Module 6 · Particles and Medical Physics

Ultrasound & Acoustic Impedance

Specification: OCR A H556 | Section: 6.5.2 Medical Physics | Teaching time: ~3 hours

By the end of this topic you should be able to…

Explain what ultrasound is and why it is used in medical imaging
Describe the piezoelectric effect and how transducers generate and detect ultrasound
Calculate acoustic impedance Z = ρc for different materials
Calculate the intensity reflection coefficient at a boundary between two media
Explain why coupling gel is needed and perform impedance matching calculations
Compare A-scan and B-scan techniques and evaluate ultrasound against other imaging methods

Part 1

What is Ultrasound?

Ultrasound refers to sound waves with frequencies above the upper limit of human hearing (above 20 kHz). In medical applications, frequencies in the range 1–15 MHz are used — much higher than in industrial or cleaning applications.

Ultrasound is a longitudinal mechanical wave: the particles of the medium oscillate parallel to the direction of energy transfer, forming compressions and rarefactions. Unlike X-rays, ultrasound is non-ionising, making it safe for imaging foetuses and soft tissues.

⚡ Key Point

Medical ultrasound uses pulsed waves — a short pulse is sent, then the transducer waits to receive echoes before sending the next pulse. The time delay between sending and receiving tells you the depth of the reflecting surface.

Ultrasound scanning relies on two key facts:

Ultrasound is reflected at boundaries between materials of different densities (e.g. tissue–organ, tissue–bone)
Ultrasound is pulsed — the transducer sends a pulse, then listens for returning echoes

Because ultrasound cannot pass through air (nearly all of it reflects at an air–tissue boundary), a coupling gel must be used between the transducer and the skin. More on this later.

Part 2

The Piezoelectric Effect

The heart of any ultrasound machine is the piezoelectric transducer. It contains a piezoelectric crystal (often made from lead zirconate titanate, PZT) that can convert electrical signals into mechanical vibrations and vice versa.

How it works

Transmitting (generating ultrasound): When a high-frequency alternating potential difference is applied across the piezoelectric crystal, the crystal expands and contracts at the same frequency as the AC signal. These rapid mechanical deformations produce longitudinal ultrasound waves.

Receiving (detecting echoes): When an ultrasound echo reaches the crystal, the pressure of the compression causes the crystal to deform. This deformation generates a p.d. across the crystal (the reverse piezoelectric effect). This small voltage is amplified and processed to build the image.

⚡ Key Point

The same crystal acts as both transmitter and receiver. The transducer cannot transmit and receive at the same time — it pulses, then listens.

Part 3

Acoustic Impedance

Acoustic impedance is a property of a material that describes how much it resists the passage of sound waves. It depends on two things: how dense the material is, and how fast sound travels through it.

Acoustic Impedance Z = ρc

Where:

Z = acoustic impedance, measured in kg m⁻² s⁻¹ (also called the Rayl)
ρ = density of the medium (kg m⁻³)
c = speed of sound in the medium (m s⁻¹)

A material with high density and a high speed of sound will have a high acoustic impedance. Bone, for example, has a very high Z because it is dense and sound travels quickly through it. Air has an extremely low Z because it has very low density and sound travels relatively slowly.

Typical Values of Acoustic Impedance

Material	Density ρ (kg m⁻³)	Speed of sound c (m s⁻¹)	Acoustic impedance Z (kg m⁻² s⁻¹)
Air	1.3	330	430
Water	1000	1480	1.48 × 10⁶
Fat	920	1450	1.38 × 10⁶
Blood	1060	1570	1.66 × 10⁶
Muscle	1050	1580	1.71 × 10⁶
Soft tissue (average)	1060	1540	1.63 × 10⁶
Lung	400	650	0.26 × 10⁶
Bone	1900	4000	7.6 × 10⁶

⚡ Key Point

Notice the enormous difference between air (430) and soft tissue (~1.6 × 10⁶) — they differ by a factor of nearly 4000. This is why virtually no ultrasound would pass from air into the body without coupling gel.

Part 4

Worked Examples: Calculating Acoustic Impedance

Worked Example 1

Calculate the acoustic impedance of muscle given its density is 1050 kg m⁻³ and the speed of sound in muscle is 1580 m s⁻¹.

