Attenuation of Ultrasound in Matter (Cambridge (CIE) A Level Physics): Revision Note

The thickness x of the layer of fat on an animal, as shown in the diagram, is to be investigated using ultrasound.

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 information in the table, 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

$\alpha =\frac{{\left(Z_2-Z_1\right)}^2}{{\left(Z_2+Z_1\right)}^2}$

Step 2: Calculate the intensity reflection coefficient

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

• 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 $0.012I$.

• Therefore, the intensity is $0.018I$ 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 $0.012I$

incident intensity = intensity of the reflected pulse

$I_0=0.018I{e}^{-\mu x}$

• Transmitted intensity = $0.012I$

• Absorption coefficient, $\mu$ = 48 m^-1

• Thickness of fat = $x$

Step 2: Write out the equation for attenuation

$I=I_0{e}^{-\mu x}$

Step 3: Substitute in values for intensity and simplify

$0.012I=\left(0.018I\times e^{-\mu x}\right)\times e^{-\mu x}$

$0.012=0.018e^{-2\mu x}$

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

$\frac{0.012}{0.018}=e^{-2\mu x}$

$\ln \left(\frac{0.012}{0.018}\right)=-2\mu x$

Step 5: Rearrange and calculate the thickness x

$x=\frac{\ln \left({\displaystyle \frac{0.012}{0.018}}\right)}{-2\mu }=\frac{\ln \left({\displaystyle \frac{0.012}{0.018}}\right)}{-2\times 48}=4.22\times {10}^{-3}\mathrm{m}=0.42\mathrm{cm}$

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