, Question 1
A transducer emits a pulse with a spatial pulse length of 1.2 mm. If the
wavelength is 0.4 mm, what is the axial resolution, and how would it change if
the frequency were doubled while the number of cycles per pulse remained
constant?
A. Axial resolution = 0.6 mm; doubling frequency improves it to 0.3 mm
B. Axial resolution = 1.2 mm; doubling frequency improves it to 0.6 mm
C. Axial resolution = 0.6 mm; doubling frequency worsens it to 1.2 mm
D. Axial resolution = 1.2 mm; doubling frequency does not change it
Correct Answer: A - Axial resolution = 0.6 mm; doubling
frequency improves it to 0.3 mm
RATIONALE
Axial resolution = spatial pulse length / 2 = 0.6 mm. Doubling
frequency halves the wavelength (if cycles constant), halving SPL,
thus improving axial resolution to 0.3 mm. Options B and D misapply
the formula; C incorrectly states worsening.
Question 2
During a quality assurance test, the measured propagation speed in a phantom
is 1540 m/s, but the system assumes 1540 m/s. If the actual speed in a patient's
tissue is 1580 m/s, how does this affect the displayed depth of a reflector
actually located at 6 cm?
A. Displayed depth is greater than 6 cm
B. Displayed depth is less than 6 cm
C. Displayed depth is exactly 6 cm
D. Displayed depth is unpredictable without knowing attenuation
Correct Answer: B - Displayed depth is less than 6 cm
Page 2
, RATIONALE
The system calculates depth using assumed speed (1540 m/s). If actual
speed is higher (1580 m/s), the echo returns sooner, so the system
underestimates depth (displays <6 cm). Options A and C are incorrect;
D is false because speed mismatch directly affects depth.
Question 3
A 5 MHz transducer with a 6 mm diameter element is used in a medium with
speed of sound 1540 m/s. What is the near field length (Fresnel zone) and how
does it change if the frequency is increased to 10 MHz while diameter remains
constant?
A. Near field = 2.9 cm; increasing frequency to 10 MHz doubles it to 5.8
cm
B. Near field = 5.8 cm; increasing frequency to 10 MHz halves it to 2.9
cm
C. Near field = 2.9 cm; increasing frequency to 10 MHz quadruples it to
11.6 cm
D. Near field = 1.45 cm; increasing frequency to 10 MHz doubles it to
2.9 cm
Correct Answer: A - Near field = 2.9 cm; increasing frequency to
10 MHz doubles it to 5.8 cm
RATIONALE
Near field length = (diameter² × frequency) / (4 × speed). For 5 MHz:
(0.006² × 5e6)/(4×1540) 0.029 m = 2.9 cm. Doubling frequency
doubles near field to 5.8 cm. Options B, C, D miscalculate the
relationship.
Question 4
Which of the following best explains why a 1.5D array transducer can improve
elevational resolution compared to a conventional 1D array?
Page 3
A transducer emits a pulse with a spatial pulse length of 1.2 mm. If the
wavelength is 0.4 mm, what is the axial resolution, and how would it change if
the frequency were doubled while the number of cycles per pulse remained
constant?
A. Axial resolution = 0.6 mm; doubling frequency improves it to 0.3 mm
B. Axial resolution = 1.2 mm; doubling frequency improves it to 0.6 mm
C. Axial resolution = 0.6 mm; doubling frequency worsens it to 1.2 mm
D. Axial resolution = 1.2 mm; doubling frequency does not change it
Correct Answer: A - Axial resolution = 0.6 mm; doubling
frequency improves it to 0.3 mm
RATIONALE
Axial resolution = spatial pulse length / 2 = 0.6 mm. Doubling
frequency halves the wavelength (if cycles constant), halving SPL,
thus improving axial resolution to 0.3 mm. Options B and D misapply
the formula; C incorrectly states worsening.
Question 2
During a quality assurance test, the measured propagation speed in a phantom
is 1540 m/s, but the system assumes 1540 m/s. If the actual speed in a patient's
tissue is 1580 m/s, how does this affect the displayed depth of a reflector
actually located at 6 cm?
A. Displayed depth is greater than 6 cm
B. Displayed depth is less than 6 cm
C. Displayed depth is exactly 6 cm
D. Displayed depth is unpredictable without knowing attenuation
Correct Answer: B - Displayed depth is less than 6 cm
Page 2
, RATIONALE
The system calculates depth using assumed speed (1540 m/s). If actual
speed is higher (1580 m/s), the echo returns sooner, so the system
underestimates depth (displays <6 cm). Options A and C are incorrect;
D is false because speed mismatch directly affects depth.
Question 3
A 5 MHz transducer with a 6 mm diameter element is used in a medium with
speed of sound 1540 m/s. What is the near field length (Fresnel zone) and how
does it change if the frequency is increased to 10 MHz while diameter remains
constant?
A. Near field = 2.9 cm; increasing frequency to 10 MHz doubles it to 5.8
cm
B. Near field = 5.8 cm; increasing frequency to 10 MHz halves it to 2.9
cm
C. Near field = 2.9 cm; increasing frequency to 10 MHz quadruples it to
11.6 cm
D. Near field = 1.45 cm; increasing frequency to 10 MHz doubles it to
2.9 cm
Correct Answer: A - Near field = 2.9 cm; increasing frequency to
10 MHz doubles it to 5.8 cm
RATIONALE
Near field length = (diameter² × frequency) / (4 × speed). For 5 MHz:
(0.006² × 5e6)/(4×1540) 0.029 m = 2.9 cm. Doubling frequency
doubles near field to 5.8 cm. Options B, C, D miscalculate the
relationship.
Question 4
Which of the following best explains why a 1.5D array transducer can improve
elevational resolution compared to a conventional 1D array?
Page 3
