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← JEE Advanced 2021 · Paper 2
PhysicsMultiple correct+4 / -2Question 3 of 57

JEE Advanced 2021 · Paper 2, Physics question 3

For a prism of prism angle θ=60∘\theta = 60^\circ, the refractive indices of the left half and the right half are, respectively, n1n_1 and n2n_2 (n2≥n1)(n_2 \ge n_1) as shown in the figure. The angle of incidence ii is chosen such that the incident light rays will have minimum deviation if n1=n2=n=1.5n_1 = n_2 = n = 1.5. For the case of unequal refractive indices, n1=nn_1 = n and n2=n+Δnn_2 = n + \Delta n (where Δn≪n\Delta n \ll n), the angle of emergence e=i+Δee = i + \Delta e. Which of the following statement(s) is(are) correct?

An equilateral prism (apex angle θ = 60°) split by a vertical dashed line from the apex into a left half of index n1 and a right half of index n2. A ray strikes the left face at angle of incidence i to the normal.
Choose one or more options
Answer and worked solution

Official answer: B, C

Uniform prism at minimum deviation. With n=1.5n=1.5, the ray inside is parallel to the base, r1=r2=30∘r_1=r_2=30^\circ, and
sin⁡i=1.5sin⁡30∘=0.75,cos⁡i≈0.661.\sin i=1.5\sin30^\circ=0.75,\qquad\cos i\approx0.661.

The internal interface. The interface is the vertical line from the apex. The ray inside is horizontal, so it meets the interface along the normal and is not bent, whatever n2n_2 is. It therefore reaches the second face at 30∘30^\circ, now inside n2n_2:
sin⁡e=(n+Δn)sin⁡30∘.\sin e=(n+\Delta n)\sin30^\circ.

Linearise. With e=i+Δee=i+\Delta e, sin⁡i+cos⁡i Δe=n2+Δn2\sin i+\cos i\,\Delta e=\dfrac n2+\dfrac{\Delta n}2, so
Δe=Δn2cos⁡i≈0.756 Δn.\Delta e=\frac{\Delta n}{2\cos i}\approx0.756\,\Delta n.

Check each option.

  • (A) Δe<Δn\Delta e<\Delta n. False.
  • (B) Δe∝Δn\Delta e\propto\Delta n. True.
  • (C) For Δn=2.8×10−3\Delta n=2.8\times10^{-3}, Δe≈2.12\Delta e\approx2.12 mrad, which is between 2.0 and 3.0. True.
  • (D) False.

(B), (C)

Topics:Prism and dispersionRefraction at plane surfaces
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