A Level Physics 24.1 Production and Use of Ultrasound MCQs – 30 Practice Questions with Answers (Cambridge 9702)

These 30 original practice questions cover Cambridge AS & A Level Physics (9702) Section 24.1, Production and Use of Ultrasound, an A Level (A2) topic in Medical Physics. They test the piezoelectric transducer, the pulse-echo technique with A-scans and B-scans, and specific acoustic impedance with reflection at boundaries.

Formula recap

Depth of a boundary: d = c t / 2 (t = time between pulse and echo, the factor of 2 is for the journey there and back)

Specific acoustic impedance: Z = ρ c, measured in kg m−2 s−1

Fraction of intensity reflected at a boundary (normal incidence): ((Z2 − Z1) / (Z2 + Z1))2

Speed of ultrasound in soft tissue ≈ 1540 m/s. See the full A Level Physics Formula Sheet for every equation in this topic.

Producing & Receiving Ultrasound

Q1. Ultrasound is defined as sound waves with a frequency

  • (A) higher than about 20 kHz, the upper limit of human hearing
  • (B) lower than about 20 Hz
  • (C) between 20 Hz and 20 kHz
  • (D) higher than 20 MHz only
Show Answer

Answer: A – Ultrasound is any sound above the top of the audible range, about 20 kHz. Frequencies below 20 Hz are infrasound, and medical imaging typically uses 1-15 MHz, but ultrasound itself starts at 20 kHz.

Q2. In a medical ultrasound transducer, the ultrasound is generated because

  • (A) a loudspeaker-style coil vibrates in a magnetic field
  • (B) a metal filament is heated until it emits sound
  • (C) an alternating p.d. applied across a piezoelectric crystal makes the crystal vibrate at the frequency of the supply
  • (D) a rotating disc with holes chops a jet of air
Show Answer

Answer: C – This is the reverse piezoelectric effect: the applied alternating p.d. makes the crystal repeatedly change shape, and these vibrations launch the ultrasound wave.

Q3. The same transducer crystal detects the returning echoes because

  • (A) the echo heats the crystal and changes its resistance
  • (B) the pressure of the echo squeezes the crystal, generating a p.d. across its faces
  • (C) the echo magnetises the crystal
  • (D) the echo makes the crystal emit light
Show Answer

Answer: B – This is the direct piezoelectric effect: mechanical pressure on the crystal produces an electrical signal, so the tiny echo pressure gives a small p.d. that can be amplified and timed.

Q4. The alternating p.d. driving the crystal is set at the crystal’s resonant frequency so that

  • (A) attenuation in tissue is reduced
  • (B) the wavelength in tissue becomes longer
  • (C) reflection at boundaries is prevented
  • (D) the crystal vibrates with maximum amplitude, giving the strongest possible pulses
Show Answer

Answer: D – At resonance the amplitude of the crystal’s vibrations is a maximum, so the transducer produces its most intense ultrasound and also responds most strongly to echoes.

Q5. Which material is a piezoelectric material used in ultrasound transducers?

  • (A) Quartz (and ceramics such as PZT)
  • (B) Copper
  • (C) Aluminium
  • (D) Soft iron
Show Answer

Answer: A – Quartz is the classic piezoelectric crystal; modern medical probes usually use the ceramic PZT (lead zirconate titanate), which has a stronger piezoelectric effect.

Q6. Ultrasound is sent out in short pulses rather than as a continuous wave mainly because

  • (A) a continuous wave would damage the tissue
  • (B) the echoes from boundaries must return and be timed before later pulses confuse the measurement
  • (C) a piezoelectric crystal cannot produce a continuous wave
  • (D) short pulses travel faster through tissue
Show Answer

Answer: B – Depth is found from the time delay of each echo, so the pulse must be short and the system must listen in the quiet gaps between pulses; continuous waves would make the echoes overlap and impossible to time.

