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JEE Advanced Physics Ray Optics 2021 JEE Advanced 2021 (Paper 2)

JEE Advanced Physics Question (2021) — Solution

Question

For a prism of prism angle  θ = 60 ° , the refractive indices of the left half and the right half are, respectively,  n 1  and  n 2 ,  n 2 ≥ n 1  as shown in the figure. The angle of incidence  i  is chosen such that the incident light rays will have minimum deviation if  n 1 = n 2 = n = 1 . 5 .  For the case of unequal refractive indices,  n 1 = n  and  n 2 = n + ∆ n  (where ∆ n < < n ), the angle of emergence  e = i + ∆ e . Which of the following statement(s) is(are) correct?

Options

  1. A. The value of  ∆ e  (in radians) is greater than that of  ∆ n .
  2. B. ∆ e  is proportional to  ∆ n .
  3. C. ∆ e  lies between  2 . 0  and  3 . 0  milliradian, if  ∆ n = 2 . 8 × 10 - 3 .
  4. D. ∆ e  lies between  1 . 0  and  1 . 6  milliradians, if  ∆ n = 2 . 8 × 10 - 3

Answer

C. ∆ e  lies between  2 . 0  and  3 . 0  milliradian, if  ∆ n = 2 . 8 × 10 - 3 .

Step-by-step solution

Given,  n 1 = n 2 = n = 3 2 In case of minimum deviation, r m = A 2 = 60 ° 2 = 30 ° and  i = e Applying expression for minimum deviation, n = sin δ min + A 2 sin A 2 ⇒ 3 2 = sin ( i ) sin 60 ° 2 ⇒ sin i = 3 4 ,    cos i = 7 4 If  n 1 = n = 3 2  and  n 2 = n + ∆ n : Angle of emergence,  e = i + Δ e Applying Snell's law, we get, 1 × sin i = n sin 30 ° Then, n + Δ n sin 30 ° = 1 sin i + Δ e Solving both equations we get, 1 2 Δ n = sin i + Δ e - sin i Δ n 2 = sin i + Δ e - sin i Δ e × Δ e Now using the definition of derivative, d sin i d i = cos i ,  For small change,  ∆ e ≈ ∆ i . Therefore, Δ n 2 = cos i Δ e ⇒ Δ n 2 = 7 4 Δ e ⇒ Δ n Δ e = 7 2 = 2 . 64 2 Δ n Δ e = 1 . 34 > 1 ⇒ Δ n > Δ e , Hence, option (A) is incorrect. Δ n = 1 . 34 Δe ⇒ Δe ∝ Δ n Hence, option (B) is correct. Δ n Δ e = 7 2 If   Δ n = 2 . 8 × 10 - 3 , ⇒ 2 . 8 × 10 - 3 Δ e = 7 2 ⇒ Δ e = 5 . 6 × 10 - 3 7 = 7 × 0 . 8 × 10 - 3 ⇒ Δ e = 2 . 64 × 0 . 8 × 10 - 3 = 2 . 11 × 10 - 3   rad Option (C) is correct.

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