A beam of polychromatic light passes through a thin prism of prism angle 6°. The refractive index of the material of the prism varies with wavelength \((\lambda)\) as \(n(\lambda)=\alpha \lambda+\frac{\beta}{\lambda^{2}}\), where \(\alpha=3 \mu \mathrm{~m}^{-1}\) and \(\beta=0.096 \mu \mathrm{~m}^{2}\). If \(\lambda_{\text{min}}\) is the wavelength at which the angle of minimum deviation \(D_{m}\) is smallest, then the correct value of \(D_{m}\) at \(\lambda_{\text{min}}\) is

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Geometrical Optics

A beam of polychromatic light passes through a thin prism of prism angle 6°. The refractive index of the material of the prism varies with wavelength

(λ)

n (λ) = α λ + \frac{β}{λ ^{2}}

, where

α = 3 μ m^{- 1}

and

β = 0.096 μ m^{2}

. If

λ_{min}

is the wavelength at which the angle of minimum deviation

D_{m}

is smallest, then the correct value of

D_{m}

λ_{min}

Answer: (B)

For a thin prism,

δ = A (μ - 1) = A (α λ + \frac{β}{λ ^{2}} - 1)

.

For minimum deviation:

\frac{d δ}{d λ} = 0

, giving

α - \frac{2 β}{λ ^{3}} = 0

, so

λ^{3} = \frac{2 β}{α} = \frac{2 \times 0.096}{3} = 0.064

, thus

λ = 0.4 μ m

.

Substituting back:

n = α λ + \frac{β}{λ ^{2}} = \frac{2 β}{λ ^{3}} λ + \frac{β}{λ ^{2}} = \frac{3 β}{λ ^{2}} = \frac{3 \times 0.096}{0.16} = 1.8

.

Therefore,

δ_{m i n} = 6° (1.8 - 1) = 4.8°

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Single correct option