Question

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 ?
\([^an_g = \frac{3}{2}, ^an_{\ell} = \frac{1}{3}]\)

• A. \(δ_2 = 3\,δ_1\)
• B. \(δ_2 = 5\,δ_1\)
• C. \(δ_2 = 7\,δ_1\)
• D. \(δ_2 = 9\,δ_1\)

Hint:

For a thin prism, deviation = (n - 1) A where n is the refractive index relative to the surrounding medium.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a thin prism of angle A, the deviation is \(\delta = (n - 1)A\), where n is the refractive index of the prism relative to its surroundings.

Step 2: In air
\[ \delta_1 = \left(\frac32 - 1\right)A = \frac A2 \]

Step 3: In the liquid
The relative index is \(n = \dfrac{n_g}{n_\ell} = \dfrac{3/2}{1/3} = \dfrac92\).
\[ \delta_2 = \left(\frac92 - 1\right)A = \frac{7A}{2} = 7\delta_1 \]
So \(\delta_2 = 7\delta_1\), option (C).

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