnel zone and wavelength

Answer 13

13-A. Snell’s law describes the angle of refraction
that occurs when an ultrasound beam passes
from one medium to another.

Question 14

14. An ultrasound beam traveling through tissue
cannot be:
(A) Absorbed
(B) Amplified
(C) Scattered
(D) Reflected
(E) Refracted

Answer 14

14-B. There is no mechanism for amplifying ultrasound
beams in patients. Echoes from tissue interfaces.
however, can be amplified electronically.

Question 15

15. Higher-frequency transducers have increased:
(A) Thickness
(B) Intensity
(C) Attenuation
(D) Velocity
(E) Wavelength

Answer 15

15-C. The attenuation in tissue is about 1 dB/cm
at 1 MHz and increases approximately linearly
with frequency (there is no direct relationship between
intensity and frequency).

Question 16

16. The Q factor of a transducer refers to:
(A) Coupling efficiency
(B) Minimum intensity
(C) Maximum intensity
(D) Purity of the frequency
(E) Transducer dead time

Answer 16

16-D. Q is defined as the operating frequency
(MHz) divided by the bandwidth, so that high Q
values correspond to a pure frequency and vice
versa.

Question 17

17. The damping material behind the crystal
transducer reduces:
(A) Pulse frequency
(B) Ring-down time
(C) Echo amplitude
(D) Lateral resolution
(E) PRF

Answer 17

17-B. The ring-down time is reduced and very
short pulses of only two or three wavelengths are
generated.

Question 18

18. The Fresnel zone length of an ultrasound
beam increases with increasing:
(A) Transducer diameter
(B) Transducer thickness
(C) Wavelength
(D) Intensity
(E) TGC

Answer 18

18-A. The Fresnel zone is given by r2/λ. where r

is the transducer radius and λ is the wavelength

Question 19

19. TGC corrects for which of the following?
    (A) Signal losses at skin interface
    (B) Velocity of moving objects
    (C) Intensity decrease with tissue penetration
    (D) Transducer damping material
    (E) Image fading on cathode ray tube

Answer 19

19-C. TGC corrects for normal attenuation in tissue,
and is generally assumed to be 1 dB/cm per 1
MHz.

Question 20

20. An echo received 65 microseconds after the
signal is sent is from what depth?
(A) 2 cm
(B) 5 cm
(C) 7 cm
(D) 10 cm
(E) 15 cm

Answer 20

20-B. The equation to use is d = c x t, where d is
the total travel distance, c is the speed (1,540
m/second), and t is the time (65 microseconds); we
obtain a round trip distance of 10 cm, which corresponds
to depth of 5 cm.

Question 21

21. In B-mode ultrasound, the PRF does not affect:
(A) Pulses per second
(B) Frame rate
(C) Number of lines per frame
(D) Maximum penetration depth
(E) Ultrasound frequency

Answer 21

21-E. Ultrasound frequency has no relationship
to the PRF but is determined by the transducer
crystal thickness.

Question 22

22. Choice of frequency in ultrasound is most
    likely a trade off between patient penetration and:
    (A) Contrast
    (B) PRF
    (C) Noise
    (D) Lateral resolution
    (E) Axial resolution

Answer 22

22-E. As frequency increases, the wavelength is
reduced, which improves resolution but reduces
patient penetration.

Question 23

23. Ultrasound signals are converted from digital
data to a video monitor display using a:
(A) Log amplifier
(B) Photomultiplier tube
(C) Photocathode
(D) Scan converter
(E) Array processor

Answer 23

23-D. Scan converters convert ultrasound data

into an image that is displayed on a video monitor

Question 24

24. The best axial resolution is obtained using:
(A) 5.0 MHz phased array
(B) 5.0 MHz linear array
(C) 5.0 MHz continuous-wave Doppler
(D) 10 MHz sector scanner
(E) 10 MHz continuous-wave Doppler

Answer 24

24-D. Highest frequency normally gives the best
axial resolution. However, continuous-wave
Doppler provides little spatial information and has
the “worst” axial resolution.

Question 25

25. Lateral resolution in ultrasound imaging
would most likely be improved by:
(A) Increasing transducer focusing
(B) Imaging in the Fraunhofer zone
(C) Using fewer scan lines
(D) Increasing the frequency
(E) Reducing the pulse length

Answer 25

25-A The lateral resolution improves with focussing

Question 26

26. Below a structure, a very faint image of the
structure is probably owing to:
(A) Reverberation artifact
(B) Side lobes
(C) Specular reflection
(D) Nonspecular reflection
(E) Incorrect TCG

Answer 26

26-A. Reverberation artifact is produced by the
beam bouncing off the posterior interface, then
off the anterior interface, then back off the posterior
interface, and finally being recorded as a
faint image further away from the actual structure.

Question 27

27. All of the following may cause significant ultrasound
artifacts except:
(A) Reverberation
(B) Side lobes
(C) Nonspecular reflections
(D) Refraction
(E) Speed displacement

Answer 27

27-C. Nonspecular reflection will result in the
beam being scattered in all directions and is unlikely
to be the direct cause of image artifacts.

Question 28

28. Clinical ultrasound beams normally have an
intensity of:
(A) 0.5 mW/cm2
(B) 5 mW/cm2
(C) 50 mW/cm2
(D) 0.5 W/cm2
(E) 5 W/cm2

Answer 28

28-B. Diagnostic ultrasound uses 1 to 10

mW/cm2.

Question 29

29. The Doppler shift from a moving object depends
on all of the following except:
(A) Speed of ultrasound beam
(B) Frequency
(C) Angle between beam and object
(D) Object depth
(E) Object speed

Answer 29

29-D. The depth of the moving object is immaterial.

Question 30

30. Continuous-wave Doppler uses:
(A) One transducer
(B) High frequency transducers
(C) A low Q factor
(D) A scan converter
(E) Little spatial information

Answer 30

30-E. Continuous-wave Doppler provides little
spatial information because the beam is continuously
on, which permits detection of the frequency
differences between emitted and reflected
signals.