Chapter 18, Problem 100A

To determine

The direction that the refracted ray will travel above the surface of the water. The depth oftank bottom appeared.

To compare: The ratio of apparent depth to true depth with the index of refraction.

Answer to Problem 100A

Therefracted ray will travel above the surface of the water at $6.7°$ .

The depth at which the bottom of the tank appears is $9.0\text{cm}$ .

The ratio of apparent depth to true depth is equal to the index of refraction and is equal to $0.75$ .

Explanation of Solution

Given:

The aquarium along with ray diagram is shown below.

  

Formula used:

The expression for Snell’s law of refraction is,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, $n_1$ is the index of refraction of the first medium, ${\theta }_1$ is the angle of incidence of the first medium, $n_2$ is the index of refraction of the second medium, and ${\theta }_2$ is the angle of incidence of the second medium.

Calculation:

Consider the given figure.The angle of incidence of the water is ${\theta }_1=5.0°$ .

Consider the index of refraction of the water $n_{\text{1}}=1.33$ and index of refraction of the air $n_{\text{2}}=1.00$ .

The angle of refraction by using Snell’s law of refraction is,

  $\begin{array}{l}{n}_{\text{1}}\sin {\theta }_1=n_{\text{2}}\sin {\theta }_2\\ \sin {\theta }_2=\frac{n_{\text{1}}\sin {\theta }_1}{n_{\text{2}}}\\ {\theta }_2={\sin }^{-1}\left(\frac{\left(1.33\right)\sin 5.0°}{1}\right)\\ {\theta }_2=6.7°\end{array}$

Hence, the refracted ray will travel above the surface of the water at ${\theta }_2=6.7°$ .

Use the geometry of right angle triangle.

  $\left(\text{Actual depth}\right)\tan {\theta }_1=\left(\text{Apparent depth}\right)\tan {\theta }_2$

The depth at which the bottom of the tank to be appears is,

  $\begin{array}{l}\left(\text{Actual depth}\right)\tan {\theta }_1=\left(\text{Apparent depth}\right)\tan {\theta }_2\\ \text{Apparent depth}=\left(\text{Actual depth}\right)\frac{\tan {\theta }_1}{\tan {\theta }_2}\\ \text{Apparent depth}=\left(\text{12 cm}\right)\frac{\tan 5.0°}{\tan 6.67°}\\ \text{Apparent depth}=9.0\text{ cm}\end{array}$

The ratio of apparent depth to true depth is,

  $\begin{array}{l}\frac{\text{Apparent depth}}{\text{True depth}}=\frac{9.0\text{ cm}}{12.0\text{ cm}}\\ \frac{\text{Apparent depth}}{\text{True depth}}=0.75\end{array}$

The ratio of apparent depth to true depth is $0.75$ .

The index of refraction is

  $\begin{array}{l}\text{Index of refraction}=\frac{n_{\text{Air}}}{n_{\text{Water}}}\\ \frac{n_{\text{Air}}}{n_{\text{Water}}}=\frac{1.00}{1.33}\\ \frac{n_{\text{Air}}}{n_{\text{Water}}}=0.75\end{array}$

It is observed that the ratio of apparent depth to true depth is equal to the index of refraction.

Conclusion:

Thus, the refracted ray will travel above the surface of the water at $6.7°$ , the depth at which the bottom of the tank to be appears is $9.0\text{ cm}$ . Also, the ratio of apparent depth to true depth is equal to the index of refraction and is equal to $0.75$ .