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A LevelPhysicsCambridge (CIE)Revision Notes24. Medical PhysicsProduction & Use of UltrasoundAttenuation of Ultrasound in Matter

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

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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 space equals space I subscript 0 e to the power of negative mu x end exponent

  • 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 0.012 I.

Medium

Z / kg m–2 s–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 space equals space fraction numerator open parentheses Z subscript 2 space minus space Z subscript 1 close parentheses squared space over denominator open parentheses Z subscript 2 space thin space plus thin space Z subscript 1 close parentheses squared end fraction

Step 2: Calculate the intensity reflection coefficient

alpha space equals space fraction numerator open parentheses 1.7 space cross times space 10 to the power of 6 space minus space 1.3 space cross times space 10 to the power of 6 close parentheses squared space over denominator open parentheses 1.7 space cross times space 10 to the power of 6 space space plus thin space 1.3 space cross times space 10 to the power of 6 close parentheses squared end fraction space equals space fraction numerator open parentheses 0.4 close parentheses squared space over denominator open parentheses 3 close parentheses squared end fraction space equals space 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 = intensity of the reflected pulse

I subscript 0 space equals space 0.018 I e to the power of negative mu x end exponent

  • Transmitted intensity = 0.012 I

  • Absorption coefficient, mu = 48 m-1

  • Thickness of fat = x

Step 2: Write out the equation for attenuation

I space equals space I subscript 0 e to the power of negative mu x end exponent

Step 3: Substitute in values for intensity and simplify

0.012 I space equals space open parentheses 0.018 I space cross times space e to the power of negative mu x end exponent close parentheses space cross times space e to the power of negative mu x end exponent

0.012 space equals space 0.018 e to the power of negative 2 mu x end exponent

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

fraction numerator 0.012 over denominator 0.018 end fraction space equals space e to the power of negative 2 mu x end exponent

ln open parentheses fraction numerator 0.012 over denominator 0.018 end fraction close parentheses space equals space minus 2 mu x

Step 5: Rearrange and calculate the thickness x

x space equals space fraction numerator ln open parentheses fraction numerator 0.012 over denominator 0.018 end fraction close parentheses space over denominator negative 2 mu end fraction space equals space fraction numerator ln open parentheses fraction numerator 0.012 over denominator 0.018 end fraction close parentheses space over denominator negative 2 space cross times space 48 end fraction space equals space 4.22 space cross times space 10 to the power of negative 3 end exponent space straight m space equals space 0.42 space cm

Examiner Tips and Tricks

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

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    Ashika

    Author: Ashika

    Expertise: Physics Project Lead

    Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.