The refractive index of a prism with apex angle A is $\cot \mathrm{A} / 2$. The angle of minimum deviation is :

(D) $\delta_{m}=180^{\circ}-2 \mathrm{~A}$

(A) $\delta_{\mathrm{m}}=180^{\circ}-\mathrm{A}$

(B) $\delta_{\mathrm{m}}=180^{\circ}-3 \mathrm{~A}$

(C) $\delta_{\mathrm{m}}=180^{\circ}-4 \mathrm{~A}$

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Geometrical Optics

The refractive index of a prism with apex angle A is

cot A /2

. The angle of minimum deviation is :

(A) $δ_{m} = 18 0^{\circ} - A$
(B) $δ_{m} = 18 0^{\circ} - 3 A$
(C) $δ_{m} = 18 0^{\circ} - 4 A$
(D) $δ_{m} = 18 0^{\circ} - 2 A$ ✓ Correct

Answer: (D)

Solution

The angle of minimum deviation is related to the refractive index by

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

Given

μ = cot \frac{A}{2} = \frac{cos ( A /2 )}{sin ( A /2 )}

, so

sin (\frac{A + δ _{m}}{2}) = cos \frac{A}{2} = sin (9 0^{\circ} - \frac{A}{2})

\frac{A + δ _{m}}{2} = 9 0^{\circ} - \frac{A}{2} ⟹ δ_{m} = 18 0^{\circ} - 2 A

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option

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