A parallel sides slab ABCD of refractive index 2 is sandwiched between two slabs of refractive indices $\sqrt{2}$ and $\sqrt{3}$ as shown in the Fig. The minimum value of angle $\mathrm{\theta }$ (in degrees) such that the ray PQ suffers total internal reflection at both the surfaces AB and CD is

Refer to Fig.

For total internal reflection at surface AB, angle $\mathrm{\theta }$ must be greater than or equal to the critical angle i₁ given by

$\sin {\mathrm{i}}_1=\frac{{\mathrm{\mu }}_2}{{\mathrm{\mu }}_1}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

which gives ${\mathrm{i}}_1={45}^{\circ }$

For total internal reflection at surface CD, angle $\mathrm{\theta }$ must be greater than or equal to the critical angle i₂ given by

$\sin {\mathrm{i}}_2=\frac{{\mathrm{\mu }}_3}{{\mathrm{\mu }}_1}=\frac{\sqrt{3}}{2}$

which gives ${\mathrm{i}}_2={60}^{\circ }$

Hence, for total internal reflection at both the surfaces AB and CD, the minimum value of $\mathrm{\theta }$ = 60°.