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A2 · Practice questions · Production and use of ultrasound

Time the echo, read the depth.

Six original Cambridge-style questions on the piezoelectric transducer, specific acoustic impedance Z = ρc, the intensity reflection coefficient at a boundary, coupling gel, and pulse-echo depth measurement with d = ½ct.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

The mismatch makes the echo.

01
Analysis
[3 marks]

Explain how a piezoelectric transducer is used both to generate a pulse of ultrasound and to detect the echoes, and explain why the ultrasound is sent out as short pulses rather than continuously.

  • An alternating p.d. applied across the piezoelectric crystal makes it expand and contract at the same frequency, so it vibrates and emits ultrasound. ✓
  • When a returning echo strikes the crystal it is deformed by the pressure variations, and this produces an alternating e.m.f. across it, which is amplified and recorded. ✓
  • The same crystal cannot transmit and receive at once, and the depth is found from the delay between sending the pulse and receiving the echo, so each pulse must end before its echoes return. ✓
02
Calculation
[3 marks]

Bone has a density of 1900 kg m⁻³ and the speed of ultrasound in it is 4000 m s⁻¹. The acoustic impedance of soft tissue is 1.63 × 10⁶ kg m⁻² s⁻¹. Calculate the acoustic impedance of bone, the fraction of the incident intensity reflected at a soft tissue-bone boundary, and the fraction transmitted into the bone.

  • Z = ρc = 1900 × 4000 = 7.6 × 10⁶ kg m⁻² s⁻¹. ✓
  • IR/I₀ = (7.6 − 1.63)² / (7.6 + 1.63)² = 35.6 / 85.2 = 0.42. ✓
  • Fraction transmitted = 1 − 0.42 = 0.58. ✓
03
Analysis
[2 marks]

State what is meant by the specific acoustic impedance of a medium. The acoustic impedances of liver and kidney are 1.65 × 10⁶ and 1.62 × 10⁶ kg m⁻² s⁻¹. Explain why the boundary between them gives only a very faint echo.

  • The specific acoustic impedance is the product of the density of the medium and the speed of sound in it, Z = ρc. ✓
  • The reflected fraction depends on the difference (Z₁ − Z₂)², which here is tiny compared with (Z₁ + Z₂)²: the coefficient is (0.03)² / (3.27)², less than 0.01%, so almost all the intensity is transmitted. ✓
04
Calculation
[3 marks]

In an A-scan, echoes from the front and back surfaces of an organ are received 52 µs and 96 µs after the pulse is transmitted. The speed of ultrasound in the tissue is 1540 m s⁻¹. Calculate the depth of the front surface below the probe, and the thickness of the organ.

  • The pulse travels to the surface and back, so d = ½ct = ½ × 1540 × 52 × 10⁻⁶. ✓
  • d = 0.040 m, so the front surface is 4.0 cm below the probe. ✓
  • Back surface: ½ × 1540 × 96 × 10⁻⁶ = 0.074 m, so the thickness is 7.4 − 4.0 = 3.4 cm (or ½ × 1540 × 44 × 10⁻⁶ = 3.4 cm). ✓
05
Calculation
[2 marks]

The acoustic impedance of air is 430 kg m⁻² s⁻¹ and that of skin is 1.63 × 10⁶ kg m⁻² s⁻¹. Calculate the fraction of the incident ultrasound intensity that would be reflected by a thin layer of air between the probe and the skin, and hence explain the purpose of the coupling gel.

  • IR/I₀ = (1.63 × 10⁶ − 430)² / (1.63 × 10⁶ + 430)² = 0.999, so 99.9% is reflected and almost nothing enters the body. ✓
  • The gel has an impedance close to that of skin and fills the gap, excluding the air, so most of the pulse is transmitted into the body and the echoes can return. ✓
06
Calculation
[4 marks]

A 2.5 MHz ultrasound probe is placed against the front of an eye. The speed of ultrasound in the eye is 1530 m s⁻¹. The echo from the retina at the back of the eye arrives 32 µs after the pulse is sent.

(a) Calculate the distance from the front of the eye to the retina. (b) Calculate the wavelength of the ultrasound in the eye. (c) State why a higher frequency would give finer detail in the image.

  • (a) The pulse travels the distance twice, so d = ½ct = ½ × 1530 × 32 × 10⁻⁶. ✓
  • d = 0.024 m, about 2.4 cm. ✓
  • (b) λ = c / f = 1530 / 2.5 × 10⁶ = 6.1 × 10⁻⁴ m (0.61 mm). ✓
  • (c) Detail smaller than about one wavelength cannot be resolved, so a higher frequency, with a shorter wavelength, resolves finer structure (at the cost of greater attenuation). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.