Answer
none
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.
Let $\theta_3$ be the refracted angle in air.
Note that the incident angle at the point of first reflection is $90^{\circ}-37.0^{\circ}$ which is $\theta_i = 53.0^{\circ}$
We can use Snell's law to find $\theta_3$:
$n_3~sin~\theta_3 = n_2~sin~\theta_i$
$sin~\theta_3 = \frac{n_2~sin~\theta_i}{n_3}$
$sin~\theta_3 = \frac{1.56~sin~53.0^{\circ}}{1.00}$
$sin~\theta_3 = 1.25$
Since there is no angle such that $sin~\theta_3 = 1.25$, there is no refraction at the point of first reflection.
