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: Understanding the Concept:
For a thin prism of angle \(A\), the angle of minimum deviation is \(\delta=(\mu-1)A\). In a liquid, \(\mu\) is replaced by the relative index \(\mu_g/\mu_w\).

Step 2: In air:
\(\delta=\left(\dfrac32-1\right)A=\dfrac A2\).

Step 3: In water:
\(\delta'=\left(\dfrac{3/2}{4/3}-1\right)A=\left(\dfrac98-1\right)A=\dfrac A8\).

Step 4: Compare:
\(\dfrac{\delta'}{\delta}=\dfrac{A/8}{A/2}=\dfrac14\), so \(\delta'=\dfrac\delta4\). Option C.

Step 5: Why the other options are wrong.
\(\delta\) would apply with no medium change, and \(\dfrac\delta2\) and \(2\delta\) come from wrong relative indices (for example using \(\mu_w-1\) as the factor).

Final Answer:
The deviation in water is delta / 4. \[ \boxed{\text{(C) }\dfrac\delta4} \]