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Question 18M.3.HL.TZ1.14d.ii

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Date May 2018 Marks available [Maximum mark: 4] Reference code 18M.3.HL.TZ1.14d.ii
Level HL Paper 3 Time zone TZ1
Command term Explain Question number d.ii Adapted from N/A
d.ii.
[Maximum mark: 4]
18M.3.HL.TZ1.14d.ii

The table shows the speed of ultrasound and the acoustic impedance for different media.

M18/4/PHYSI/HP3/ENG/TZ1/14.d

The fraction F of the intensity of an ultrasound wave reflected at the boundary between two media having acoustic impedances Z1 and Z2 is given by F =  ( Z 1 − Z 2 ) 2 ( Z 1 + Z 2 ) 2 .

(d.ii)

Explain, with appropriate calculations, why a gel is used between the transducer and the skin.

[4]

Markscheme

F =  ( 1.99 × 10 6 − 4.3 × 10 2 ) 2 ( 1.99 × 10 6 + 4.3 × 10 2 ) 2 = 1.0

F =  ( 1.48 × 10 6 − 1.99 × 10 6 ) 2 ( 1.48 × 10 6 + 1.99 × 10 6 ) 2 = 0.02

almost 100% of the ultrasound will be reflected from the air-skin surface OR almost none is transmitted

whereas only 2% will be reflected from the gel-skin surface and so a much greater proportion is transmitted

 

Need to explain that more is transmitted through gel-skin surface for MP4.

[4 marks]

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