Ray Optics

JEE Main level, single correct, medium

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Question 1 of 295

A ray of light undergoes minimum deviation through an equilateral glass prism of refractive index μ=3\mu = \sqrt{3}. The angle of minimum deviation δm\delta_m and angle of incidence ii are respectively:

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Answer: (A) δm=60∘,i=60∘\delta_m = 60^\circ, i = 60^\circ

At minimum deviation, r=A/2=30∘r = A/2 = 30^\circ. By Snell's Law, sin⁡i=μsin⁡r=3×12=32  ⟹  i=60∘\sin i = \mu \sin r = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2} \implies i = 60^\circ. Then δm=2i−A=2(60∘)−60∘=60∘\delta_m = 2i - A = 2(60^\circ) - 60^\circ = 60^\circ.

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