A2 Daily A Level Physics question

2025-09-01 OCR A Medical imaging — Ultrasound A-scan, Acoustic impedance and boundary reflections, Coupling gel and impedance matching, Attenuation via half-value thickness, Time-of-flight and echo positioning 6.5.3(c) 6.5.3(e) 6.5.3(f)

A pulsed ultrasound A-scan is used to locate a small bone 3.0 cm beneath the skin. Trial 1 is performed with a thin air gap between the transducer and skin; at an air–skin boundary about 99% of incident intensity is reflected each time the wave hits it. Trial 2 is performed with coupling gel so reflections at the front surface are negligible. In both trials the same pulse power is used, and any attenuation within the tissue and reflection at the bone are unchanged. Which statement must be true when switching from Trial 1 (no gel) to Trial 2 (with gel)?

A The near-surface echo becomes larger because the gel introduces an extra boundary, while the bone echo is unchanged; the A-scan trace shape otherwise stays the same.
B The bone echo increases by only about a factor of 100 because the gel affects just the outgoing path, and the echo times shift significantly.
C The display would effectively become a B-scan so depth appears as a bright area; the bone echo is limited by a linear fall of intensity with depth.
D The early surface echo becomes much smaller, and the later bone echo becomes roughly 10,000 times larger; their arrival times stay essentially the same. (correct)

Answer

The correct answer is D.

Correct: D — The early surface echo becomes much smaller, and the later bone echo becomes roughly 10,000 times larger; their arrival times stay essentially the same. With gel, the two front-surface transmissions are ~100% instead of 1% each, so the bone echo gains a factor ≈1/(0.01×0.01)=10,000, while the depth (hence echo time) is unchanged to a first approximation. A is wrong because the gel reduces the front-surface reflection rather than adding one, and the bone echo cannot remain unchanged when coupling improves by orders of magnitude. B is wrong because it counts only one pass (giving ×100) and ignores that the wave must cross the front surface twice; the echo timings are set by path length, not surface coupling. C is wrong because an A-scan is a 1D time–amplitude trace (not a 2D brightness map), and attenuation with depth is not linear. D is the only option that applies two-way surface coupling correctly and recognises that echo timing depends on path length, which is essentially unchanged.

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