The angle of deviation through a prism is minimum when
(A) Incident ray and emergent ray are symmetric to the prism.
(B) The refracted ray inside the prism becomes parallel to its base
(C) Angle of incidence is equal to that of the angle of emergence
(D) When angle of emergence is double the angle of incidence
Choose the correct answer from the options given below :

Solution

Physics sir · 3 steps

At minimum deviation the ray goes through the prism symmetrically, so i equals e and r1 equals r2.

1. Deviation formula. For a prism of angle A, the deviation is δ = i + e - A, with r₁ + r₂ = A. As i increases, δ first falls, reaches a minimum, and then rises again.
2. Symmetric path. If you reverse the ray, i and e swap but δ stays the same. The minimum is a single point, so it can only occur where i = e. This also makes r₁ = r₂ = A/2.

   i = e, r₁ = r₂ = A/2, δₘ = 2i - A

   Statement	At minimum deviation?	Reason
   (A) Incident and emergent rays symmetric	True	i = e, so both rays make equal angles with the two faces
   (B) Ray inside parallel to the base	True	r₁ = r₂ means the ray meets both faces at equal angles. In an isosceles prism that ray runs parallel to the base
   (C) i = e	True	This is the defining condition
   (D) e = 2i	False	This breaks the symmetry, so δ is not at its minimum

3. So. Statements (A), (B) and (C) are all true, which is option (c).

The idea to keep. Minimum deviation means a symmetric path: i = e, r₁ = r₂ = A/2, and the ray inside runs parallel to the base. Use μ = (sin(A+δₘ)/2)/(sin A/2) only at this point.

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