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Q.76PE

Expert-verifiedFound in: Page 631

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.

**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).**

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.

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}

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.

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