Key points
- Acoustic impedance Z = ρc, with unit kg m⁻² s⁻¹.
- The reflected fraction is (Z₂ − Z₁)² / (Z₂ + Z₁)², so a bigger mismatch gives a bigger echo.
- Coupling gel removes air and matches impedance between probe and skin.
- A-scans give depth from d = ct / 2, and B-scans combine many A-scans into an image.
- Doppler shift for moving blood is Δf / f = 2v cos θ / c.
What acoustic impedance means for ultrasound
Acoustic impedance and ultrasound go together because an ultrasound image is built from echoes, and echoes only form where the acoustic impedance changes. The acoustic impedance of a material is Z = ρc, and at a boundary between materials 1 and 2 the fraction of intensity reflected is I_R/I₀ = (Z₂ − Z₁)² / (Z₂ + Z₁)².
Ultrasound is sound with a frequency above the range of human hearing, more than 20 kHz. Medical scanners use much higher frequencies, typically a few megahertz. The pulses travel into the body, partly reflect at each boundary between different materials, and return to the same probe. The time delay tells you how deep the boundary is, and the strength of the echo tells you how different the materials are.
In this guide you will meet the transducer, the equation for acoustic impedance, why coupling gel is essential, the A-scan and B-scan, and Doppler ultrasound. A short comparison with X-ray, CT and PET at the end helps you answer the synoptic questions that ask you to evaluate one technique against another.
Piezoelectric transducers, frequency and detail
The probe contains a piezoelectric crystal or ceramic. A piezoelectric material changes shape when a p.d. is applied across it, and it produces a p.d. when it is squeezed or stretched. So the same crystal can work as both a transmitter and a receiver.
- Transmit: a short burst of alternating p.d. at the crystal's natural frequency makes it vibrate, so it emits a pulse of ultrasound.
- Receive: the returning echo squeezes the crystal, which generates a small alternating p.d. that is amplified and processed.
- Timing: the probe is mostly listening. It sends a short pulse, then waits to receive echoes before sending the next one.
Frequency matters because it sets the wavelength through λ = c / f. In soft tissue, where c is about 1540 m s⁻¹, a 5.0 MHz beam has a wavelength of 1540 / 5.0 × 10⁶ = 3.08 × 10⁻⁴ m, about 0.3 mm. Details smaller than about a wavelength cannot be resolved, so higher frequency gives sharper images.
There is a trade-off. Higher-frequency ultrasound is absorbed and scattered more strongly, so it does not penetrate as far. A scan of a deep organ uses a lower frequency, and a scan of a surface structure can use a higher one for finer detail. Short pulses help too, because two nearby boundaries give separate echoes only if the pulse is shorter than the time between them.
Z = ρc and the intensity reflection coefficient
Z = ρc
- Z = acoustic impedance / kg m⁻² s⁻¹
- ρ = density / kg m⁻³
- c = speed of sound in the material / m s⁻¹
I_R / I₀ = (Z₂ − Z₁)² / (Z₂ + Z₁)²
- I₀ = incident intensity
- I_R = reflected intensity
- Z₁, Z₂ = impedances of the first and second materials
Intensity transmitted (ignoring absorption) = I₀ − I_R, so the fraction transmitted is 1 − I_R/I₀.
If the two impedances are equal the numerator is zero, so there is no reflection and no echo. The bigger the mismatch, the stronger the echo, and I_R/I₀ approaches 1 for a huge mismatch. Because the difference is squared, it does not matter which material comes first: tissue to bone gives the same fraction as bone to tissue.
| Material | ρ / kg m⁻³ | c / m s⁻¹ | Z / kg m⁻² s⁻¹ |
|---|---|---|---|
| Air | 1.2 | 330 | 4.0 × 10² |
| Water | 1000 | 1480 | 1.48 × 10⁶ |
| Soft tissue | 1060 | 1540 | 1.63 × 10⁶ |
| Bone | 1900 | 4080 | 7.75 × 10⁶ |
Why coupling gel is used
Put a bare probe on dry skin and almost no ultrasound gets in. There is a thin layer of air between the probe and the skin, and air has an acoustic impedance thousands of times smaller than tissue. Almost all the intensity is reflected at the air-skin boundary, and the little that gets through must also come back through the same barrier.
The gel is an oil-based or water-based jelly with an impedance close to that of skin and soft tissue. Spread between the probe and the skin, it fills the air gap. The impedances are now nearly matched, so a very small fraction is reflected and nearly all of the ultrasound enters the body. The worked examples below put numbers on this.
