Ray Optics

JEE Main level, single correct, medium

237 of 295

Question 237 of 295

The exit surface of a prism with refractive index nn is coated with a material having refractive index n2\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 ____\_\_\_\_.

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Answer: (D) 60∘60^{\circ}

Let the refractive index of the prism be n1=nn_1 = n and the refractive index of the coating on the exit surface be n2=n2n_2 = \frac{n}{2}.
At the condition of minimum deviation, the angle of refraction at the first surface r1r_1 and the angle of incidence at the second surface r2r_2 are equal, i.e., r1=r2=A2r_1 = r_2 = \frac{A}{2}, where AA is the prism angle.
The problem states that at the exit surface, the light meets the condition of critical angle. Therefore, r2=θcr_2 = \theta_c.
The critical angle θc\theta_c for the interface between the prism and the coating is given by sin⁡θc=n2n1\sin \theta_c = \frac{n_2}{n_1}.
Substituting the values, sin⁡θc=n/2n=12\sin \theta_c = \frac{n/2}{n} = \frac{1}{2}.
Thus, θc=30∘\theta_c = 30^\circ.
Since r2=θcr_2 = \theta_c, we have r2=30∘r_2 = 30^\circ.
Using the relation r1=r2=A2r_1 = r_2 = \frac{A}{2} for minimum deviation, we get A2=30∘\frac{A}{2} = 30^\circ.
Therefore, A=60∘A = 60^\circ.

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