Answer
The angle of refraction at the point of second reflection is $~~70^{\circ}$
Work Step by Step
Let $\theta_2$ be the refracted angle inside the block.
We can use Snell's law to find $\theta_2$:
$n_2~sin~\theta_2 = n_1~sin~\theta_1$
$sin~\theta_2 = \frac{n_1~sin~\theta_1}{n_2}$
$\theta_2 = sin^{-1}~(\frac{n_1~sin~\theta_1}{n_2})$
$\theta_2 = sin^{-1}~(\frac{1.00~sin~70^{\circ}}{1.56})$
$\theta_2 = sin^{-1}~(0.60237)$
$\theta_2 = 37.0^{\circ}$
If the ray travels a vertical distance of $H = 2.00~cm$, we can find the horizontal distance $x$ that the ray travels:
$\frac{H}{x} = tan~\theta_2$
$x = \frac{H}{tan~\theta_2}$
$x = \frac{2.00~cm}{tan~37.0^{\circ}}$
$x = 2.65~cm$
Since $x \lt W$, then the point of first reflection is on side 2.
Since $2x \gt W$, then the reflected ray meets side 3 before it is able to cross the vertical distance of $H = 2.00~cm$ down to side 4.
The point of second reflection is on side 3.
Let $\theta_4$ be the refracted angle in air at the point of second reflection.
By symmetry, the incident angle at the point of second reflection is $37.0^{\circ}$
We can use Snell's law to find $\theta_4$:
$n_4~sin~\theta_4 = n_2~sin~\theta_2$
$sin~\theta_4 = \frac{n_2~sin~\theta_2}{n_4}$
$sin~\theta_4 = \frac{1.56~sin~37.0^{\circ}}{1.00}$
$sin~\theta_4 = 0.9388$
$\theta_4 = sin^{-1}~(0.9388)$
$\theta_4 = 70^{\circ}$
The angle of refraction at the point of second reflection is $~~70^{\circ}$