A-scans and B-scans
In an A-scan (amplitude scan), a single pulse is sent along one line, and the echo amplitude is plotted against time. Each peak represents a boundary. Because the speed of ultrasound in the tissue is known, the time of each peak gives the depth of the boundary. A-scans are used when you need an accurate distance along a line, for example the length of the eye.
d = ct / 2
- d = depth of the boundary / m
- c = speed of ultrasound in tissue / m s⁻¹
- t = time between the pulse and its echo / s
The factor of 2 appears because the pulse travels to the boundary and back.
A B-scan (brightness scan) builds a two-dimensional image. The probe contains an array of transducers, or the beam is swept across the body, so many A-scans are recorded along neighbouring lines. Each echo is shown as a dot whose brightness represents its amplitude, and whose position comes from the depth and the direction of the line. Together the dots form a slice through the body. A fast B-scan is shown in real time, which is why you can watch a foetus move.
The weaker echoes from deeper boundaries are corrected by amplifying later signals more, so equally reflecting boundaries look equally bright whatever their depth.
Doppler ultrasound
Doppler ultrasound measures the speed of moving structures such as blood. The waves reflect from moving red blood cells, so the echo has a different frequency from the transmitted beam. Blood moving towards the probe gives a higher frequency, and blood moving away gives a lower one. The size of the shift tells you the speed.
Δf / f = 2v cos θ / c
- Δf = change in frequency / Hz
- f = transmitted frequency / Hz
- v = speed of the blood / m s⁻¹
- θ = angle between the beam and the direction of motion
- c = speed of ultrasound in tissue / m s⁻¹
The factor 2 is there because the moving blood acts first as a moving receiver and then as a moving source.
The beam is angled to the blood vessel, and cos θ takes care of the fact that only the component of velocity along the beam changes the frequency. At θ = 90° there would be no shift. Doppler imaging is often combined with a B-scan, so the colour-coded blood flow is overlaid on the anatomy.
Comparing ultrasound with X-ray, CT and PET
Synoptic questions ask you to choose a technique for a given situation and justify it. Compare them on what they show, the radiation used, risk, and practicality.
| Technique | How it works | Best for | Main limits |
|---|---|---|---|
| Ultrasound | Pulses of high-frequency sound, echoes at impedance boundaries | Soft tissue, foetus, blood flow; real-time, portable, non-ionising | Poor through bone and gas; limited depth; less detail at depth |
| X-ray | X-ray photons are absorbed differently by different tissues, forming a shadow image | Bones, chest; quick and cheap; contrast from attenuation | Ionising; flat 2D image; weak soft tissue contrast |
| CT scan | X-ray source and detectors rotate around the patient, and a computer builds slice images | Detailed 3D images of bone and soft tissue | Much larger dose than a single X-ray; expensive, not portable |
| PET scan | A positron-emitting tracer is injected. Each annihilation produces two gamma photons in opposite directions, detected in a ring | Function and activity of tissue, such as metabolism | Ionising from the tracer; low spatial detail; expensive; tracer must be made on site or nearby |
PET connects directly to the nuclear physics in this module. A positron meets an electron, and both are annihilated, converting their rest mass into energy. Each photon has an energy of about 0.511 MeV, from E = mc² for one electron mass. See binding energy and mass defect for E = mc², and radioactive decay and half-life for why the tracer's half-life must be short enough to limit the dose but long enough for the scan.
Ultrasound uses non-ionising mechanical waves, and at diagnostic intensities it is regarded as low risk, which is why it is the first choice for scanning a developing baby. X-rays, CT and PET all use ionising radiation, so the benefit to the patient must outweigh the risk of the dose.
Worked examples
Acoustic impedance
Soft tissue has a density of 1060 kg m⁻³ and ultrasound travels through it at 1540 m s⁻¹. Calculate its acoustic impedance.
- Z = ρc.
- Z = 1060 × 1540 = 1.63 × 10⁶ kg m⁻² s⁻¹.
Answer Z = 1.63 × 10⁶ kg m⁻² s⁻¹
Without and with gel
Air has Z = 4.0 × 10² kg m⁻² s⁻¹ and soft tissue has Z = 1.63 × 10⁶ kg m⁻² s⁻¹. The gel has Z = 1.50 × 10⁶ kg m⁻² s⁻¹. Calculate the fraction of the intensity reflected at (a) an air-tissue boundary and (b) a gel-tissue boundary.
