Chapter 17, Problem 76PE

To determine

(a)

The intensity reflection coefficient between transducer material and air.

Answer to Problem 76PE

The intensity reflection coefficient between transducer material and air is $1$.

Explanation of Solution

Given:

Refer Table [17.8]

The value of acoustic impedance of transducer material is, $Z_t=30.8\times {10}^6\text{kg}/{\text{m}}^{\text{2}}\cdot \text{s}$.

The value of acoustic impedance of air is, $Z_a=429\text{kg}/{\text{m}}^{\text{2}}\cdot \text{s}$.

Formula used:

The intensity reflection coefficient betweenthe transducer material and air is given by

  $a=\frac{{\left(Z_t-Z_a\right)}^2}{{\left(Z_t+Z_a\right)}^2}$

Calculation:

The intensity reflection coefficient between transducer material and air is calculated as follows:

  $\begin{array}{c}a=\frac{{\left(Z_t-Z_a\right)}^2}{{\left(Z_t+Z_a\right)}^2}\\ =\frac{{\left(30.8\times {10}^6\text{kg}/{\text{m}}^{\text{2}}\cdot \text{s}-429\text{kg}/{\text{m}}^{\text{2}}\cdot \text{s}\right)}^2}{{\left(30.8\times {10}^6\text{kg}/{\text{m}}^{\text{2}}\cdot \text{s}+429\text{kg}/{\text{m}}^{\text{2}}\cdot \text{s}\right)}^2}\\ =0.999\\ \approx 1\end{array}$

Conclusion:

The intensity reflection coefficient between transducer material and air is $1$.

To determine

(b)

The intensity reflection coefficient between the transducer material and gel which is identical to water.

Answer to Problem 76PE

The intensity reflection coefficient between the transducer material and gel which is identical to water is $0.823$.

Explanation of Solution

Given:

Refer Table [17.8]

The value of acoustic impedance of gel which is identical to water is, $Z_w=1.5\times {10}^6\text{kg}/{\text{m}}^{\text{2}}\cdot \text{s}$.

Formula used:

The intensity reflection coefficient between the transducer material and gel which is identical to water is given by

  $a=\frac{{\left(Z_t-Z_w\right)}^2}{{\left(Z_t+Z_w\right)}^2}$

Calculation:

The intensity reflection coefficient between the transducer material and gel which is identical to water is calculated as follows:

  $\begin{array}{c}a=\frac{{\left(Z_t-Z_w\right)}^2}{{\left(Z_t+Z_w\right)}^2}\\ =\frac{{\left(30.8\times {10}^6\text{kg}/{\text{m}}^{\text{2}}\cdot \text{s}-1.5\times {10}^6\text{kg}/{\text{m}}^{\text{2}}\cdot \text{s}\right)}^2}{{\left(30.8\times {10}^6\text{kg}/{\text{m}}^{\text{2}}\cdot \text{s}+1.5\times {10}^6\text{kg}/{\text{m}}^{\text{2}}\cdot \text{s}\right)}^2}\\ =0.823\end{array}$

Conclusion:

The intensity reflection coefficient between transducer material and gel which is identical to water is $0.823$.

To determine

(c)

The reason behind the use of gel.

Explanation of Solution

Introduction:

In a medical application, during ultrasound a transducer is used to emit the ultrasonic waves which enter into the body due to which high vibration are produced by the piezoelectric effect. The entered waves produce voltage pulses and are recorded for examination.

There is air between the transducer and the body, but during examination, the air is replaced by the gel. This is replaced due to the fact that the reflections also reduced with the air. The gel produces minimum reflection during ultrasound and produce accurate result.

Conclusion:

The gel is used to minimize the reflections.