# Ultrasound imaging

> CIE A-Level Physics · Unit 28: Medical imaging
> Source: https://www.owlsprep.com/study/cie-9702-u28-ultrasound-imaging/

This subtopic covers the physics of ultrasound for medical diagnostic imaging, including production, detection, acoustic impedance, reflection, scanning techniques, and safety considerations, a key topic for CIE A-Level Papers 2 and 4.

**Prerequisites:** [Longitudinal wave properties](https://www.owlsprep.com/study/cie-9702-u13-longitudinal-waves/); [Wave reflection and intensity](https://www.owlsprep.com/study/cie-9702-u13-wave-intensity/)

## Learning objectives

- Explain how ultrasound is produced and detected using piezoelectric crystals
- Calculate acoustic impedance and intensity reflection coefficient at material boundaries
- Compare A-scan and B-scan imaging techniques and their uses
- Explain the purpose of impedance matching with ultrasound gel
- Evaluate advantages and limitations of ultrasound imaging

## Production and Detection of Ultrasound

Ultrasound is defined as sound with frequency above 20 kHz, the upper limit of human hearing. For medical imaging, frequencies between 1 MHz and 10 MHz are typically used: higher frequencies give better resolution but are absorbed more quickly, so cannot penetrate as deep.

**Piezoelectric Effect** — An effect observed in certain crystals (e.g. quartz) where: (1) mechanical deformation of the crystal induces an e.m.f. across it, and (2) an applied alternating e.m.f. causes mechanical deformation/vibration of the crystal.

*Example:* An applied a.c. voltage of 2 MHz will make the crystal vibrate at 2 MHz, emitting 2 MHz ultrasound waves.

**Worked example:** Explain why a single piezoelectric crystal can be used for both production and detection of ultrasound.

1. For production: When an alternating voltage matching the required ultrasound frequency is applied across the crystal, it vibrates at that frequency, emitting ultrasound waves.
2. For detection: When reflected ultrasound waves hit the crystal, they cause the crystal to deform mechanically.
3. This deformation induces an alternating e.m.f. across the crystal, matching the frequency of the reflected wave. This signal can be processed by electronic equipment.

> **tip**
>
> CIE examiners often ask why piezoelectric crystals are suitable for both roles: remember the reversible nature of the effect.

## Acoustic Impedance and Reflection

When an ultrasound beam reaches a boundary between two different media, part of the intensity is reflected and part is transmitted. The proportion reflected depends on the difference in acoustic impedance of the two media.

**Acoustic Impedance** — The product of the density ($\rho$) of the medium and the speed ($c$) of ultrasound in the medium: $Z = \rho c$. Units are kg m⁻² s⁻¹.

*Notation:* Z

*Example:* Z for air is ~4 × 10² kg m⁻² s⁻¹, and Z for soft tissue/skin is ~1.7 × 10⁶ kg m⁻² s⁻¹.

The intensity reflection coefficient $\alpha$ is given by the formula:

$$\alpha = \frac{I_r}{I_0} = \frac{(Z_2 - Z_1)^2}{(Z_2 + Z_1)^2}$$

**Worked example:** Calculate the fraction of incident ultrasound intensity reflected at an air-skin boundary. $Z_{air} = 4.3 \times 10^2$ kg m⁻² s⁻¹, $Z_{skin} = 1.7 \times 10^6$ kg m⁻² s⁻¹. Comment on your result.

1. Substitute values into the formula for $\alpha$:
2. $$\alpha = \frac{(1.7 \times 10^6 - 4.3 \times 10^2)^2}{(1.7 \times 10^6 + 4.3 \times 10^2)^2}$$
3. Since $Z_{air} << Z_{skin}$, the difference and sum are approximately equal to $Z_{skin}$, so $\alpha \approx 1.0$ (calculated value ≈ 0.999).
4. Comment: Almost all incident intensity is reflected at an air-skin boundary, which is why we use ultrasound gel for impedance matching, to eliminate the air gap between transducer and skin.

