JEE Advanced 2021 Paper 2 · Q03 · Geometrical Optics
For a prism of prism angle $\theta = 60^\circ$, the refractive indices of the left half and the right half are, respectively, $n_1$ and $n_2$ ($n_2 \geq n_1$) as shown in the figure. The angle of incidence $i$ is chosen such that the incident light rays will have minimum deviation if $n_1 = n_2 = n = 1.5$. For the case of unequal refractive indices, $n_1 = n$ and $n_2 = n + \Delta n$ (where $\Delta n \ll n$), the angle of emergence $e = i + \Delta e$. Which of the following statement(s) is(are) correct?
[Figure: Triangular prism of apex angle theta, divided vertically into two halves with refractive indices n1 (left) and n2 (right). Ray incident at angle i.]
Reveal answer + step-by-step solution
Correct answer:B, C
Solution
At minimum deviation with $n_1 = n_2 = 1.5$ and $\theta = 60^\circ$, $r_1 = r_2 = 30^\circ$. Snell at exit face: $n \sin 30^\circ = \sin e$, so $\sin e = 0.75$, $\cos e = \sqrt{7}/4$. When $n_2 = n + \Delta n$ with $n_1 = n$ unchanged, the path inside the prism is unaltered (deviation at internal interface is zero since $n_1 = n_2$ at that boundary path — only the exit face has new index). Differentiating $n_2 \sin 30^\circ = \sin e$: $\Delta n \cdot (1/2) = \cos e \cdot \Delta e = (\sqrt{7}/4)\Delta e$, so $\Delta e = (2/\sqrt{7})\Delta n \approx 0.756\,\Delta n$. Hence $\Delta e < \Delta n$ (A wrong); $\Delta e \propto \Delta n$ (B correct). For $\Delta n = 2.8\times 10^{-3}$: $\Delta e = 2 \times 2.8\times 10^{-3}/\sqrt{7} \approx 2.12\times 10^{-3}$ rad $= 2.12$ mrad, which lies in $[2.0, 3.0]$ — (C) correct, (D) wrong.
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