JEE Main 2026 · 28 January · evening · Single correct · +4 / −1

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index n of the prism satisfies.

  1. (a) n≥ 2
  2. (b) √2<n<2
  3. (c) 1<n<2
  4. (d) √2<n<2√2
Answer and solution

Answer (b) √2<n<2

Solution

Physics sir · 4 steps

Put δₘ = A into the prism formula and n becomes 2 cos(A/2). Then ask what values A can take.

  1. Prism formula. At minimum deviation, n = sin((A+δₘ)/2)/sin(A/2). With δₘ = A, the top becomes sin A = 2 sin A/2 cos A/2.

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

  2. Limit on A. At minimum deviation the angle of incidence is i = (A+δₘ)/2 = A. An angle of incidence cannot exceed 90°, so 0 < A ≤ 90° and 0 < A/2 ≤ 45°.
  3. Range of n. Over this range cos A/2 goes from just below 1 down to 1/√2. So 2 cos A/2 lies between √2 and 2. The end points belong to limiting cases: A = 90° means grazing incidence, and A → 0 means no prism at all.
  4. So. The refractive index must satisfy √2 < n < 2, which is option (b).

The idea to keep. Whenever a prism question fixes δₘ, also find the angle of incidence i = (A+δₘ)/2. The condition i ≤ 90° is what gives you the hidden bound.

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 2026.