The angle of minimum deviation produced by a thin prism in air is δ₁. What will be the minimum deviation ( δ₂) if the prism is immersed in liquid? [a ng = 3/2, a n = 1/3] (a) δ₂ = 3δ₁ (b) δ₂ = 5δ₁ (c) δ₂ = 7δ₁ (d) δ₂ = 9δ₁

Answer: (c). Thin prism: δ = (ngnmedium - 1)A In air: δ₁ = (3/2 - 1)A = A/2 In liquid: δ₂ = ((3/2)/(1/3) - 1)A = (9/2 - 1)A = 7A/2 So the answer is (c) δ₂ = 7δ₁.

The angle of minimum deviation produced by a thin prism in air is δ₁… | MHT CET 2026 (15 Apr, Shift 2)

Q9MHT CET 2026 · 15 APR · SHIFT 2Medium

Prism, Dispersion & Scattering

The angle of minimum deviation produced by a thin prism in air is $\delta_1$ . What will be the minimum deviation ( $\delta_2$ ) if the prism is immersed in liquid?

$\left[{}^a n_g = \frac{3}{2},\ {}^a n_\ell = \frac{1}{3}\right]$

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Answer: (c)

Thin prism: $\delta = \left(\frac{n_g}{n_{\text{medium}}} - 1\right)A$
In air: $\delta_1 = \left(\frac{3}{2} - 1\right)A = \frac{A}{2}$
In liquid: $\delta_2 = \left(\frac{3/2}{1/3} - 1\right)A = \left(\frac{9}{2} - 1\right)A = \frac{7A}{2}$
So the answer is (c) $\delta_2 = 7\delta_1$ .

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