Chapter 18, Problem 49A

To determine

The angle of the light reaching the edge of the pool leaves the water.

To identify: Whether the light viewed from this angle appear deeper or shallower than its actual size.

Answer to Problem 49A

The angle of the light is $53°$ .

The light viewed from this angle will appear 1.1 m shallower than its actual size.

Explanation of Solution

Given:

The thickness of the plastic sheet is $25\text{ mm}$ .

The angle of the light day strikes and leaves the plastic sheet is $45°$

The given 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 index of refraction for water is $n_{\text{w}}=1.33$ and the index of refraction for air is $n_{\text{A}}=1.00$ .

Sketch the situation as shown below.

  

The angle of the light source with the edge is,

  $\begin{array}{l}\tan {\theta }_{\text{i}}=\frac{1.5\text{ m}}{2.0\text{ m}}\\ {\theta }_{\text{i}}={\tan }^{-1}\left(0.75\right)\\ {\theta }_{\text{i}}=37°\end{array}$

The angle of the light reaching the edge of the pool while leaving the water is,

  $\begin{array}{l}{n}_{\text{A}}\sin {\theta }_1=n_{\text{w}}\sin {\theta }_{\text{i}}\\ {\theta }_1={\sin }^{-1}\left(\frac{n_{\text{w}}\sin {\theta }_{\text{i}}}{n_{\text{A}}}\right)\\ {\theta }_1={\sin }^{-1}\left(\frac{1.33\times \sin 37°}{1.00}\right)\\ {\theta }_1=53°\end{array}$

The adjacent side of the light viewed from the angle of $53°$ is,

  $\begin{array}{l}\tan {\theta }_1=\frac{\text{opposite side}}{\text{adjacent side}}\\ \text{adjacent side}=\frac{\text{1}\text{.5 m}}{\tan 53°}\\ \text{adjacent side}=1.1\text{ m}\end{array}$

The light viewed from this angle will appear $1.1\text{ m}$ shallower than its actual size.

Conclusion:

Thus, the angle of the light is $53°$ . Also, the light will appear shallower than its actual size.