Physics (AS & A Level) · Feb / March 2016 · Paper 42 · Question 6
9702
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The question as text
A parallel beam of ultrasound is incident normally on the surface of a layer of fat of thickness 1.1 cm, as shown in Fig. 6.1. I1 I2 I4 I3 1.1 cm muscle fat ultrasound beam Fig. 6.1 For the ultrasound, I1 is the intensity just after entering the surface of the fat layer, I2 is the intensity incident on the fat-muscle boundary, I3 is the intensity reflected from the fat-muscle boundary, I4 is the intensity received back at the surface of the fat layer. Some data for the fat are given in Fig. 6.2. specific acoustic impedance Z 1.4 × 106 kg m–2 s–1 density ρ 940 kg m–3 absorption (attenuation) coefficient μ 48 m–1 Fig. 6.2
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6(a)[3]
speed = Z / ρ = 1.4 × 106 / 940 (=1490) time = (1.1 × 10–2 × 2) / 1490 = 1.5 × 10–5 s [3]
(time of 7.4 × 10–6 s is one way only and scores 2 / 3 marks) (use of speed of light is wrong physics and scores 0 / 3 marks)
6(b)[2]
I = I0 exp (–µx) or I2 = I1 exp(–µx) ratio = exp (– 48 × 1.1 × 10–2) = 0.59 [2]
6(c)[2]
0.33 / 100 = 0.59 × (I3 / I2) × 0.59 ratio = 9.5 × 10–3 or 0.33 / 100 = exp (– 48 × 2.2 × 10–2) × (I3 / I2) (C1) ratio = 9.5 × 10–3 (A1) [2]
6(d)[1]
ratio I3 / I2 increases [1]
(accept: “there is an increase in the proportion of the intensity that is reflected”)
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