The angle of minimum deviation produced by a thin prism in air is $δ_1$. What will be the minimum deviation ($δ_2$) if the prism is immersed in liquid ?
$[^an_g = \frac{3}{2}, ^an_{\ell} = \frac{1}{3}]$

Question

A light ray is passing from air (refractive index $ \mu_1 = 1.0 $) into water (refractive index $ \mu_2 = 1.33 $). If the angle of incidence in air is 30°, what is the angle of refraction in water?

Answer 1

Step 1: Relative index
In liquid, $n = (3/2)/(1/3) = 9/2$.

Step 2: Deviation
$\delta_2 = (9/2 - 1)A = 7A/2$, while $\delta_1 = A/2$.

Step 3: Ratio
$\delta_2/\delta_1 = 7$. Option (C).

Final Answer:
Option (C). \[ \boxed{\delta_2 = 7\delta_1} \]

The angle of minimum deviation produced by a thin prism in air is \(δ_1\). What will be the minimum deviation (\(δ_2\)) if the prism is immersed in liquid ?
\([^an_g = \frac{3}{2}, ^an_{\ell} = \frac{1}{3}]\)

Show Hint

The Correct Option is C

Solution and Explanation

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The angle of minimum deviation produced by a thin prism in air is \(δ_1\). What will be the minimum deviation (\(δ_2\)) if the prism is immersed in liquid ?\([^an_g = \frac{3}{2}, ^an_{\ell} = \frac{1}{3}]\)