Z = ρc = 1050 × 1580 = 1,659,000

Z = 1.66 × 10⁶ kg m⁻² s⁻¹ (to 3 s.f.)

Worked Example 2

Calculate the acoustic impedance of bone given its density is 1900 kg m⁻³ and the speed of sound in bone is 4000 m s⁻¹.

Z = ρc = 1900 × 4000 = 7,600,000

Z = 7.6 × 10⁶ kg m⁻² s⁻¹ (to 2 s.f.)

Worked Example 3

Calculate the acoustic impedance of air given its density is 1.3 kg m⁻³ and the speed of sound in air is 330 m s⁻¹.

Z = ρc = 1.3 × 330 = 429

Z = 430 kg m⁻² s⁻¹ (to 2 s.f.)

Worked Example 4

Blood has an acoustic impedance of 1.66 × 10⁶ kg m⁻² s⁻¹. If the density of blood is 1060 kg m⁻³, calculate the speed of sound in blood.

c = Z / ρ = (1.66 × 10⁶) / 1060 = 1566

c = 1570 m s⁻¹ (to 3 s.f.)

Worked Example 5

The speed of sound in fat is 1450 m s⁻¹. If the acoustic impedance of fat is 1.38 × 10⁶ kg m⁻² s⁻¹, calculate the density of fat.

ρ = Z / c = (1.38 × 10⁶) / 1450 = 951.7

ρ = 952 kg m⁻³ (to 3 s.f.)

Part 5

Intensity Reflection Coefficient

When ultrasound reaches a boundary between two materials with different acoustic impedances, some energy is reflected and some is transmitted. The fraction reflected depends on the difference between the two impedances.

Intensity Reflection Coefficient Iᵣ / I₀ = ((Z₂ − Z₁) / (Z₂ + Z₁))²

Where:

Iᵣ = reflected intensity (W m⁻²)
I₀ = incident intensity (W m⁻²)
Z₁ = acoustic impedance of the first medium
Z₂ = acoustic impedance of the second medium

The result is always between 0 (no reflection, perfect transmission) and 1 (total reflection, nothing transmitted). The larger the difference between Z₁ and Z₂, the greater the fraction reflected.

⚡ Exam Tip

The order of Z₁ and Z₂ doesn't matter for the ratio — the result is the same whether ultrasound goes from medium 1 → 2 or 2 → 1. The fraction (Z₂ − Z₁)/(Z₂ + Z₁) is squared, so the sign cancels out.

Part 6

Worked Examples: Reflection Coefficient

Worked Example 6 — Soft tissue to bone

Calculate the fraction of ultrasound intensity reflected at a boundary between soft tissue (Z = 1.63 × 10⁶ kg m⁻² s⁻¹) and bone (Z = 7.6 × 10⁶ kg m⁻² s⁻¹).

Iᵣ/I₀ = ((Z₂ − Z₁)/(Z₂ + Z₁))²

= ((7.6 × 10⁶ − 1.63 × 10⁶)/(7.6 × 10⁶ + 1.63 × 10⁶))²

= (5.97/9.23)²

= (0.647)²

Iᵣ/I₀ = 0.419 — so about 42% of the intensity is reflected

Worked Example 7 — Muscle to fat

Calculate the fraction of ultrasound intensity reflected at a boundary between muscle (Z = 1.71 × 10⁶ kg m⁻² s⁻¹) and fat (Z = 1.38 × 10⁶ kg m⁻² s⁻¹).

Iᵣ/I₀ = ((1.38 × 10⁶ − 1.71 × 10⁶)/(1.38 × 10⁶ + 1.71 × 10⁶))²

= (−0.33/3.09)²

= (−0.107)²

Iᵣ/I₀ = 0.0114 — so only about 1.1% is reflected (98.9% transmitted)

Worked Example 8 — Blood to muscle

Calculate the fraction of ultrasound intensity reflected at a boundary between blood (Z = 1.66 × 10⁶ kg m⁻² s⁻¹) and muscle (Z = 1.71 × 10⁶ kg m⁻² s⁻¹).

