Intensity Reflection Coefficient

The intensity reflection coefficient α is defined as:

The ratio of the intensity of the reflected wave relative to the incident (transmitted) wave

This can be calculated using the fraction:

$α = \frac{I_{R}}{I_{0}} = \frac{{(Z_{2} - Z_{1})}^{2}}{{(Z_{2} + Z_{1})}^{2}}$

Where:
- α = intensity reflection coefficient
- I_R = intensity of the reflected wave (W m^-2)
- I₀ = intensity of the incident wave (W m^-2)
- Z₁ = acoustic impedance of one material (kg m^-2 s^-1)
- Z₂ = acoustic impedance of a second material (kg m^-2 s^-1)

This equation will be provided on the datasheet for your exam
This ratio shows:
- If there is a large difference between the impedance of the two materials, then most of the energy will be reflected
- If the impedance is the same, then there will be no reflection

Coupling Medium

When ultrasound is used in medical imaging, a coupler is needed between the transducer and the body
The soft tissues of the body are much denser than air
If air is present between the transducer and the body, then almost all the ultrasound energy will be reflected
The coupling gel is placed between the transducer and the body, as skin and the coupling gel have a similar density, so little ultrasound is reflected
- This is an example of impedance matching

Worked example

A beam of ultrasound is incident at right-angles to a boundary between two materials as shown in the diagram. WE - Intensity reflection coefficient question image, downloadable AS & A Level Physics revision notes The materials have acoustic impedances of Z₁ and Z₂. The intensity of the transmitted ultrasound beam is I_T, and the reflected intensity is I_R.

a) What is the relationship between I, I_T and I_R?

b) Use the data from the table to determine the reflection coefficient α for a boundary between

(i) gel and soft tissue

(ii) air and soft tissue

c) Explain why gel is usually put on the skin during medical diagnosis using ultrasound.

medium	speed of ultrasound / m s^–1	acoustic impedance / kg m^–2 s^–1
air	330	4.3 × 10²
gel	1500	1.5 × 10⁶
soft tisuse	1600	1.6 × 10⁶
bone	4100	7.0 × 10⁶

Answer:

Part (a)

Incident intensity = Transmitted intensity + Reflected intensity

$I = I_{T} + I_{R}$

Part (b)(i)

Step 1: Write down the equation for intensity reflection coefficient α

$α = \frac{{(Z_{2} - Z_{1})}^{2}}{{(Z_{2} + Z_{1})}^{2}}$

Step 2: Write down the acoustic impedances for gel and soft tissue

Gel, Z₁ = 1.5 × 10⁶ kg m^-2 s^-1
Soft tissue, Z₂ = 1.6 × 10⁶ kg m^-2 s^-1

Step 3: Calculate the intensity reflection coefficient

$α = \frac{{(1.6 \times 10^{6} - 1.5 \times 10^{6})}^{2}}{{(1.6 \times 10^{6} + 1.5 \times 10^{6})}^{2}} = \frac{{(0.1)}^{2}}{{(3.1)}^{2}} = 0.001$

Intensity Reflection Coefficient Worked Example equation 2

This result means that only 0.1% of the incident intensity will be reflected, with the remaining being transmitted

Part (b)(ii)

Step 1: Write down the acoustic impedances for air and soft tissue

Air, Z₁ = 4.3 × 10² kg m^-2 s^-1
Soft tissue, Z₂ = 1.6 × 10⁶ kg m^-2 s^-1

Step 2: Calculate the intensity reflection coefficient

$α = \frac{{(1.6 \times 10^{6} - 4.3 \times 10^{2})}^{2}}{{(1.6 \times 10^{6} + 4.3 \times 10^{2})}^{2}} \approx \frac{{(1.6 \times 10^{6})}^{2}}{{(1.6 \times 10^{6})}^{2}} \approx 1$

This result means that 100% of the incident intensity will be reflected, with none being transmitted

Part (c)

At the air-soft tissue boundary, the intensity reflection coefficient is α ≈ 1
- Therefore, without gel, there is almost complete reflection - no ultrasound is transmitted through the skin
Therefore, the gel enables almost complete transmission of the ultrasound through the skin, with very little reflection
- At the gel-soft tissue boundary, the intensity reflection coefficient is α = 0.001

Intensity Reflection Coefficient (CIE A Level Physics)

Revision Note