A composite transparent block consists of two media, Medium 1 and...

Applying Snell's law at the first horizontal interface (air to Medium 1):

$1 \cdot \sin 45^\circ = \mu_1 \sin r_1$

$1 \cdot \dfrac{1}{\sqrt{2}} = \sqrt{2} \sin r_1$

$\sin r_1 = \dfrac{1}{2} \Rightarrow r_1 = 30^\circ$

The angle of refraction $r_1$ is with respect to the vertical normal.

The ray then strikes the vertical interface between Medium 1 and Medium 2. The normal to this vertical interface is horizontal.

The angle of incidence $i_2$ at this second interface is $90^\circ - r_1$:

$i_2 = 90^\circ - 30^\circ = 60^\circ$

Applying Snell's law at the second vertical interface (Medium 1 to Medium 2):

$\mu_1 \sin i_2 = \mu_2 \sin r_2$

$\sqrt{2} \sin 60^\circ = \sqrt{6} \sin r_2$

$\sqrt{2} \cdot \dfrac{\sqrt{3}}{2} = \sqrt{6} \sin r_2$

$\dfrac{\sqrt{6}}{2} = \sqrt{6} \sin r_2$

$\sin r_2 = \dfrac{1}{2} \Rightarrow r_2 = 30^\circ$

The angle of refraction in Medium 2 is $30^\circ$.

Answer: $30^\circ$