Answer
The angle of refraction is $40^{\circ}$
Work Step by Step
Let $\theta_2$ be the refracted angle inside the block.
We can use Snell's law to find $\theta_2$:
$n_2~sin~\theta_2 = n_1~sin~\theta_1$
$sin~\theta_2 = \frac{n_1~sin~\theta_1}{n_2}$
$\theta_2 = sin^{-1}~(\frac{n_1~sin~\theta_1}{n_2})$
$\theta_2 = sin^{-1}~(\frac{1.00~sin~40^{\circ}}{1.56})$
$\theta_2 = sin^{-1}~(0.41204)$
$\theta_2 = 24.3^{\circ}$
If the ray travels a horizontal distance of $W = 3.00~cm$, we can find the vertical distance $y$ that the ray travels:
$\frac{y}{W} = tan~\theta_2$
$y = W~tan~\theta_2$
$y = (3.00~cm)~tan~24.3^{\circ}$
$y = 1.36~cm$
Since $y \lt H$, then the point of first reflection is on side 3.
Let $\theta_3$ be the refracted angle in the air.
We can use Snell's law to find $\theta_3$:
$n_3~sin~\theta_3 = n_2~sin~\theta_2$
$sin~\theta_3 = \frac{n_2~sin~\theta_2}{n_3}$
$\theta_3 = sin^{-1}~(\frac{n_2~sin~\theta_2}{n_3})$
$\theta_3 = sin^{-1}~(\frac{1.56~sin~24.3^{\circ}}{1.00})$
$\theta_3 = sin^{-1}~(0.6420)$
$\theta_3 = 40^{\circ}$
The angle of refraction is $40^{\circ}$
