The angle of minimum deviation produced by a thin prism in air is δ₁. If it is immersed in water the angle of minimum deviation is [ag = 3/2, aw = 4/3] (a) 2δ₁ (b) (δ₁)/(2) (c) (δ₁)/(3) (d) (δ₁)/(4)

Answer: (d). Thin prism: δ = (μrel - 1)A. In air: δ₁ = (3/2 - 1)A = A/2 In water: μrel = (3/2)/(4/3) = 9/8, so δ₂ = A/8 (δ₂)/(δ₁) = 1/4 ⇒ δ₂ = (δ₁)/(4) So the answer is (d).

The angle of minimum deviation produced by a thin prism in air is δ₁… | MHT CET 2024 (15 May, Shift 2)

Q31MHT CET 2024 · 15 MAY · SHIFT 2Easy

Prism, Dispersion & Scattering

The angle of minimum deviation produced by a thin prism in air is $\delta_1$ . If it is immersed in water the angle of minimum deviation is

$\left[a_g = \frac{3}{2}, a_w = \frac{4}{3}\right]$

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Answer: (d)

Thin prism: $\delta = (\mu_{rel} - 1)A$ .
In air: $\delta_1 = \left(\frac{3}{2} - 1\right)A = \frac{A}{2}$
In water: $\mu_{rel} = \frac{3/2}{4/3} = \frac{9}{8}$ , so $\delta_2 = \frac{A}{8}$
$\frac{\delta_2}{\delta_1} = \frac{1}{4} \Rightarrow \delta_2 = \frac{\delta_1}{4}$
So the answer is (d).

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Prism, Dispersion & Scattering question from MHT CET 2024 (15 May, Shift 2)

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