Question

A 4.0-m-wide swimming pool is filled to the top. The bottom of the pool becomes completely shaded in the afternoon when the sun is 20° above the horizon. How deep is the pool?

Short answer

The depth of the pool is 4.0 m

Step-by-step solution

Step 1. Given information is :Width of the swimming pool = 4.0 mBottom of the pool becomes completely shaded in the afternoon when sun is 20o above the horizonRefractive index of air n1 = 1Refractive index of water n2 = 1.33

We need to find the depth of the pool

Step 2. Using Snell's Law

As the Bottom of the pool becomes completely shaded in the afternoon when sun is 20^o above the horizon, the ray refracting from the top edge of the pool do not reach the bottom of the pool after refraction as shown below :

Using snell's Law,

$$\begin{gathered} n_1\sin {\theta }_1=n_2\sin {\theta }_2 \\ \sin {\theta }_2=\frac{n_1}{n_2}\sin {\theta }_1 \\ {\theta }_2={\sin }^{-1}\left(\frac{n_1}{n_2}\sin {\theta }_1\right) \\ {\theta }_2={\sin }^{-1}\left(\frac{1.00}{1.33}\sin 70°\right) \\ {\theta }_2=44.95° \end{gathered}$$

Let d be the depth of the pool. Then,

$$\begin{gathered} \tan {\theta }_2=\frac{4.0m}{d} \\ d=\frac{4.0m}{\tan \left(44.95\right)°}=4.0m \end{gathered}$$