Iᵣ/I₀ = ((1.71 − 1.66)/(1.71 + 1.66))² (all in units of 10⁶)

= (0.05/3.37)²

= (0.0148)²

Iᵣ/I₀ = 2.20 × 10⁻⁴ — so only 0.022% is reflected (nearly all transmitted)

⚡ Key Insight

When two materials have similar acoustic impedances (e.g. blood and muscle, or muscle and fat), very little ultrasound is reflected. This means most of the energy continues deeper into the body, allowing imaging of structures beyond the first boundary.

Part 7

Coupling Gel & Impedance Matching

Now let's answer the critical question: why is gel essential?

Worked Example 9 — Why you MUST use gel (air → skin boundary)

Calculate the fraction of ultrasound reflected at an air–skin boundary. Air has Z = 430 kg m⁻² s⁻¹ and soft tissue (skin) has Z = 1.63 × 10⁶ kg m⁻² s⁻¹.

Iᵣ/I₀ = ((Z₂ − Z₁)/(Z₂ + Z₁))²

= ((1.63 × 10⁶ − 430)/(1.63 × 10⁶ + 430))²

Since 430 is negligible compared to 1.63 × 10⁶:

≈ (1.63 × 10⁶ / 1.63 × 10⁶)² = (1)² = 1

Iᵣ/I₀ ≈ 0.9999 — virtually 100% of the ultrasound is reflected!

Worked Example 10 — With coupling gel (gel → skin boundary)

Ultrasound gel has Z ≈ 1.5 × 10⁶ kg m⁻² s⁻¹ (similar to soft tissue). Calculate the fraction reflected at a gel–skin boundary.

Iᵣ/I₀ = ((1.63 × 10⁶ − 1.5 × 10⁶)/(1.63 × 10⁶ + 1.5 × 10⁶))²

= (0.13/3.13)²

= (0.0415)²

Iᵣ/I₀ = 0.00173 — only 0.17% reflected (99.8% transmitted into the body)

⚡ Why gel matters — the numbers

Without gel (air gap): ~99.99% reflected — almost nothing enters the body.

With gel: ~0.17% reflected — nearly all the ultrasound enters the body.

This is called impedance matching: the gel provides an acoustic bridge between the transducer and the skin, eliminating the air gap that would otherwise cause total reflection.

The coupling gel also serves other practical purposes:

Eliminates tiny air pockets between the transducer and skin
Allows the transducer to glide smoothly across the skin
Has similar acoustic impedance to soft tissue (Z ≈ 1.5 × 10⁶ kg m⁻² s⁻¹)

🎮 Interactive: Reflection Intensity Calculator

Select two materials to see how much ultrasound reflects at their boundary. Try air → soft tissue vs gel → soft tissue to see why gel matters.

Medium 1 (Z₁)

→

Medium 2 (Z₂)

Reflected: — Transmitted: —

—

      Select two media above to see the calculation.
    

Part 8

Ultrasound Scanning Techniques

A-Scan (Amplitude Scan)

In an A-scan, a single ultrasound pulse is sent along one beam direction. The transducer receives echoes from boundaries at different depths and displays them as vertical spikes on a cathode ray oscilloscope (CRO). The horizontal position of each spike gives the time (and therefore depth), while the height gives the amplitude of the echo.

The depth of a reflecting surface is calculated from the time delay using:

Depth from echo time d = ½ct

Where d is depth, c is the speed of sound in the medium, and t is the time for the pulse to travel to the boundary and back (hence the factor of ½).

Worked Example 11 — Calculating depth

An ultrasound pulse is sent into the body and an echo is received 4.0 μs later. If the speed of sound in the tissue is 1540 m s⁻¹, calculate the depth of the reflecting surface.

d = ½ct = ½ × 1540 × 4.0 × 10⁻⁶

d = 3.08 × 10⁻³ m = 3.1 mm

B-Scan (Brightness Scan)

A B-scan is much more common in clinical practice. Instead of a single beam, the transducer is swept across the patient's body (or an array of transducers is used). Each beam direction produces a line of brightness values, and these are combined to form a 2D cross-sectional image.

In a B-scan, the brightness of each pixel corresponds to the amplitude of the reflected echo at that point. Strong reflections (large impedance mismatch, e.g. at bone boundaries) appear as bright spots, while weak reflections (similar impedances) appear dim.

