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

Step 1: Understanding the Concept:
A thin prism deviates a light ray by an angle that depends on its refracting angle and the relative refractive index of the prism material with respect to the surrounding medium.
When immersed in a liquid, the relative refractive index decreases, resulting in a smaller angle of deviation.
Step 2: Key Formulas or Approach:
Angle of deviation for a thin prism: \( \delta = (\mu_{\text{relative}} - 1)A \).
Relative refractive index: \( {}_{\text{med}}\mu_{\text{prism}} = \frac{\mu_{\text{prism}}}{\mu_{\text{med}}} \).
Step 3: Detailed Explanation:
Let \( A \) be the refracting angle of the prism.
Case 1: Prism in air
The relative refractive index is \( {}_{\text{a}}\mu_{\text{g}} = \frac{3}{2} \).
The deviation in air, \( \delta_1 \), is:
\[ \delta_1 = ({}_{\text{a}}\mu_{\text{g}} - 1)A \] \[ \delta_1 = \left(\frac{3}{2} - 1\right)A = \left(\frac{3 - 2}{2}\right)A = \frac{1}{2}A \] Case 2: Prism immersed in water
The surrounding medium is now water. The relative refractive index of glass with respect to water is:
\[ {}_{\text{w}}\mu_{\text{g}} = \frac{\mu_{\text{g}}}{\mu_{\text{w}}} = \frac{{}_{\text{a}}\mu_{\text{g}}}{{}_{\text{a}}\mu_{\text{w}}} \] Given \( {}_{\text{a}}\mu_{\text{g}} = \frac{3}{2} \) and \( {}_{\text{a}}\mu_{\text{w}} = \frac{4}{3} \):
\[ {}_{\text{w}}\mu_{\text{g}} = \frac{3/2}{4/3} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8} \] The deviation in water, \( \delta_2 \), is:
\[ \delta_2 = ({}_{\text{w}}\mu_{\text{g}} - 1)A \] \[ \delta_2 = \left(\frac{9}{8} - 1\right)A = \left(\frac{9 - 8}{8}\right)A = \frac{1}{8}A \] Finding the ratio:
Now, calculate the ratio \( \frac{\delta_2}{\delta_1} \):
\[ \frac{\delta_2}{\delta_1} = \frac{\frac{1}{8}A}{\frac{1}{2}A} \] Cancel \( A \) and simplify the fraction:
\[ \frac{\delta_2}{\delta_1} = \frac{1/8}{1/2} = \frac{1}{8} \times \frac{2}{1} = \frac{2}{8} = \frac{1}{4} \] Therefore, \( \delta_2 : \delta_1 = 1 : 4 \).
Step 4: Final Answer:
The ratio of deviation in water to air is \( 1 : 4 \).