JEE Main 2024 · 31 January · morning · Single correct · +4 / −1

The refractive index of a prism with apex angle A is cot A / 2. The angle of minimum deviation is :

  1. (a) δₘ=180°-3 A
  2. (b) δₘ=180°-4 A
  3. (c) δₘ=180°-2 A
  4. (d) δₘ=180°-A
Answer and solution

Answer (c) δₘ=180°-2 A

Solution

Physics sir · 4 steps

Write cot A/2 as cos A/2 over sin A/2; then match it to the prism formula.

  1. Prism formula. At minimum deviation, the refractive index of a prism is given by the formula below.

    μ = sin((A+δₘ)/2)/(sin A/2) = cot A/2 = (cos A/2)/(sin A/2)

  2. Cancel the denominator. Both sides have sin A/2 in the denominator, so sin((A+δₘ)/2) = cos A/2 = sin(90° - A/2).
  3. Equate the angles. (A+δₘ)/2 = 90° - A/2. Multiply by 2 to get A + δₘ = 180° - A.
  4. So. Moving A to the right side gives δₘ = 180° - 2A, which is option (c).

The idea to keep. When μ is given as a trig function of A/2, write it as a ratio over sin A/2. Then sin(A+δₘ)/2 equals the numerator directly.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Ray Optics

Asked about 200 times in JEE Main since 2020. This question is from 2024.