A-Scan

Single beam direction
Displays amplitude spikes on a CRO
Gives depth measurements
No image produced
Used for measuring distances (e.g. eye length before cataract surgery)

B-Scan

Multiple beam directions (swept or array)
Produces a 2D cross-sectional image
Brightness = echo amplitude
Real-time imaging possible
Used for foetal scanning, abdominal imaging, echocardiography

Part 9

Ultrasound vs Other Imaging Techniques

Advantages of ultrasound

Non-ionising — no radiation dose, safe for foetuses and repeated use
Real-time imaging — can observe moving structures (heartbeat, blood flow)
Relatively inexpensive compared to CT and MRI
Portable — machines can be brought to the patient
No known harmful effects at diagnostic power levels
Good contrast between different soft tissues

Disadvantages of ultrasound

Cannot penetrate bone — high impedance mismatch causes strong reflection
Cannot image through air — lungs and gas-filled organs are problematic
Lower resolution than CT or MRI for fine structural detail
Operator-dependent — image quality depends heavily on the skill of the sonographer
Limited depth of penetration — higher frequencies give better resolution but less penetration

⚡ Key Point

The choice of ultrasound frequency involves a trade-off: higher frequency gives better resolution (shorter wavelength) but less penetration depth (more attenuation). Lower frequencies penetrate deeper but produce less detailed images.

Check your understanding

Knowledge Check

Define acoustic impedance. State its units and the equation used to calculate it.

2 marks

Z = ρc where ρ is the density of the medium and c is the speed of sound in the medium (1 mark)
Units: kg m⁻² s⁻¹ (or Rayl) (1 mark)

Explain why coupling gel must be applied to the skin before an ultrasound scan. Refer to acoustic impedance in your answer.

3 marks

Air has a very low acoustic impedance (~430) compared to skin (~1.6 × 10⁶) (1 mark)
This large mismatch means nearly all ultrasound would be reflected at the air–skin boundary (1 mark)
Gel has a similar impedance to soft tissue, providing impedance matching so most ultrasound enters the body (1 mark)

Explain how a piezoelectric transducer acts as both a transmitter and receiver of ultrasound.

3 marks

Transmit: An alternating p.d. applied across the crystal causes it to vibrate at the same frequency, emitting ultrasound (1 mark)
Receive: When an ultrasound echo reaches the crystal, the pressure causes it to deform, generating a p.d. across it (1 mark)
The same crystal can perform both functions because the piezoelectric effect is reversible (1 mark)

State two differences between an A-scan and a B-scan.

2 marks

A-scan uses a single beam; B-scan uses multiple beams (or a swept transducer) (1 mark)
A-scan produces a voltage-time trace (no image); B-scan produces a 2D image (1 mark)
Accept also: A-scan measures depth; B-scan shows structure. Or: A-scan uses amplitude; B-scan uses brightness.

The speed of sound in a certain tissue is 1600 m s⁻¹ and the tissue density is 1050 kg m⁻³. Calculate the acoustic impedance of the tissue.

2 marks

Z = ρc = 1050 × 1600 = 1,680,000 (1 mark)
Z = 1.68 × 10⁶ kg m⁻² s⁻¹ (1 mark for correct standard form and units)

Exam Practice

Exam-Style Questions

(a) Explain what is meant by the term acoustic impedance.

(b) Ultrasound is directed at a boundary between muscle (Z = 1.71 × 10⁶ kg m⁻² s⁻¹) and bone (Z = 7.6 × 10⁶ kg m⁻² s⁻¹). Calculate the fraction of the incident intensity that is reflected at this boundary.

(c) State and explain one reason why ultrasound is preferred to X-rays for imaging a foetus.

8 marks

(a) [2 marks]

The product of the density of a medium and the speed of sound in that medium (1 mark)
Z = ρc with units kg m⁻² s⁻¹ (1 mark)

(b) [3 marks]

Iᵣ/I₀ = ((Z₂ − Z₁)/(Z₂ + Z₁))² (1 mark for formula)
= ((7.6 − 1.71) × 10⁶ / (7.6 + 1.71) × 10⁶)² (substitution)
= (5.89/9.31)² = (0.633)²
= 0.400 or 40.0% (1 mark for answer)
Reasonable attempt at calculation with correct method (1 mark for method)

(c) [3 marks]

Ultrasound is non-ionising / X-rays are ionising (1 mark)
Ionising radiation can damage DNA / cause mutations / harm developing foetal cells (1 mark)
Ultrasound has no known harmful effects at diagnostic power levels, making it safe for repeated use during pregnancy (1 mark)

(a) An ultrasound transducer emits a pulse that is reflected from a boundary 5.0 cm below the skin surface. The speed of sound in the tissue is 1540 m s⁻¹. Calculate the time between the pulse being emitted and the echo being received.

