JEE Advanced 2026 Paper 2, Physics Section 1 Q3: Prism

JEE Advanced2026Paper 2Physics Section 1

A beam of polychromatic light passes through a thin prism of prism angle $6^{\circ}$ . The refractive index of the material of the prism varies with wavelength $(λ)$ as $n (λ) = α λ + \frac{β}{λ ^{2}}$ , where $α = 3 μ m^{- 1}$ and $β = 0.096 μ m^{2}$ . If $λ_{m i n}$ is the wavelength at which the angle of minimum deviation $D_{m}$ is smallest, then the correct value of $D_{m}$ at $λ_{m i n}$ is

A
$6. 4^{\circ}$
B
$4. 8^{\circ}$
C
$3. 2^{\circ}$
D
$2. 4^{\circ}$

Solution

Step 1 — Thin-prism deviation. For a thin prism, $D_{m} = (n - 1) A$ with $A = 6^{\circ}$ , so $D_{m} = 6^{\circ} (α λ + β / λ^{2} - 1)$ .

Step 2 — Minimise with respect to $λ$ .

$\frac{d D _{m}}{d λ} = 6^{\circ} (α - \frac{2 β}{λ ^{3}}) = 0 \Rightarrow λ^{3} = \frac{2 β}{α} = \frac{2 ( 0.096 )}{3} = 0.064$ .

$∴ λ_{m i n} = 0.4 μ m$ . The second derivative $\frac{6 β}{λ ^{4}} > 0$ confirms a minimum.

Step 3 — Refractive index at $λ_{m i n}$ . Using $α = \frac{2 β}{λ ^{3}}$ ,

$n = α λ + \frac{β}{λ ^{2}} = \frac{2 β}{λ ^{2}} + \frac{β}{λ ^{2}} = \frac{3 β}{λ ^{2}} = \frac{3 ( 0.096 )}{0.16} = 1.8$ .

Step 4 — Deviation. $D_{m} = (1.8 - 1) (6^{\circ}) = 4. 8^{\circ}$ , i.e. option (B).

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JEE Advanced 2026 Paper 2, Physics Section 1 Q3: Prism

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