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Date	May 2018	Marks available	4	Reference code	18M.3.HL.TZ1.14
Level	Higher level	Paper	Paper 3	Time zone	1
Command term	Explain	Question number	14	Adapted from	N/A

Question

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 = $\frac{{{{({{\text{Z}}_1} - {{\text{Z}}_2})}^2}}}{{{{({{\text{Z}}_1} + {{\text{Z}}_2})}^2}}}$ .

Outline how ultrasound is generated for medical imaging.

[2]

Describe one advantage and one disadvantage of using high frequencies ultrasound over low frequencies ultra sound for medical imaging.

[2]

Suggest one reason why doctors use ultrasound rather than X-rays to monitor the development of a fetus.

[1]

Calculate the density of skin.

[1]

d.i.

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

[4]

d.ii.

Markscheme

crystal vibration /piezo-electric effect

caused by an alternating potential difference is applied across a crystal

[2 marks]

ADVANTAGES

the wavelength must be less than the size of the object being imaged to avoid diffraction effects

the frequency must be high to ensure several full wavelengths in the pulse

DISADVANTAGES

the depth of the organ being imaged must be considered (no more than 200 wavelengths)

attenuation increases at higher frequencies

[1] for advantages, [1] for disadvantages.

[2 marks]

X-rays are an ionizing radiation and so might cause harm to the developing fetus.

there are no known harmful effects when using ultrasound

Ignore “moving images by ultrasound”.

[1 mark]

ρ = $\frac{{1.99 \times {{10}^6}}}{{1.73 \times {{10}^3}}}$ = 1.15 × 10³ «kgm^–3»

[1 mark]

d.i.

F = $\frac{{{{(1.99 \times {{10}^6} - 4.3 \times {{10}^2})}^2}}}{{{{(1.99 \times {{10}^6} + 4.3 \times {{10}^2})}^2}}}$ = 1.0

F = $\frac{{{{(1.48 \times {{10}^6} - 1.99 \times {{10}^6})}^2}}}{{{{(1.48 \times {{10}^6} + 1.99 \times {{10}^6})}^2}}}$ = 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]

d.ii.

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

Syllabus sections

Option C: Imaging » Option C: Imaging (Additional higher level option topics) » C.4 – Medical imaging (HL only)

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Option C: Imaging » Option C: Imaging (Additional higher level option topics)

Option C: Imaging

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