The exit surface of a prism with refractive index n is coated with a material having refractive index n2. When this prism is set for minimum angle of deviation, it exactly meets the condition of crit…

Question

The exit surface of a prism with refractive index n is coated with a material having refractive index n2. When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is \_ \_ \_ \_

Accepted Answer

Correct answer: D. (d) Let the refractive index of the prism be n_1=n and the refractive index of the coating on the exit surface be n_2= n2. At the condition of minimum deviation, the angle of refraction at the first surface r_1 and the angle of incidence at the second surface r_2 are equal, i.e., r_1=r_2= A2, where A is the prism angle. The problem states that at the exit surface, the light meets the condition of critical angle. There fore, r_2= _c. The critical angle _c for the interface between the prism and the coating is given by sin ⁡ _c= n_2n_1. Substituting the values, sin ⁡ _c= n/2n= 12. Thus, _c=30^ . Since r_2= _c, we have r_2=30^ . Using the relation r_1=r_2= A2 for minimum deviation, we get A2=30^ . Therefore, A=60^ .

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

Solution

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