A light ray is incident on a transparent slab of refractive index $\mu =\sqrt{2}$, at an angle of incidence $\pi /4$. Find the ratio of the lateral displacement suffered by the light ray to the maximum value which it could have suffered.

$\mathrm{\mu }=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}$

lateral shift is given by $\mathrm{x}=\frac{\mathrm{tsin}\left(\mathrm{i}-\mathrm{r}\right)}{\cos \mathrm{r}}$

Maximum lateral shift possible is t. 

Ratio of the lateral displacement suffered by the light ray to the maximum value which it could have suffered is 

$=\frac{\sin \left(\mathrm{i}-\mathrm{r}\right)}{\mathrm{cosr}}=\frac{\sin (45-30)}{\cos 30}$

$=\frac{\sqrt{3}-1}{\sqrt{6}}$