Answer
The distance of the goggles from the edge of the pool is $~~4.7~m$
Work Step by Step
We can find the angle from the normal where the light ray meets the water surface:
$tan~\theta_1 = \frac{2.0~m}{1.0~m}$
$\theta_1 = tan^{-1}~(\frac{2.0~m}{1.0~m})$
$\theta_1 = 63.435^{\circ}$
We can find the angle from the normal the light ray makes after entering the water:
$n_2~sin~\theta_2 = n_1~sin~\theta_1$
$sin~\theta_2 = \frac{n_1~sin~\theta_1}{n_2}$
$\theta_2 = sin^{-1}~(\frac{n_1~sin~\theta_1}{n_2})$
$\theta_2 = sin^{-1}~(\frac{1.00~sin~63.435^{\circ}}{1.33})$
$\theta_2 = 42.26^{\circ}$
We can find the horizontal distance $x$ the light ray travels in the water:
$\frac{x}{3.0~m} = tan~42.26^{\circ}$
$x = (3.0~m)(tan~42.26^{\circ})$
$x = 2.7~m$
The total distance of the goggles from the edge of the pool is $~2.7~m+2.0~m~~$ which is $~~4.7~m$
