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MCQEasyJEE 2026Prisms & Total Internal Reflection

JEE Physics 2026 Question with Solution

The exit surface of a prism with refractive index nn is coated with a material having refractive index n2\dfrac{n}{2}. When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is _____.

  • A

    30∘30^\circ

  • B

    60∘60^\circ

  • C

    15∘15^\circ

  • D

    45∘45^\circ

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The prism has refractive index nn and the coating has refractive index n2\dfrac{n}{2}. The prism is set for minimum deviation and the emergent face is at the critical-angle condition.

Find: The prism angle AA.

At minimum deviation, the ray travels symmetrically inside the prism, so the refraction angles at the two faces are equal.

A=2rA = 2r

For the prism-to-coating interface, the critical angle CC satisfies

sin⁡C=n2n=12\sin C = \dfrac{\dfrac{n}{2}}{n} = \dfrac{1}{2}

Hence,

C=30∘C = 30^\circ

Since the prism exactly meets the critical-angle condition at the exit surface,

r=C=30∘r = C = 30^\circ

Therefore,

A=2r=2×30∘=60∘A = 2r = 2 \times 30^\circ = 60^\circ

So, the prism angle is 60∘60^\circ and the correct option is B.

Common mistakes

  • Assuming the critical angle is for air instead of the coating. That is wrong because the exit surface is coated with a material of refractive index n2\dfrac{n}{2}, so the interface is prism-to-coating. Use sin⁡C=n/2n\sin C = \dfrac{n/2}{n}.

  • Using A=rA = r at minimum deviation. That is wrong because at minimum deviation the internal path is symmetric and the two refraction angles are equal, giving A=r1+r2=2rA = r_1 + r_2 = 2r.

  • Taking the critical angle condition at the first face. That is wrong because the question states the exit surface exactly meets the condition of critical angle. Apply the critical-angle relation at the emergent face.

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