Back to Daily Question archive

A2 Daily A Level Physics question

2025-08-24 OCR A A Level Physics 6.5.1(c) 6.5.1(d)

A chest radiograph is taken at 80 kVp with exposure settings that give good detector signal. For a second patient of the same composition but 3.0 cm thicker along the beam path, the radiographer wants the same detector exposure. At 80 kVp the half-value thickness (HVT) of soft tissue is 3.0 cm. Which statement must be true to keep image appearance comparable?

  1. A The mAs must be doubled; using a higher kVp instead to achieve the same detector signal would reduce soft‑tissue contrast. (correct)
  2. B The mAs must be increased by 3.0%; using a higher kVp instead would improve soft‑tissue contrast.
  3. C The mAs must be halved because more tissue means more scatter reaching the detector; using a higher kVp would leave contrast unchanged.
  4. D The mAs can be left unchanged if the modality is switched to B‑scan ultrasound, which displays one‑dimensional amplitude spikes giving equivalent contrast.

Answer

The correct answer is A.

Correct: A — The mAs must be doubled; using a higher kVp instead to achieve the same detector signal would reduce soft‑tissue contrast. Adding one HVT halves transmitted intensity, so doubling mAs restores the detector exposure; raising kVp to achieve the same exposure lowers differential attenuation and thus reduces soft‑tissue contrast. A This is correct as explained. B Incorrect: 3.0 cm is one HVT, not a 3% change; attenuation is not linear with thickness, and higher kVp tends to reduce (not improve) soft‑tissue contrast. C Incorrect: thicker patients transmit less primary X‑ray intensity; halving mAs would further reduce detector signal, and kVp does affect contrast. D Incorrect: changing to ultrasound is a different modality; B‑scan displays a 2D brightness map (dots), not one‑dimensional spikes (that is A‑scan), and it does not provide equivalent X‑ray contrast.

Try today's guest Daily MCQ A Level Physics MCQ practice Problem-solving previews