JEE Advanced 2023 Paper 1 Q02 Physics Waves & Optics Ray Optics Hard

JEE Advanced 2023 Paper 1 · Q02 · Ray Optics

A plane polarized blue light ray is incident on a prism such that there is no reflection from the surface of the prism. The angle of deviation of the emergent ray is $\delta = 60°$ (see Figure-1). The angle of minimum deviation for red light from the same prism is $\delta_{\min} = 30°$ (see Figure-2). The refractive index of the prism material for blue light is $\sqrt{3}$. Which of the following statement(s) is(are) correct?

  1. A. The blue light is polarized in the plane of incidence
  2. B. The angle of the prism is $45°$
  3. C. The refractive index of the material of the prism for red light is $\sqrt{2}$
  4. D. The angle of refraction for blue light in air at the exit plane of the prism is $60°$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C, D

Solution

No reflection at incidence means Brewster's condition: $\mu = \tan i_p \Rightarrow \sqrt{3} = \tan i_p \Rightarrow i_p = 60°$. Reflected light is polarized perpendicular to plane of incidence; for no reflection, incident wave is polarized in the plane of incidence — (A) correct. Using $\delta = i + e - A$ with $i = 60°$ and $\delta = 60°$ gives $e = A$. Snell at entry: $\sin 60° = \sqrt{3}\sin r_1 \Rightarrow r_1 = 30°$, so $r_2 = A - 30°$ and at exit $\sin A = \sqrt{3}\sin(A - 30°) \Rightarrow \tan A = \sqrt{3} \Rightarrow A = 60°$. (B) is wrong. $e = 60°$ — (D) correct. For red light at minimum deviation: $\mu_R = \sin((A+\delta_{\min})/2)/\sin(A/2) = \sin 45°/\sin 30° = \sqrt{2}$ — (C) correct.

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