Specification: AQA A-level Physics (7408) 3.10.4.1 Medical physics option | Paper: 3 Section B | Data and formulae booklet: Z = ρc and Ir/Ii = ((Z2 − Z1)/(Z2 + Z1))² are given in the exam; depth = ct ÷ 2 for an A-scan is not, so learn it
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Acoustic impedance: Z = density × speed of sound in the material, Z = ρc, measured in kg m⁻² s⁻¹.
Reflection at a boundary: when ultrasound meets a boundary between materials of different acoustic impedance, the fraction of the intensity reflected is Ir/Ii = ((Z2 − Z1)/(Z2 + Z1))². The bigger the difference in Z, the greater the fraction reflected.
Coupling gel: a gel with an acoustic impedance close to that of skin is spread between the transducer and the skin to remove the air gap. Without it almost all the ultrasound would be reflected at the air–skin boundary.
The equation
acoustic impedance = density × speed of sound Z = ρc and fraction reflected Ir/Ii = ((Z2 − Z1)/(Z2 + Z1))²
| Quantity | Symbol | Unit |
|---|---|---|
| acoustic impedance | Z | kg m⁻² s⁻¹ |
| density of the material | ρ | kg m⁻³ |
| speed of sound in the material | c | m s⁻¹ |
| reflected and incident intensity | Ir, Ii | W m⁻² |
Rearranged: ρ = Z ÷ c | c = Z ÷ ρ | fraction transmitted = 1 − Ir/Ii (ignoring absorption) | Convert g cm⁻³ to kg m⁻³ (× 1000) first. The squared bracket is always between 0 and 1, so it does not matter which material you call Z1.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) Soft tissue has a density of 1060 kg m⁻³ and the speed of ultrasound in it is 1540 m s⁻¹. Calculate its acoustic impedance, to 3 significant figures.
Show solution
F: Z = ρc
I: Z = 1060 × 1540
F: Z is already the subject
A: 1 632 400 kg m⁻² s⁻¹ = 1.63 × 106 kg m⁻² s⁻¹ (3 s.f.)
Q2 (Core) Ultrasound travels at 4000 m s⁻¹ in bone of density 1900 kg m⁻³. Calculate the acoustic impedance of the bone.
Show solution
F: Z = ρc
I: Z = 1900 × 4000
F: Z is already the subject
A: 7.6 × 106 kg m⁻² s⁻¹
Q3 (Core) Fat has an acoustic impedance of 1.38 × 106 kg m⁻² s⁻¹ and a density of 0.95 g cm⁻³. Calculate the speed of ultrasound in fat, to 3 significant figures.
Show solution
Convert first: 0.95 g cm⁻³ = 950 kg m⁻³
F: Z = ρc
I: 1.38 × 106 = 950 × c
F: c = 1.38 × 106 ÷ 950 = 1452.6… m s⁻¹
A: 1450 m s⁻¹ (3 s.f.)
Watch out: using 0.95 as the density gives a speed of over a million m s⁻¹. Always convert g cm⁻³ to kg m⁻³.
Q4 (Core) Ultrasound passes from soft tissue (Z = 1.63 × 106 kg m⁻² s⁻¹) into bone (Z = 7.80 × 106 kg m⁻² s⁻¹). Calculate the percentage of the incident intensity that is reflected, to 2 significant figures.
Show solution
F: Ir/Ii = ((Z2 − Z1)/(Z2 + Z1))²
I: Ir/Ii = ((7.80 − 1.63)/(7.80 + 1.63))² (the 106 cancels)
F: Ir/Ii = (6.17 ÷ 9.43)² = 0.428…
A: 43% (2 s.f.)
Watch out: forgetting to square the bracket gives 65%.
Q5 (Exam) The acoustic impedance of air is 430 kg m⁻² s⁻¹, of skin is 1.7 × 106 kg m⁻² s⁻¹ and of a coupling gel is 1.5 × 106 kg m⁻² s⁻¹. (a) Calculate the percentage of ultrasound intensity reflected at an air–skin boundary, to 3 significant figures. (b) Calculate the percentage reflected at a gel–skin boundary, to 2 significant figures. (c) Explain why gel is used.
Show solution
F: Ir/Ii = ((Z2 − Z1)/(Z2 + Z1))²
Step 1 – (a) air to skin
I: ((1.7 × 106 − 430)/(1.7 × 106 + 430))² = 0.9989…
A: 99.9% (3 s.f.)
Step 2 – (b) gel to skin
I: ((1.7 − 1.5)/(1.7 + 1.5))² = (0.2 ÷ 3.2)² = 0.003906…
A: 0.39% (2 s.f.)
(c) The gel displaces the air between the transducer and the skin. Its impedance is close to that of skin, so almost all the ultrasound is transmitted into the body instead of being reflected at the surface.
Q6 (Exam) A beam of ultrasound of intensity 2.0 × 102 W m⁻² meets a boundary between muscle (Z = 1.70 × 106 kg m⁻² s⁻¹) and bone (Z = 7.80 × 106 kg m⁻² s⁻¹). Calculate the intensity transmitted into the bone, to 3 significant figures. Ignore absorption.
