Answer
$d = 4.0~m$
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:
$\frac{4.0~m}{d} = tan~44.95^{\circ}$
$d = \frac{4.0~m}{tan~44.95^{\circ}}$
$d = 4.0~m$
