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JEE Advanced 2025 Paper 2 · Physics · Ray and Wave Optics · medium

An equilateral prism of apex angle 60^° is made of glass whose refractive index varies with wavelength as n(λ)=1.50+(B)/(λ^2, with B=5.0×10^3 nm^2. A…

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An equilateral prism of apex angle 60∘60^\circ is made of glass whose refractive index varies with wavelength as n(λ)=1.50+Bλ2n(\lambda)=1.50+\dfrac{B}{\lambda^{2}}, with B=5.0×103 nm2B=5.0\times10^{3}\ \mathrm{nm^{2}}. A parallel beam containing the wavelengths 400 nm400\ \mathrm{nm} and 600 nm600\ \mathrm{nm} is incident on the prism from air, and each wavelength is rotated into its own position of minimum deviation. Find the difference, in degrees, between the minimum deviations of the two wavelengths. Give the answer correct to two decimal places.
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Answer: 1.53

The refractive indices are n1=1.50+50004002=1.53125n_1=1.50+\dfrac{5000}{400^2}=1.53125 for 400 nm400\ \mathrm{nm} and n2=1.50+50006002=1.51389n_2=1.50+\dfrac{5000}{600^2}=1.51389 for 600 nm600\ \mathrm{nm}. At minimum deviation the ray passes symmetrically, so n=sin⁡A+δm2sin⁡A2n=\dfrac{\sin\frac{A+\delta_m}{2}}{\sin\frac A2}, giving δm=2sin⁡−1 ⁣(nsin⁡30∘)−A\delta_m=2\sin^{-1}\!\left(n\sin30^\circ\right)-A. For n1n_1: δ1=2sin⁡−1(0.765625)−60∘=39.93∘\delta_1=2\sin^{-1}(0.765625)-60^\circ=39.93^\circ. For n2n_2: δ2=2sin⁡−1(0.756944)−60∘=38.39∘\delta_2=2\sin^{-1}(0.756944)-60^\circ=38.39^\circ. The difference is δ1−δ2≈1.53∘\delta_1-\delta_2\approx1.53^\circ. The shorter wavelength has the larger index and is deviated more. (Using the thin-prism formula δ=(n−1)A\delta=(n-1)A would give (n1−n2)A=0.0174×60∘≈1.04∘(n_1-n_2)A=0.0174\times60^\circ\approx1.04^\circ, which is not valid for a 60∘60^\circ prism.)
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