- (a) I_R/I₀ = (Z₂ − Z₁)² / (Z₂ + Z₁)² with Z₁ = 396 (from ρ = 1.2 and c = 330 for air) and Z₂ = 1.6324 × 10⁶.
- Z₂ − Z₁ = 1.6320 × 10⁶ and Z₂ + Z₁ = 1.6328 × 10⁶. The ratio is 0.99951, and squared it is 0.9990.
- So 99.9% of the intensity is reflected and only about 0.1% enters.
- (b) Z₂ − Z₁ = 1.6324 × 10⁶ − 1.50 × 10⁶ = 1.324 × 10⁵ and Z₂ + Z₁ = 3.1324 × 10⁶.
- Ratio = 0.0423, and squared it is 0.0018. So 0.18% is reflected and 99.8% is transmitted.
Answer (a) 0.999 (b) 0.0018. With gel, almost all the intensity enters the body.
Tissue to bone
Soft tissue has Z = 1.63 × 10⁶ kg m⁻² s⁻¹ and bone has Z = 7.75 × 10⁶ kg m⁻² s⁻¹. What fraction of the intensity is transmitted into bone?
- I_R/I₀ = (7.75 × 10⁶ − 1.63 × 10⁶)² / (7.75 × 10⁶ + 1.63 × 10⁶)².
- = (6.12 × 10⁶)² / (9.38 × 10⁶)² = (0.6525)² = 0.426.
- Transmitted fraction = 1 − 0.426 = 0.574.
Answer About 0.57 (57%) is transmitted. A strong echo from bone is why bone shows up clearly and hides what is behind it.
Depth from an A-scan
An A-scan shows an echo 65 μs after the pulse. Take the speed of ultrasound in tissue as 1540 m s⁻¹. Calculate the depth of the boundary.
- The pulse travels there and back, so 2d = ct.
- t = 65 × 10⁻⁶ s, so 2d = 1540 × 65 × 10⁻⁶ = 0.1001 m.
- d = 0.0500 m.
Answer d = 5.0 cm
Doppler shift for blood
A 5.0 MHz beam meets blood moving at 0.40 m s⁻¹ along a vessel. The beam makes an angle of 30° with the direction of flow. The speed of ultrasound in tissue is 1540 m s⁻¹. Calculate the frequency shift.
- Δf / f = 2v cos θ / c.
- Δf = 5.0 × 10⁶ × 2 × 0.40 × cos 30° / 1540 = 5.0 × 10⁶ × 0.6928 / 1540.
- Δf = 2.25 × 10³ Hz.
Answer Δf = 2.2 × 10³ Hz (about 2.2 kHz)
Common mistakes
- Forgetting the factor of 2 in depth. The pulse travels to the boundary and back, so d = ct / 2.
- Using the wrong values for c. The speed of ultrasound in tissue is about 1540 m s⁻¹, not the speed of light or the speed of sound in air.
- Not squaring the ratio. The reflection coefficient is the whole bracket squared, so check it is not above 1.
- Saying gel 'lubricates' the probe. Its main purpose is impedance matching, to remove air and reduce reflection.
- Mixing up transmitted and reflected. Transmitted fraction is 1 − I_R/I₀, ignoring absorption.
- Forgetting that bone and gas give big echoes. Large impedance differences produce strong reflections and poor imaging behind them.
- Treating B-scan and A-scan as the same. A-scan is amplitude against time along one line, B-scan is a brightness image of many lines.
- Calling ultrasound radiation harmless in every case. It is non-ionising and low risk at diagnostic levels, which is more precise than 'safe'.
How OCR A examines this topic
Ultrasound sits within Medical imaging in the Particles and medical physics module. Expect short calculations (Z = ρc, depth from time, reflection coefficient), explanations (gel, piezoelectric action, A- and B-scans) and evaluation questions comparing techniques. Check the current specification for which techniques and equations are listed.
- Calculate: impedance, reflection coefficient and transmitted intensity, depth from echo time.
- Explain: why gel is used, how the transducer both sends and receives, why higher frequency gives better resolution but less penetration.
- Describe: an A-scan trace and how a B-scan image is formed.
- Evaluate: choose between ultrasound, X-ray, CT and PET for a given patient, with reasons about radiation, detail and cost.
Synoptic links include wave properties, such as intensity and the wave equation in stationary waves, exponential attenuation which links to radioactive decay and half-life, and E = mc² for PET annihilation in binding energy and mass defect. Be ready to move between topics within one question.
