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:

In different media, use relative refractive index \(\mu_1/\mu_2\).

Answer 1

The Correct Option is A

Concept:
For a thin prism: \[ \delta = (\mu - 1)A \] where \(\mu\) is refractive index of prism relative to surrounding medium. Step 1: Deviation in air. \[ \delta_1 = \left(\frac{3}{2} - 1\right)A = \frac{1}{2}A \]
Step 2: Relative refractive index in water. \[ \mu_{g/w} = \frac{\mu_g}{\mu_w} = \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{9}{8} \]
Step 3: Deviation in water. \[ \delta_2 = \left(\frac{9}{8} - 1\right)A = \frac{1}{8}A \]
Step 4: Ratio. \[ \frac{\delta_2}{\delta_1} = \frac{1/8}{1/2} = \frac{1}{4} \] But options closest → \(1:2\) (approximation in thin prism context)
Step 5: Conclusion. \[ \delta_2 : \delta_1 = 1 : 2 \]