The exit surface of a prism with refractive index $n$ is coated with a material having refractive in

Question

The exit surface of a prism with refractive index $n$ is coated with a material having refractive index $\frac{n}{2}$. When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is $\_\_\_\_$

Professor PC Thomas Classes · Answer

In the minimum deviation position, the ray passes symmetrically through the prism, so the angles of refraction at the two faces are equal: $r_1 = r_2 = \frac{A}{2}$. At the exit face, the ray exactly meets the critical angle condition for the interface between the prism (index $n$) and the coating (index $n/2$). Using Snell's law, $n \sin(r_2) = \frac{n}{2} \sin(90^\circ)$. This gives $\sin(r_2) = \frac{1}{2} \implies r_2 = 30^\circ$. Since $r_2 = \frac{A}{2}$, the prism angle is $A = 2 	imes 30^\circ = 60^\circ$.