Ray Optics and Optical Instruments Part 3

Concept:At the exit surface, the light ray meets the critical angle condition. For a prism at minimum deviation, the internal angles of refraction are equal.Explanation:The refractive index of the prism is $n$. The refractive index of the coating is $\frac{n}{2}$.The critical angle $\theta_C$ at the exit surface is given by:$\sin \theta_C = \frac{\text{refractive index of rare medium}}{\text{refractive index of dense medium}} = \frac{n/2}{n} = \frac{1}{2}$.Therefore, $\theta_C = \sin^{-1}(1/2) = 30^{\circ}$.This critical angle equals the angle of incidence at the second face, so $r_2 = 30^{\circ}$.For minimum deviation, the ray passes symmetrically through the prism, giving $r_1 = r_2 = 30^{\circ}$.The prism angle $A$ is the sum of the two internal angles: $A = r_1 + r_2 = 30^{\circ} + 30^{\circ} = 60^{\circ}$.Answer:The prism angle is $60^{\circ}$, which corresponds to option A.