Question

A ray of light incident normally on face AB of an isosceles prism travels as shown in figure. The least value of the refractive index the prism must be:

Answer 1

The ray Pq is incident normally incident on the face AB, So it is normally reflected and the ray QR strikes the face AC at an angle of incidence ${45}^0$. Doesn't go under refraction but is normally reflected at the face AC.

This gives a maximum value of the critical angle is ${45}^{\circ }$

Thus, $\sin c=\sin {45}^{\circ }=\frac{1}{\sqrt{2}}\Rightarrow \mu =\frac{1}{\sin c}=\sqrt{2}\Rightarrow {\mu }_{min}=\sqrt{2}$