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MHT CET 2025 (19 Apr, Shift 1) · Prism, Dispersion & Scattering

Prism, Dispersion & Scattering question from MHT CET 2025 (19 Apr, Shift 1)

Q24MHT CET 2025 · 19 APR · SHIFT 1Medium

Prism, Dispersion & Scattering

For a thin prism, $\delta_1$ is the angle of deviation produced, when prism is placed in air. When the prism is immersed in water, the angle of deviation produced is $\delta_2$ . Given ${}_a\mu_g = \frac{3}{2}$ and ${}_a\mu_w = \frac{4}{3}$ . The ratio $\delta_2 : \delta_1$ is

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Answer: (b)

Thin prism: $\delta = (n_{\text{rel}} - 1)A$
$\delta_1 = \left(\frac{3}{2} - 1\right)A = \frac{A}{2}$ ; $\delta_2 = \left(\frac{3/2}{4/3} - 1\right)A = \frac{A}{8}$
$\delta_2 : \delta_1 = \frac{1}{8} : \frac{1}{2} = 1 : 4$ , so the answer is (b).

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