Ultrasound Principles: Acoustic Impedance and Wave Behaviour Revision Notes | A Level AQA Physics

1. Why Boundaries Matter in Ultrasound Imaging

Picture a pulse of ultrasound travelling through the body. Every time it crosses from one tissue type into another — say, from muscle into bone — it hits a boundary. At that boundary, some of the wave bounces back towards the transducer, and some carries on deeper into the tissue. The balance between those two outcomes is what makes ultrasound imaging possible.

The key question is: what determines how much reflects and how much transmits? The answer is a property called acoustic impedance, $Z$. Understanding $Z$ is the foundation of everything else in this topic — the gel on the probe, the bright spots on the scan, and the limits of how deep ultrasound can see.

2. Acoustic Impedance: The Property That Controls Boundaries

Think of acoustic impedance as a measure of how 'resistant' a material is to being disturbed by a sound wave. A dense material through which sound travels quickly will have a high $Z$. Bone, for example, has a much higher acoustic impedance than soft tissue — and that contrast is precisely what creates a visible boundary on an ultrasound scan.

Notice that $Z$ depends on two independent properties: how dense the material is, and how fast sound moves through it. A material can have a high $Z$ because it is very dense, or because sound travels very fast within it, or both. This is why bone ($Z \approx 7.0 \times 10^6$ kg m⁻² s⁻¹) and air ($Z \approx 430$ kg m⁻² s⁻¹) sit at opposite extremes — they differ in both density and sound speed.

Scan that table and you will see something immediately useful: soft tissue, blood, and coupling gel all have very similar values of $Z$. Bone and air sit far outside that cluster. That pattern will matter enormously when we look at what happens at boundaries.

3. Reflection and Transmission at Tissue Boundaries

When an ultrasound pulse reaches a boundary between two media with acoustic impedances $Z_1$ and $Z_2$, the wave splits: part reflects back and part transmits forward. The fraction that reflects is described by the intensity reflection coefficient, $\frac{I_r}{I_i}$, where $I_r$ is the reflected intensity and $I_i$ is the incident intensity.

The equation tells a clear story: when $Z_1$ and $Z_2$ are nearly equal, the numerator $(Z_2 - Z_1)$ is close to zero, so almost nothing reflects and nearly all the intensity transmits. When $Z_1$ and $Z_2$ are very different, the numerator is large relative to the denominator, and a large fraction reflects back.

This is why the boundary between soft tissue and bone produces a strong, bright echo on a scan — the large impedance mismatch means most of the ultrasound reflects. Conversely, the boundary between blood and soft tissue produces only a faint echo, because their impedances are so close. Both outcomes are clinically useful: strong reflections locate hard structures, while weaker reflections reveal soft-tissue detail.

4. Worked Calculation: Applying the Reflection Coefficient

The reflection coefficient equation looks straightforward, but exam questions often test whether you can handle the squaring step correctly and interpret the result physically. Work through the following example carefully, paying attention to how the numbers combine.

Worked example: Calculating the Intensity Reflection Coefficient at a Tissue–Bone Boundary

An ultrasound pulse travels from soft tissue ($Z_1 = 1.63 \times 10^6$ kg m⁻² s⁻¹) into bone ($Z_2 = 6.84 \times 10^6$ kg m⁻² s⁻¹). Calculate the intensity reflection coefficient at this boundary. [3 marks]

1. $$\frac{I_r}{I_i} = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2$$ — [M1] Stating the correct equation earns the method mark. Always write it out in full before substituting.
2. $Z_2 - Z_1 = 6.84 \times 10^6 - 1.63 \times 10^6 = 5.21 \times 10^6$ kg m⁻² s⁻¹ $Z_2 + Z_1 = 6.84 \times 10^6 + 1.63 \times 10^6 = 8.47 \times 10^6$ kg m⁻² s⁻¹ — Keep the powers of ten consistent. Both values share the factor $10^6$, so it cancels in the ratio — a useful check.
3. $$\frac{5.21 \times 10^6}{8.47 \times 10^6} = \frac{5.21}{8.47} = 0.6152...$$ — The $10^6$ factors cancel, leaving a dimensionless ratio. Keep at least 4 significant figures here to avoid rounding errors in the next step.
4. $(0.6152)^2 = 0.6152 \times 0.6152 = 0.3785...$ — [A1] Squaring is the step most often done incorrectly. Write out the multiplication explicitly: $0.6152 \times 0.6152$, not just $0.6152^2$ without checking.
5. $$\frac{I_r}{I_i} \approx 0.38$$ — [A1] Approximately 38% of the incident intensity is reflected at this boundary. The remaining 62% transmits into the bone. This large reflection explains why bone appears as a bright, high-contrast feature on ultrasound scans.

