Chapter 22, Problem 63E

(a)

To determine

The percentage of intensity being reflected.

Answer to Problem 63E

The percentage of intensity being reflected is $\underset{\_}{99.89\%}$.

Explanation of Solution

Write the expression for ratio of reflected intensity to incident intensity.

    $\frac{I_r}{I_i}={\left(\frac{{\rho }_1{c}_1-{\rho }_2{c}_2}{{\rho }_1{c}_1+{\rho }_2{c}_2}\right)}^2$        (I)

Here, $I_r$ is the intensity of the reflected wave, $I_i$ is the intensity of the incident wave, ${\rho }_1$ is the density of air medium, $c_1$ is the speed of the sound wave in air medium 1, ${\rho }_2$ is the density of water medium, and $c_2$ is the speed of sound in water medium.

Write the expression for intensity reflected.

    $p=\frac{I_r}{I_i}\times 100\%$        (II)

Here, $p$ is the percentage of intensity reflected.

Conclusion:

Substitute $1.20\text{kg}/{\text{m}}^3$ for ${\rho }_1$, $344\text{m}/\text{s}$ for $c_1$, $998\text{kg}/{\text{m}}^3$ for ${\rho }_2$, and $1498\text{m}/\text{s}$ for $c_2$ in equation (I), to find $\frac{I_r}{I_i}$.

    $\begin{array}{c}\frac{I_r}{I_i}={\left(\frac{\left(1.20\text{kg}/{\text{m}}^3\right)\left(344\text{m}/\text{s}\right)-\left(998\text{kg}/{\text{m}}^3\right)\left(1498\text{m}/\text{s}\right)}{\left(1.20\text{kg}/{\text{m}}^3\right)\left(344\text{m}/\text{s}\right)+\left(998\text{kg}/{\text{m}}^3\right)\left(1498\text{m}/\text{s}\right)}\right)}^2\\ =\left(\frac{412.8-1495004}{412.8+1495004}\right)\\ =0.998896\end{array}$

Substitute $0.998896$ for $\frac{I_r}{I_i}$ in equation (II), to find $p$.

    $\begin{array}{c}p=\left(0.998896\right)\times 100\%\\ =99.8896\%\\ \approx 99.89\%\end{array}$

Therefore, the percentage of intensity being reflected is $\underset{\_}{99.89\%}$.

(b)

To determine

The percentage of intensity being transmitted.

Answer to Problem 63E

The percentage of intensity being transmitted is $\underset{\_}{11\%}$.

Explanation of Solution

Write the expression for transmitted intensity.

    $I_t=I_i-I_r$

Here, $I_t$ is the transmitted intensity.

Write the expression for transmitted intensity.

    $p=\frac{\left(I_i-I_r\right)}{I_i}\times 100\%$

Here, $p$ is the percentage of transmitted intensity.

Rearrange the above equation.

    $\begin{array}{c}p=\left(\frac{I_i}{I_i}-\frac{I_r}{I_i}\right)100\%\\ =\left(1-\frac{I_r}{I_i}\right)100\%\end{array}$        (III)

From subpart (a), the ratio of $\frac{I_r}{I_i}$ is obtained as $0.998896$.

Use the above condition in equation (III).

    $\begin{array}{c}p=\left(1-0.998896\right)100\%\\ =\left(0.001104\right)100\%\\ =11.04\%\\ \approx 11\%\end{array}$

Conclusion:

Therefore, The percentage of intensity being transmitted is $\underset{\_}{11\%}$.