## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 34 - Ray Optics - Exercises and Problems - Page 990: 13

#### Answer

The sun's angle above the horizon is $~~31^{\circ}$

#### Work Step by Step

The sun's rays travel through the water at an angle of $40^{\circ}$ from the normal. We can use Snell's Law to find the ray's angle in the air with respect to the normal: $n_2~sin~\theta_2 = n_1~sin~\theta_1$ $sin~\theta_2 = \frac{n_1~sin~\theta_1}{n_2}$ $sin~\theta_2 = \frac{(1.33)~(sin~40^{\circ})}{1.0}$ $sin~\theta_2 = 0.8549$ $\theta_2 = sin^{-1}~(0.8549)$ $\theta_2 = 59^{\circ}$ We can find the ray's angle with respect to the horizontal: $90^{\circ}-59^{\circ} = 31^{\circ}$ The sun's angle above the horizon is $~~31^{\circ}$

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