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MHT CET 2026 (11 Apr, Shift 2) · Prism, Dispersion & Scattering

Prism, Dispersion & Scattering question from MHT CET 2026 (11 Apr, Shift 2)

Q10MHT CET 2026 · 11 APR · SHIFT 2Medium

Prism, Dispersion & Scattering

A thin prism in air produces the angle of minimum deviation $\delta$ . If the prism is immersed in water, the angle of minimum deviation for the same ray is (refractive index of water is $4/3$ and that of glass prism is $3/2$ )

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Answer: (c)

Thin prism: $\delta = (n_{\text{rel}} - 1)A$
In air: $\delta = \left(\frac{3}{2} - 1\right)A = \frac{A}{2}$
In water: $n_{\text{rel}} = \frac{3/2}{4/3} = \frac{9}{8}$ , so $\delta' = \frac{A}{8}$
$\frac{\delta'}{\delta} = \frac{1}{4}$ , so the answer is (c) $\frac{\delta}{4}$ .

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