(b) Explain why a coupling gel is applied between the transducer and the skin. Your answer should include a calculation to support your explanation.

(c) In a B-scan, the brightness of a pixel on the image depends on a particular property of the reflected ultrasound. State what this property is.

9 marks

(a) [3 marks]

Distance travelled = 2 × 0.050 = 0.100 m (round trip) (1 mark)
t = d/c = 0.100 / 1540 = 6.49 × 10⁻⁵ s (1 mark)
t = 65 μs (1 mark for correct answer with appropriate unit)

(b) [4 marks]

Without gel, there is an air gap between transducer and skin (1 mark)
Air (Z ≈ 430) and tissue (Z ≈ 1.6 × 10⁶) have very different impedances (1 mark)
Calculation: Iᵣ/I₀ ≈ ((1.6 × 10⁶ − 430)/(1.6 × 10⁶ + 430))² ≈ 1.0, so nearly all ultrasound is reflected (1 mark)
Gel has impedance similar to tissue, so most ultrasound enters the body (impedance matching) (1 mark)

(c) [2 marks]

The amplitude of the reflected echo (2 marks)
Accept "intensity" — amplitude and intensity are related (amplitude² ∝ intensity)

(a) The table below shows acoustic impedance values for various body materials. Use the data to calculate the fraction of ultrasound intensity reflected at each of the following boundaries:
    (i) fat → muscle
    (ii) soft tissue → bone
    (iii) soft tissue → lung

Data: fat Z = 1.38 × 10⁶, muscle Z = 1.71 × 10⁶, soft tissue Z = 1.63 × 10⁶, bone Z = 7.6 × 10⁶, lung Z = 0.26 × 10⁶ (all kg m⁻² s⁻¹)

(b) At which boundary is the most ultrasound reflected? Explain the significance of this for medical imaging.

9 marks

(a)(i) fat → muscle [2 marks]

Iᵣ/I₀ = ((1.71 − 1.38)/(1.71 + 1.38))² = (0.33/3.09)² = (0.107)²
= 0.0114 (1.1%) (2 marks: 1 for method, 1 for answer)

(a)(ii) soft tissue → bone [2 marks]

Iᵣ/I₀ = ((7.6 − 1.63)/(7.6 + 1.63))² = (5.97/9.23)² = (0.647)²
= 0.419 (41.9%) (2 marks: 1 for method, 1 for answer)

(a)(iii) soft tissue → lung [2 marks]

Iᵣ/I₀ = ((0.26 − 1.63)/(0.26 + 1.63))² = (−1.37/1.89)² = (−0.725)²
= 0.525 (52.5%) (2 marks: 1 for method, 1 for answer)

(b) [3 marks]

The most ultrasound is reflected at the soft tissue → lung boundary (52.5%) (1 mark)
This means ultrasound cannot effectively image structures behind air-filled lungs (1 mark)
Similarly, the large reflection at the tissue → bone boundary (42%) means ultrasound struggles to image structures behind bone (1 mark)

Topic Summary

Acoustic Impedance

Z = ρc — a measure of how much a medium resists the passage of sound. Depends on density and speed of sound in the medium.

Reflection Coefficient

Iᵣ/I₀ = ((Z₂ − Z₁)/(Z₂ + Z₁))² — the fraction of ultrasound reflected depends on the impedance mismatch at a boundary.

Coupling Gel

Eliminates the air gap (Z ≈ 430) between transducer and skin (Z ≈ 1.6 × 10⁶). Without it, ~100% of ultrasound reflects.

A-Scan vs B-Scan

A-scan: single beam, amplitude spikes, depth measurements. B-scan: swept beam, 2D image, brightness = echo amplitude.

Equations to Know

Z = ρc

Iᵣ/I₀ = ((Z₂ − Z₁)/(Z₂ + Z₁))²

d = ½ct