# Chapter 34 - Ray Optics - Exercises and Problems - Page 992: 51

$d = 4.0~m$

#### Work Step by Step

We can find the angle the light rays make with the normal after entering the water: $n_2~sin~\theta_2 = n_1~sin~\theta_1$ $sin~\theta_2 = \frac{n_1~sin~\theta_1}{n_2}$ $\theta_2 = sin^{-1}~(\frac{n_1~sin~\theta_1}{n_2})$ $\theta_2 = sin^{-1}~(\frac{1.00~sin~70^{\circ}}{1.33})$ $\theta_2 = 44.95^{\circ}$ We can find the depth $d$ of the pool: $\frac{4.0~m}{d} = tan~44.95^{\circ}$ $d = \frac{4.0~m}{tan~44.95^{\circ}}$ $d = 4.0~m$

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