JEE Main 2021 · 25 July · evening · Single correct · +4 / −1

A prism of refractive index μ and angle of prism A is placed in the position of minimum angle of deviation. If minimum angle of deviation is also A, then in terms of refractive index

  1. (a) 2cos ⁻¹( μ/2 )
  2. (b) sin ⁻¹( μ/2 )
  3. (c) sin ⁻¹( √((μ - 1)/2) )
  4. (d) cos ⁻¹( μ/2 )
Answer and solution

Answer (a) 2cos ⁻¹( μ/2 )

Solution

Physics sir · 4 steps

Put δₘ = A in the prism formula, and the double-angle identity hands you A directly.

  1. Prism formula. At minimum deviation the ray passes through the prism symmetrically. The refractive index is then μ = sin((A+δₘ)/2)/sin(A/2).
  2. Set deviation to A. With δₘ = A, the numerator becomes sin(2A/2) = sin A. Write sin A = 2 sin A/2 cos A/2.

    μ = (2 sin A/2 cos A/2)/(sin A/2) = 2 cos A/2

  3. Solve for A. cos A/2 = μ/2, so A/2 = cos⁻¹(μ/2) and A = 2 cos⁻¹(μ/2).
  4. So. The prism angle is A = 2 cos⁻¹(μ/2), which is option (a).

The idea to keep. Whenever δₘ is given in terms of A, substitute it into the prism formula first, then use sin A = 2 sin A/2 cos A/2 to cancel the denominator.

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

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Asked about 200 times in JEE Main since 2020. This question is from 2021.