Q7. Medical scanners use frequencies of a few MHz rather than, say, 50 kHz, mainly because

  • (A) MHz waves travel faster through tissue
  • (B) MHz waves are attenuated less in tissue
  • (C) the much shorter wavelength gives finer detail (better resolution) in the image
  • (D) MHz waves are not reflected at the skin
Show Answer

Answer: C – Structures smaller than about one wavelength cannot be resolved. At 5 MHz the wavelength in tissue is only about 0.3 mm, so fine detail is possible; in fact MHz waves are attenuated MORE, which is the price paid for the detail.

Q8. Ultrasound of frequency 5.0 MHz travels through soft tissue at 1540 m/s. Its wavelength in the tissue is

  • (A) 3.1 × 10−1 m
  • (B) 3.1 × 10−2 m
  • (C) 3.1 × 10−3 m
  • (D) 3.1 × 10−4 m
Show Answer

Answer: D – λ = c / f = 1540 / (5.0 × 106) = 3.1 × 10−4 m, about 0.31 mm, which sets the scale of the smallest detail the scan can show.

Q9. How can one transducer both send the pulses and receive the echoes?

  • (A) The echoes are detected by a separate microphone placed behind the patient
  • (B) After emitting each short pulse, the crystal is used as a receiver and detects the echoes that return before the next pulse is sent
  • (C) The crystal emits and receives at exactly the same instant
  • (D) The echoes are recorded on photographic film
Show Answer

Answer: B – The transducer works in bursts: it transmits a short pulse (reverse piezoelectric effect), then switches to listening mode (direct piezoelectric effect) for the much longer gap while echoes return.

Q10. The reverse piezoelectric effect is

  • (A) an applied p.d. causing the crystal to change shape and vibrate
  • (B) pressure on a crystal generating a p.d. across it
  • (C) a crystal reversing the frequency of a wave
  • (D) an echo reversing its direction at a boundary
Show Answer

Answer: A – Applying a p.d. makes the crystal contract or expand, so an alternating p.d. makes it vibrate – this is how the pulse is produced. Option B is the direct effect, used in reception.

Pulse-Echo Technique & Scans

Q11. The depth of a boundary is calculated from the echo time using d = c t / 2. The division by 2 is needed because

  • (A) the pulse loses half its energy on the way in
  • (B) the speed of ultrasound halves inside the body
  • (C) the measured time is for the pulse to travel to the boundary AND back again
  • (D) the time is measured in microseconds instead of seconds
Show Answer

Answer: C – The total distance travelled is 2d (there and back), so d = (c × t) / 2. Forgetting this factor of 2 is the most common error in pulse-echo questions.

Q12. An echo returns 65 μs after the pulse is sent. Taking the speed of ultrasound in tissue as 1540 m/s, the reflecting boundary is at a depth of

  • (A) 5.0 cm
  • (B) 10 cm
  • (C) 2.5 cm
  • (D) 0.50 cm
Show Answer

Answer: A – d = (1540 × 65 × 10−6) / 2 = 0.100 / 2 = 0.050 m = 5.0 cm.

Q13. An echo is received 130 μs after the pulse. Taking the speed of ultrasound in tissue as 1540 m/s, the boundary is at a depth of

  • (A) 2.0 cm
  • (B) 5.0 cm
  • (C) 20 cm
  • (D) 10 cm
Show Answer

Answer: D – d = (1540 × 130 × 10−6) / 2 = 0.200 / 2 = 0.100 m = 10 cm.

Q14. In an A-scan, the horizontal position of a spike on the trace shows

  • (A) the frequency of the pulse
  • (B) the depth of the boundary that produced the echo
  • (C) the acoustic impedance of the tissue
  • (D) the speed of ultrasound in the tissue
Show Answer

Answer: B – The horizontal axis is time, and time maps directly onto depth through d = c t / 2, so each spike marks a boundary along the single line of the beam.

