JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Geometrical Optics

A ray of light passing through an equilateral prism is having velocity

2.12 \times 1 0^{8}

m/s in the prism material, then the minimum angle of deviation is ______ degrees.

(A)45
(B)30✓ Correct
(C)28
(D)58

Answer: (B)

Solution

The refractive index of the prism material is

μ = \frac{c}{v} = \frac{3 \times 1 0 ^{8}}{2.12 \times 1 0 ^{8}} = 2

. For an equilateral prism

A = 6 0^{\circ}

, and at minimum deviation

μ = \frac{sin ( \frac{A + δ _{m}}{2} )}{sin \frac{A}{2}}

. Thus

2 = \frac{sin ( \frac{6 0 ^{\circ} + δ _{m}}{2} )}{sin 3 0 ^{\circ}}

, so

sin (\frac{6 0 ^{\circ} + δ _{m}}{2}) = 2 \times \frac{1}{2} = \frac{1}{2}

. Hence

\frac{6 0 ^{\circ} + δ _{m}}{2} = 4 5^{\circ}

, giving

δ_{m} = 3 0^{\circ}

. The minimum angle of deviation is

3 0^{\circ}

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option

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