Question

For a thin prism, \(\delta_1\) is the angle of deviation produced, when prism is placed in air. When the prism is immersed in water, the angle of deviation produced is \(\delta_2\). Given \({}_{\text{a}}\mu_{\text{g}} = \frac{3}{2}\) and \({}_{\text{a}}\mu_{\text{w}} = \frac{4}{3}\) . The ratio \(\delta_2 : \delta_1\) is

• A. \(1 : 2\)
• B. \(1 : 4\)
• C. \(1 : 8\)
• D. \(4 : 1\)

Hint:

The deviation of a thin prism is proportional to the relative refractive index minus one.

Answer 1

The Correct Option is B

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

Step 2: In air
\(\delta_1=\left(\frac32-1\right)A=\frac{A}{2}\).

Step 3: In water
\(\mu_{rel}=\frac{3/2}{4/3}=\frac98\), so \(\delta_2=\left(\frac98-1\right)A=\frac{A}{8}\).

Step 4: Ratio
\(\frac{\delta_2}{\delta_1}=\frac{A/8}{A/2}=\frac14\).

Step 5: Compare the options
The ratio is \(1:4\), the second option.

Final Answer:
The ratio \(\delta_2:\delta_1\) is \(1:4\). \[ \boxed{\text{(B) }1:4} \]