Practice questions
Fat has a density of 920 kg m⁻³ and ultrasound travels through it at 1450 m s⁻¹. Calculate its acoustic impedance.2 marks
Show the answer
Answer: 1.33 × 10⁶ kg m⁻² s⁻¹
Z = ρc = 920 × 1450 = 1.334 × 10⁶ kg m⁻² s⁻¹.
A boundary separates a material with Z = 1.50 × 10⁶ kg m⁻² s⁻¹ from one with Z = 1.65 × 10⁶ kg m⁻² s⁻¹. Calculate the fraction of intensity reflected.3 marks
Show the answer
Answer: 0.0023
I_R/I₀ = (1.65 − 1.50)² / (1.65 + 1.50)² = 0.15² / 3.15² = 0.0225 / 9.9225 = 0.00227.
A boundary lies 8.0 cm below the skin surface. Calculate the time between sending a pulse and receiving its echo. Take c = 1540 m s⁻¹.3 marks
Show the answer
Answer: 1.0 × 10² μs (104 μs)
Total distance = 2 × 0.080 = 0.160 m. t = distance / c = 0.160 / 1540 = 1.04 × 10⁻⁴ s = 104 μs.
A 4.0 MHz Doppler probe detects a shift of 1.8 kHz from blood. The beam makes an angle of 45° with the flow. Calculate the speed of the blood. Take c = 1540 m s⁻¹.3 marks
Show the answer
Answer: 0.49 m s⁻¹
v = Δf c / (2f cos θ) = 1.8 × 10³ × 1540 / (2 × 4.0 × 10⁶ × cos 45°) = 2.772 × 10⁶ / (5.657 × 10⁶) = 0.490 m s⁻¹.
Explain why a coupling gel is used between an ultrasound probe and the skin.3 marks
Show the answer
Answer: The gel removes the air layer and matches acoustic impedance, so little is reflected and most of the ultrasound enters the body.
Air has a much smaller acoustic impedance than skin, so with an air gap nearly all the intensity is reflected. Gel has impedance close to skin, so the reflection coefficient is small and most of the intensity is transmitted.
Give two advantages and two disadvantages of ultrasound compared with X-rays for imaging a developing foetus.4 marks
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Answer: Advantages: non-ionising and real-time. Disadvantages: poor contrast through bone or gas and limited resolution at depth.
Advantages: it uses non-ionising waves, so the dose risk is low, and it gives real-time moving images that are portable and cheap. Disadvantages: reflections from bone and gas limit what can be seen, and fine detail is lost at greater depth.
Next steps
Practise the reflection coefficient with different materials until the algebra feels familiar, then learn the A-scan and B-scan descriptions. For the other imaging techniques, link to the nuclear physics in radioactive decay and half-life and binding energy and mass defect. The Doppler idea reappears in Hubble's law and redshift. See the full Particles and medical physics module, or practise this topic with step-by-step marking.
Frequently asked questions
What is acoustic impedance?
Acoustic impedance is Z = ρc, the density of a material multiplied by the speed of sound in it. It measures how much a material resists the passage of sound. The unit is kg m⁻² s⁻¹. Reflection of ultrasound depends on the difference in impedance between two materials.
Why is gel used in ultrasound scans?
Without gel there is a thin layer of air between the probe and skin. Air has a very low impedance compared with tissue, so almost all the ultrasound would reflect. Gel has an impedance close to skin, so it matches the two and lets most of the intensity through.
What is the difference between an A-scan and a B-scan?
An A-scan plots echo amplitude against time along a single line, so you can measure depths accurately. A B-scan combines many such lines side by side and shows echoes as dots of varying brightness, building a two-dimensional image of a slice.
How do you calculate depth from an ultrasound echo?
Use d = ct / 2, where c is the speed of ultrasound in the tissue and t is the time between sending the pulse and receiving the echo. The factor of 2 is there because the pulse travels to the boundary and back.
How does a piezoelectric transducer work?
A piezoelectric crystal changes shape when a p.d. is applied across it, so an alternating p.d. makes it vibrate and emit ultrasound. When an echo hits it, the crystal is squeezed and generates a small p.d. So one crystal can both send and receive.
Why does ultrasound not show what is behind bone or air?
Bone and air have very different impedances from soft tissue, so a large fraction of the intensity is reflected at their surfaces. Very little ultrasound passes through to reach structures behind them, so those structures give weak echoes or none.
Related guides
Written for the OCR A Level Physics A (H556) specification. Questions are original and written in exam style; they are not taken from OCR papers. Spotted a mistake? Tell us.