Q15. In an A-scan, a taller spike means that at that boundary

  • (A) the boundary is deeper
  • (B) the pulse was travelling faster
  • (C) a stronger echo was reflected (a bigger fraction of the intensity was reflected)
  • (D) the frequency of the ultrasound was higher
Show Answer

Answer: C – Spike height is proportional to the amplitude (strength) of the received echo, which depends on how much of the pulse was reflected at that boundary.

Q16. A B-scan produces a two-dimensional image because

  • (A) the brightness of each dot represents the strength of the echo from that point, and many scan lines are combined
  • (B) a single spike trace is photographed
  • (C) it measures only the speed of blood flow
  • (D) it uses a beam of X-rays instead of sound
Show Answer

Answer: A – The transducer is swept (or an array is used) to take many echo lines across a section; each echo becomes a dot whose brightness shows echo strength, building up a cross-section picture.

Q17. A boundary lies 7.7 cm below the skin. Taking the speed of ultrasound in tissue as 1540 m/s, the echo from this boundary is received after a time of

  • (A) 50 μs
  • (B) 100 μs
  • (C) 200 μs
  • (D) 20 μs
Show Answer

Answer: B – t = 2d / c = (2 × 0.077) / 1540 = 0.154 / 1540 = 1.0 × 10−4 s = 100 μs.

Q18. Pulses, rather than a continuous beam, are essential in scanning because

  • (A) continuous ultrasound cannot be reflected
  • (B) tissue absorbs a continuous beam completely
  • (C) pulses travel faster than continuous waves
  • (D) echoes from different boundaries would overlap and their arrival times, and so the depths, could not be separated
Show Answer

Answer: D – All depth information is carried by echo timing. With a continuous beam the returning echoes merge into one signal and no boundary depths can be picked out.

Q19. Two boundaries are at depths of 4.0 cm and 9.0 cm. Taking the speed of ultrasound in tissue as 1540 m/s, the time interval between the echoes from the two boundaries is

  • (A) 32 μs
  • (B) 130 μs
  • (C) 65 μs
  • (D) 26 μs
Show Answer

Answer: C – The depths differ by 5.0 cm, so the extra path is 2 × 0.050 = 0.100 m, and Δt = 0.100 / 1540 = 6.5 × 10−5 s = 65 μs.

Q20. Which change would give the finest detail (best resolution) in an ultrasound image?

  • (A) Use a lower frequency so the beam penetrates further
  • (B) Use a higher frequency, so the wavelength is shorter
  • (C) Use longer pulses
  • (D) Use a coupling medium with a much lower sound speed
Show Answer

Answer: B – Detail is limited by the wavelength: at 10 MHz the wavelength in tissue is about 0.15 mm, compared with about 0.77 mm at 2 MHz, so the higher frequency resolves smaller structures (but is attenuated more).


Acoustic Impedance & Reflection at Boundaries

Q21. The specific acoustic impedance Z of a material is defined by

  • (A) Z = c / ρ
  • (B) Z = ρ / c
  • (C) Z = ρ c2
  • (D) Z = ρ c, the product of density and the speed of ultrasound in the material
Show Answer

Answer: D – Z = ρ c. It is a property of the material alone and decides how much ultrasound is reflected when the wave meets a boundary with another material.

Q22. A material has density 1000 kg/m3 and the speed of ultrasound in it is 1500 m/s. Its specific acoustic impedance is

  • (A) 1.5 × 103 kg m−2 s−1
  • (B) 1.5 × 106 kg m−2 s−1
  • (C) 6.7 × 10−4 kg m−2 s−1
  • (D) 2.5 × 106 kg m−2 s−1
Show Answer

Answer: B – Z = ρ c = 1000 × 1500 = 1.5 × 106 kg m−2 s−1, a typical value for soft tissue and water-based materials.

Q23. The fraction of the ultrasound intensity reflected at a boundary between materials of impedances Z1 and Z2 (at normal incidence) is

  • (A) ((Z2 − Z1) / (Z2 + Z1))2
  • (B) (Z2 − Z1) / (Z2 + Z1)
  • (C) Z1 / Z2
  • (D) ((Z2 + Z1) / (Z2 − Z1))2
Show Answer

Answer: A – The unsquared ratio is the amplitude reflection coefficient; squaring it gives the fraction of the intensity (energy) reflected. If Z1 = Z2 the fraction is zero – no echo at all.

