Attenuation of Ultrasound in Matter (Cambridge International A-Level Physics)

Spec point — 24.1.6

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 1 MHz

• The intensity I of the ultrasound decreases with distance x, according to the equation:

$I=I_{0}e^{-μx}$

• 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

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.

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 $0.012I$.

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

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

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

Step 2: Calculate the intensity reflection coefficient

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

• 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 $0.012I$.

• 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

$I_{0}=0.018Ie^{-μx}$

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

Step 2: Write out the equation for attenuation

$I=I_{0}e^{-μx}$

Step 3: Substitute in values for intensity and simplify

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

$0.012=0.018e^{-2μx}$

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

$\frac{0.012}{0.018}=e^{-2μx}$

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

Step 5: Rearrange and calculate the thickness x

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

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