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Expert-verified Found in: Page 631 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # In the clinical use of ultrasound, transducers are always coupled to the skin by a thin layer of gel or oil, replacing the air that would otherwise exist between the transducer and the skin. (a) Using the values of acoustic impedance given in Table 17.5 calculate the intensity reflection coefficient between transducer material and air. (b) Calculate the intensity reflection coefficient between transducer material and gel (assuming for this problem that its acoustic impedance is identical to that of water). (c) Based on the results of your calculations, explain why the gel is used.

(a) Intensity reflection coefficient for air to gel boundaries is 0.99.

(b) Intensity reflection coefficient for air to gel boundaries is 0.823.

(c) Because of high acoustic impedance use gel.

See the step by step solution

## Step 1: Formula for the intensity reflection coefficient:

Intensity reflection coefficient (a) is given as the ratio of the intensity of the reflected wave to the intensity of an incident wave.

$a = \frac{{{{\left( {{z_2} - {z_1}} \right)}^2}}}{{{{\left( {{z_2} + {z_1}} \right)}^2}}}$ ….. (1)

Here, ${z_2}$ and ${z_1}$ are acoustic impedances of two boundaries (transmission and reflection boundaries).

## Step 2: (a) Calculation for the intensity reflection coefficient:

Consider the given data as below.

The acoustic impedance of transducer (gel or oil) is,

${z_2} = 30.8 \times {10^{6{\rm{ }}}}kg{\rm{ }}{m^{ - 2}}{\rm{ }}{s^{ - 1}}$

The acoustic impedance of air is,

${z_1} = 429{\rm{ }}kg{\rm{ }}{m^{ - 2}}{\rm{ }}{s^{ - 1}}$

Now put these values into equation (1).

\begin{aligned}a &= \frac{{{{\left( {30.8 \times {{10}^6} - 429} \right)}^2}}}{{{{\left( {30.8 \times{{10}^{6{\rm{ }}}} + 429} \right)}^2}}}\\ &= \frac{{9.486136 \times {{10}^{14}}}}{{9.48666 \times {{10}^{14}}}}\\ &= 0.99\end{aligned}

Hence, the value is closed to one tells that boundary of the gel is highly reflective.

## Step 3: (b) Calculation for the intensity reflection coefficient

Consider the given data as below.

The acoustic impedance of transducer (gel or oil) is, ${z_2} = 30.8 \times {10^{6{\rm{ }}}}kg{\rm{ }}{m^{ - 2}}{\rm{ }}{s^{ - 1}}$

The acoustic impedance of water is, ${z_1} = 1.5 \times {10^{6{\rm{ }}}}kg{\rm{ }}{m^{ - 2}}{\rm{ }}{s^{ - 1}}$

Now put these values into equation (1), and you have

\begin{aligned}a &= \frac{{{{\left( {30.8 \times {{10}^6} - 1.5 \times {{10}^{6{\rm{ }}}}} \right)}^2}}}{{{{\left( {30.8 \times {{10}^{6{\rm{ }}}} + 1.5 \times {{10}^{6{\rm{ }}}}} \right)}^2}}}\\ &= \frac{{\left( {858.49 \times {{10}^{12}}} \right)}}{{\left( {1043.29 \times {{10}^{{\rm{12}}}}} \right)}}\\ &= 0.823\end{aligned}

## Step 3: (c) Explain, why the gel is used

From the values of the final reflection coefficient, the gel has a high acoustic impedance that provides a high reflection coefficient.

Due to the high reflection coefficient, it will provide a very good reflection that’s why use gel. ### Want to see more solutions like these? 