Q24. Soft tissue has Z ≈ 1.6 × 106 kg m−2 s−1 and air has Z ≈ 4.0 × 102 kg m−2 s−1. The fraction of intensity reflected at a tissue-air boundary is about

  • (A) 50%
  • (B) 0.1%
  • (C) 99.9%
  • (D) 25%
Show Answer

Answer: C – ((1.6 × 106 − 400) / (1.6 × 106 + 400))2 ≈ (0.9995)2 ≈ 0.999, so about 99.9% is reflected and almost nothing enters or leaves through an air gap. This is why air must be excluded from the beam path.

Q25. Coupling gel is spread on the skin before a scan because

  • (A) it cools the skin during the scan
  • (B) it increases the frequency of the ultrasound
  • (C) it makes the pulses travel faster
  • (D) it fills the air gap between transducer and skin, and its impedance is close to that of tissue, so very little is reflected at the skin surface
Show Answer

Answer: D – Without gel, the thin air layer would reflect about 99.9% of the beam (see Q24). The gel replaces the air with a material of tissue-like impedance, so the beam passes into the body.

Q26. At a soft tissue-bone boundary, Ztissue = 1.6 × 106 and Zbone = 7.6 × 106 kg m−2 s−1. The fraction of the intensity reflected is

  • (A) 4.3%
  • (B) 65%
  • (C) 43%
  • (D) 90%
Show Answer

Answer: C – ((7.6 − 1.6) / (7.6 + 1.6))2 = (6.0 / 9.2)2 = (0.652)2 ≈ 0.43, so about 43% is reflected – bone gives a strong echo and shadows what lies behind it.

Q27. Fat (Z = 1.4 × 106) and muscle (Z = 1.7 × 106 kg m−2 s−1) form a boundary inside the body. Structures deeper than this boundary can still be imaged because

  • (A) only about 1% of the intensity is reflected at the boundary, so most of the beam is transmitted through it
  • (B) about 99% of the intensity is reflected at the boundary
  • (C) the boundary produces no echo at all, so nothing is detected from it and it is invisible in every way
  • (D) fat has zero acoustic impedance
Show Answer

Answer: A – ((1.7 − 1.4) / (1.7 + 1.4))2 = (0.3 / 3.1)2 ≈ 0.0094, roughly 1%. The small echo still marks the boundary, while about 99% of the beam carries on to deeper structures.

Q28. Increasing the frequency of the ultrasound used in a scan

  • (A) improves the detail AND increases the depth that can be imaged
  • (B) improves the detail but the beam is attenuated more, so the maximum imaging depth falls
  • (C) has no effect on the detail in the image
  • (D) gives coarser detail because the wavelength is longer
Show Answer

Answer: B – Higher frequency means shorter wavelength and finer resolution, but attenuation in tissue rises with frequency, so high-frequency probes are used for shallow structures and lower frequencies for deep ones.

Q29. Ultrasound passes from a medium of impedance 2.0 × 106 into one of impedance 6.0 × 106 kg m−2 s−1. The fraction of the intensity reflected at the boundary is

  • (A) 6.25%
  • (B) 50%
  • (C) 75%
  • (D) 25%
Show Answer

Answer: D – ((6.0 − 2.0) / (6.0 + 2.0))2 = (4.0 / 8.0)2 = (0.50)2 = 0.25, so 25% is reflected and 75% is transmitted.

Q30. The unit of specific acoustic impedance is

  • (A) kg m s−1
  • (B) kg m−1 s−2
  • (C) kg m−2 s−1
  • (D) kg m2 s−1
Show Answer

Answer: C – From Z = ρ c: (kg m−3) × (m s−1) = kg m−2 s−1. This unit is also called the rayl.

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