Show solution
Step 1 – fraction reflected
F: Ir/Ii = ((Z2 − Z1)/(Z2 + Z1))²
I: Ir/Ii = ((7.80 − 1.70)/(7.80 + 1.70))² = (6.10 ÷ 9.50)² = 0.4122…
F: Ir = 0.4122… × 200 = 82.45… W m⁻²
Step 2 – transmitted intensity
It = Ii − Ir = 200 − 82.45… = 117.5… W m⁻²
A: 118 W m⁻² (3 s.f.)
Q7 (Exam) Blood has a density of 1060 kg m⁻³ and ultrasound travels through it at 1570 m s⁻¹. Fat has a density of 925 kg m⁻³ and ultrasound travels through it at 1450 m s⁻¹. Calculate the percentage of the intensity reflected at a blood–fat boundary, to 2 significant figures.
Show solution
Step 1 – both impedances
F: Z = ρc
I: Zblood = 1060 × 1570 = 1 664 200 kg m⁻² s⁻¹; Zfat = 925 × 1450 = 1 341 250 kg m⁻² s⁻¹
Step 2 – fraction reflected
Ir/Ii = ((1 664 200 − 1 341 250)/(1 664 200 + 1 341 250))² = (322 950 ÷ 3 005 450)² = 0.01154…
A: 1.2% (2 s.f.)
Boundaries between soft tissues reflect only a small fraction, so ultrasound can still image the structures behind them.
Q8 (Exam) At a boundary between soft tissue (Z = 1.6 × 106 kg m⁻² s⁻¹) and a second material of higher acoustic impedance, 36% of the incident intensity is reflected. Calculate the acoustic impedance of the second material.
Show solution
F: Ir/Ii = ((Z2 − Z1)/(Z2 + Z1))²
I: 0.36 = ((Z2 − 1.6 × 106)/(Z2 + 1.6 × 106))²
F: square root: (Z2 − Z1)/(Z2 + Z1) = 0.60 (positive, as Z2 > Z1), so Z2 − Z1 = 0.60Z2 + 0.60Z1, giving 0.40Z2 = 1.60Z1 and Z2 = 4Z1
A: Z2 = 4 × 1.6 × 106 = 6.4 × 106 kg m⁻² s⁻¹ (similar to bone)
Watch out: forgetting to square root 0.36 before rearranging.
Q9 (Stretch) In an A-scan of an eye, an echo from the front of the lens arrives 4.6 μs after the pulse is sent, and an echo from the back of the lens arrives 9.8 μs after the pulse is sent. Assume the speed of ultrasound is 1540 m s⁻¹ throughout. Calculate (a) the depth of the front of the lens and (b) the thickness of the lens, each in mm to 2 significant figures.
Show solution
Convert first: 4.6 μs = 4.6 × 10−6 s; 9.8 μs = 9.8 × 10−6 s
Step 1 – (a) depth of the front surface
the pulse travels there and back, so depth = ct ÷ 2
I: depth = 1540 × 4.6 × 10−6 ÷ 2 = 3.542 × 10−3 m
A: 3.5 mm (2 s.f.)
Step 2 – (b) thickness
time difference = 9.8 − 4.6 = 5.2 μs; thickness = 1540 × 5.2 × 10−6 ÷ 2 = 4.004 × 10−3 m
A: 4.0 mm (2 s.f.)
Watch out: forgetting to halve the time. The echo time is for the journey there and back.
Q10 (Stretch) A transducer sends ultrasound through coupling gel (Z = 1.5 × 106 kg m⁻² s⁻¹) into soft tissue (Z = 1.6 × 106 kg m⁻² s⁻¹). It then reflects from bone (Z = 7.0 × 106 kg m⁻² s⁻¹) and returns through the tissue and gel to the transducer. Ignore absorption. (a) Calculate the fraction of the original intensity that returns to the transducer, to 2 significant figures. (b) Repeat with an air gap (Z = 430 kg m⁻² s⁻¹) in place of the gel, to 2 significant figures.
Show solution
Step 1 – (a) gel–tissue boundary
reflected fraction = ((1.6 − 1.5)/(1.6 + 1.5))² = (0.1 ÷ 3.1)² = 0.00104…, so fraction transmitted = 0.9989…
Step 2 – tissue–bone boundary
reflected fraction = ((7.0 − 1.6)/(7.0 + 1.6))² = (5.4 ÷ 8.6)² = 0.3942…
Step 3 – combine
the wave crosses the gel–tissue boundary twice (in and out), so fraction returned = 0.9989…² × 0.3942… = 0.3934…
A: 0.39 (39%, 2 s.f.)
Step 4 – (b) with an air gap
reflected fraction at air–tissue = ((1.6 × 106 − 430)/(1.6 × 106 + 430))² = 0.99892…, so fraction transmitted = 1.07… × 10−3
fraction returned = (1.07… × 10−3)² × 0.3942… = 4.55… × 10−7
A: 4.6 × 10−7 (2 s.f.), almost a million times smaller, which is why gel is essential
Watch out: multiplying by the transmitted fraction only once. The echo has to cross the surface boundary on the way back too.
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