Attenuation of Ultrasound in Matter

Attenuation of ultrasound is defined as:

The reduction of energy due to the absorption of ultrasound as it travels through a material

The attenuation coefficient of the ultrasound is expressed in decibels per centimetre lost for every incremental increase in megahertz frequency
- Generally, 0.5 dB cm^–1 is lost for every 1MHz
The intensity I of the ultrasound decreases with distance x, according to the equation:

$I = I_{0} e^{- μ x}$

Where:
- I₀ = the intensity of the incident beam (W m^-2)
- I = the intensity of the reflected beam (W m^-2)
- μ = the absorption coefficient (m^-1)
- x = distance travelled through the material (m)
The absorption coefficient μ, will vary from material to material
Attenuation is not a major problem in ultrasound scanning as the scan relies on the reflection of the ultrasounds at the boundaries of materials

Intensity-depth graph showing attenuation

Attenuation of Ultrasound Graph, downloadable AS & A Level Physics revision notes

When the intensity is expressed in decibels, the amplitudes of the echoes can be seen to decrease linearly

Worked example

The thickness x of the layer of fat on an animal, as shown in the diagram, is to be investigated using ultrasound. WE - Attenuation of Ultrasound question image, downloadable AS & A Level Physics revision notes The intensity of the parallel ultrasound beam entering the surface S of the layer of fat is I.

The beam is reflected from the boundary between fat and muscle.

The intensity of the reflected ultrasound detected at the surface S of the fat is 0.012I.

Using the table below,

Medium	Z / kg m^–2 s^–1	μ / m^–1
Fat	1.3 × 10⁶	48
Muscle	1.7 × 10⁶	23

calculate:

a) The intensity reflection coefficient at the boundary between the fat and the muscle.

b) The thickness x of the layer of fat.

Answer:

Part (a)

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: Calculate the intensity reflection coefficient

$α = \frac{{(1.7 \times 10^{6} - 1.3 \times 10^{6})}^{2}}{{(1.7 \times 10^{6} + 1.3 \times 10^{6})}^{2}} = \frac{{(0.4)}^{2}}{{(3)}^{2}} = 0.018$

This means that 0.018 of the intensity is reflected at the interface between fat and muscle
This reflected intensity will move back through the fat towards surface S

Part (b)

Step 1: Write out the known quantities

The intensity of the ultrasound pulse is affected 3 times:

Attenuation from the surface S to the fat-muscle boundary
Reflection at the boundary
Attenuation from the boundary back to the surface S

After being transmitted in the fat, the intensity at surface S is given to be 0.012 I.
Therefore, the intensity is 0.018 I at the fat-muscle boundary, and as the ultrasound moves through the fat, it gets attenuated and the new intensity at the surface S is now 0.012 I

Incident intensity, equal to the intensity of the reflected pulse, $I_{0} = 0.018 I e^{- μ x}$

Transmitted intensity, I = 0.012 I
Absorption coefficient, μ = 48 m^-1
Thickness of fat = x

Step 2: Write out the equation for attenuation

$I = I_{0} e^{- μ x}$

Step 3: Substitute in values for intensity and simplify

$0.012 I = (0.018 I \times e^{- μ x}) \times e^{- μ x}$

$0.012 = 0.018 e^{- 2 μ x}$

Step 4: Rearrange and take the natural log of both sides

$\frac{0.012}{0.018} = e^{- 2 μ x}$

$\ln (\frac{0.012}{0.018}) = - 2 μ x$

Step 5: Rearrange and calculate the thickness x

$x = \frac{\ln (\frac{0.012}{0.018})}{- 2 μ} = \frac{\ln (\frac{0.012}{0.018})}{- 2 \times 48} = 4.22 \times 10^{- 3} m = 0.42 cm$

Exam Tip

The intensity equation will not be provided for you on your exam datasheet, so make sure you remember this!

Attenuation of Ultrasound in Matter (CIE A Level Physics)

Revision Note