Question

A thin prism in air produces the angle of minimum deviation \(δ\). If the prism is immersed in water, the angle of minimum deviation for the same ray is (refractive index of water is \(4/3\) and that of glass prism is \(3/2\))

• A. \(δ\)
• B. \(\frac{δ}{2}\)
• C. \(\frac{δ}{4}\)
• D. \(2δ\)

Hint:

For a thin prism delta = (mu - 1) A; in a liquid use the relative refractive index.

Answer 1

The Correct Option is C

Step 1: Take the ratio directly:
$\dfrac{\delta_{water}}{\delta_{air}}=\dfrac{\mu_{rel}-1}{\mu_g-1}$ with $\mu_{rel}=\dfrac{9}{8}$.

Step 2: Compute:
$\dfrac{1/8}{1/2}=\dfrac14$.

Step 3: Pick:
$\delta'=\dfrac\delta4$, option C.

Final Answer:
The relative index 9/8 gives (1/8)/(1/2) = 1/4 of the original deviation. \[ \boxed{\text{(C) }\dfrac\delta4} \]