Answer: $\frac{I_r}{I_i} \approx 0.38$ (dimensionless). About 38% of the incident intensity is reflected.

Now compare that result with what happens at a soft tissue–blood boundary. The impedances are $Z_1 = 1.63 \times 10^6$ and $Z_2 = 1.66 \times 10^6$ kg m⁻² s⁻¹. The difference is only $0.03 \times 10^6$, giving a ratio of $\frac{0.03}{3.29} \approx 0.0091$, and squaring gives $\approx 8 \times 10^{-5}$. Less than 0.01% of the intensity reflects — almost all of it transmits. That tiny echo is still detectable by modern equipment, but it is orders of magnitude weaker than the bone echo.

5. The Air Problem and the Role of Coupling Gel

Here is a practical consequence of the reflection coefficient equation that you must be able to explain. Air has an acoustic impedance of approximately $430$ kg m⁻² s⁻¹, while soft tissue has $Z \approx 1.63 \times 10^6$ kg m⁻² s⁻¹. The mismatch is enormous — a factor of roughly 3800.

Substituting into the reflection coefficient equation gives a value extremely close to 1 — meaning virtually all the ultrasound intensity reflects at an air–tissue boundary and almost none enters the body. Without intervention, the ultrasound probe pressed against dry skin would produce no useful image at all, because the thin layer of air trapped between the probe and skin would reflect essentially the entire pulse.

Coupling gel solves this problem. The gel has an acoustic impedance very close to that of soft tissue ($Z_{\text{gel}} \approx 1.50 \times 10^6$ kg m⁻² s⁻¹). Applying it between the probe and the skin eliminates the air layer, replacing the extreme air–tissue mismatch with a near-perfect impedance match. The reflection coefficient at the probe–gel and gel–skin boundaries drops to nearly zero, allowing the ultrasound pulse to enter the body with minimal reflection loss.

6. Attenuation: Why Ultrasound Cannot See Everything

Even when ultrasound enters the body successfully, its intensity decreases as it travels deeper. This progressive loss of intensity is called attenuation. Three mechanisms contribute: absorption (the medium converts wave energy into thermal energy), scattering (the wave is redirected in multiple directions by small structures), and reflection at the many internal boundaries the pulse encounters.

Attenuation increases with frequency. Higher-frequency ultrasound (typically 5–15 MHz for medical imaging) attenuates more rapidly than lower-frequency ultrasound. This creates a direct clinical trade-off: higher frequencies produce better spatial resolution because the wavelength is shorter, but they cannot penetrate as deeply before the signal becomes too weak to detect. Lower frequencies penetrate more deeply but produce coarser images.

This is why a clinician scanning a superficial structure such as a thyroid gland uses a high-frequency probe, while imaging a deep abdominal organ such as the liver requires a lower-frequency probe. The choice of frequency is always a compromise between resolution and penetration depth — both of which are governed by attenuation.

7. Exam Technique and Common Mistakes

Ultrasound questions on AQA papers follow predictable patterns. The most common calculation task is applying the reflection coefficient equation, and the most common explanation task is justifying the use of coupling gel or comparing boundaries with different impedance mismatches. Knowing where marks are awarded — and where they are lost — makes a significant difference.

A common six-mark question style asks you to compare the ultrasound imaging of two different tissue boundaries — for example, soft tissue versus bone, and soft tissue versus a fluid-filled cyst. Structure your answer by calculating or estimating the impedance mismatch at each boundary, predicting the relative reflection coefficient, and then linking this to the brightness of the echo and the clinical information obtained.

Every concept in this topic connects back to the same central idea: the greater the difference in acoustic impedance between two media, the more ultrasound reflects at their boundary. Mastering the $Z = \rho c$ and reflection coefficient equations, and being able to explain their physical consequences for imaging, covers the full scope of what AQA requires on this topic.