Question

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 another liquid medium? \(\{{}_{a}n_{g}=3/2,\ {}_{a}n_{l}=4/3\}\)

• A. \(\delta _{2}=\frac{\delta _{1}}{2}\)
• B. \(\delta _{2}=2\delta _{1}\)
• C. \(\delta _{2}=\frac{\delta _{1}}{3}\)
• D. \(\delta _{2}=\frac{\delta _{1}}{4}\)

Hint:

For a thin prism the deviation is (n relative minus 1) times the angle; use the index relative to the surrounding medium.

Answer 1

The Correct Option is D

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

Step 2: In air
\(\delta_1=\left(\dfrac{3}{2}-1\right)A=\dfrac{A}{2}\)

Step 3: In the liquid
\(n=\dfrac{3/2}{4/3}=\dfrac{9}{8}\), so \(\delta_2=\left(\dfrac{9}{8}-1\right)A=\dfrac{A}{8}\).

Step 4: Ratio
\(\dfrac{\delta_2}{\delta_1}=\dfrac{1/8}{1/2}=\dfrac{1}{4}\), so \(\delta_2=\dfrac{\delta_1}{4}\), option (D).

Final Answer:
The minimum deviation becomes \(\dfrac{\delta_1}{4}\). \[ \boxed{\text{(D) }\delta_2=\dfrac{\delta_1}{4}} \]