## Ultrasound Scanning Techniques

Two main scanning techniques are used for diagnostic imaging:

- **A-scan (Amplitude scan):** A single transducer sends a pulse along one line into the body. Reflected pulses are displayed as peaks on an oscilloscope. The time delay of the peak gives the depth of the boundary, and peak amplitude gives the reflection intensity. Used for simple measurements like eye lens thickness.
- **B-scan (Brightness scan):** A moving transducer or array of transducers sends pulses along multiple lines. Each reflected pulse is converted to a dot on a screen, where brightness corresponds to reflection intensity. Combining all lines produces a 2D real-time image. This is the standard technique for prenatal imaging.

**Worked example:** State one use each for an A-scan and a B-scan, and one key difference between them.

1. Difference: An A-scan produces a 1-dimensional plot of amplitude against depth, while a B-scan produces a full 2-dimensional image of internal body structures.
2. Use of A-scan: Measuring the size of the eye or detecting the depth of a brain tumour.
3. Use of B-scan: Producing a real-time image of a fetus during pregnancy, or imaging gallstones or other organ abnormalities.

## Advantages and Limitations

- **Advantages:** Non-ionising (no DNA damage, safe for repeated use and for pregnant patients), produces real-time moving images, lower cost than CT or MRI, portable.
- **Limitations:** Cannot penetrate bone or air-filled organs (e.g. lungs), higher frequency ultrasound has poor depth penetration, lower resolution than CT or MRI for deep structures.

## Common pitfalls

- **Wrong:** Claiming ultrasound is electromagnetic radiation.
  - Why it fails: Ultrasound is a mechanical pressure wave, it requires a medium to travel and cannot propagate through a vacuum.
  - Correct: Always classify ultrasound as a mechanical longitudinal wave with frequency above 20 kHz.
- **Wrong:** Forgetting to square the terms in the intensity reflection coefficient formula.
  - Why it fails: $\alpha$ is a ratio of intensities, and intensity is proportional to the square of amplitude, so the difference in impedance must be squared.
  - Correct: Memorize $\alpha = \frac{(Z_2-Z_1)^2}{(Z_2+Z_1)^2}$, always check your result is between 0 and 1.
- **Wrong:** Stating the purpose of ultrasound gel is to lubricate the skin.
  - Why it fails: Lubrication is a secondary effect; the key medical purpose is impedance matching.
  - Correct: Explain that gel has a similar acoustic impedance to skin, which eliminates the air gap between transducer and skin, so most ultrasound is transmitted into the body instead of being reflected.
- **Wrong:** Confusing A-scan and B-scan, claiming A-scans produce 2D images.
  - Why it fails: A-scans only measure along one line, the 'A' stands for Amplitude not 2D Array.
  - Correct: Remember: A = Amplitude (1D), B = Brightness (2D).

## Cheatsheet

| Term | Key Formula/Info | Exam Note |
| --- | --- | --- |
| Ultrasound | f > 20 kHz, 1-10 MHz medical | Mechanical longitudinal wave |
| Acoustic impedance | $Z = \rho c$ | Unit: kg m⁻² s⁻¹ |
| Reflection coefficient | $\alpha = \frac{(Z_2-Z_1)^2}{(Z_2+Z_1)^2}$ | $0 \leq \alpha \leq 1$ |
| A-scan | 1D amplitude scan | Measures depth of boundaries |
| B-scan | 2D brightness image | Real-time imaging |
| Ultrasound gel | Impedance matching | Removes air gap between transducer and skin |

## What's next

Ultrasound imaging is a core application of wave physics in medicine, and builds on your understanding of wave interactions with different media. This knowledge will help you when learning how other imaging techniques use radiation interaction with tissue to produce diagnostic images. Mastering acoustic impedance and reflection here will also help you answer cross-topic wave questions in Paper 4.

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