Question

When a light ray is incident on a small angle prism of material of refractive index 1.5, the angle of minimum deviation is $7^\circ$. If the prism is immersed in a liquid of refractive index 1.2, then the angle of minimum deviation is:

• A. $10.5^\circ$
• B. $1.75^\circ$
• C. $3.5^\circ$
• D. $14^\circ$

Hint:

For a given small-angle prism, deviation is directly proportional to $(\mu_{\text{rel}} - 1)$.
Ratio: $\frac{\delta_2}{\delta_1} = \frac{\mu_{\text{rel}} - 1}{\mu_g - 1} = \frac{1.25 - 1}{1.5 - 1} = \frac{0.25}{0.5} = \frac{1}{2}$.
Thus, $\delta_2 = \frac{1}{2} \delta_1 = \frac{7^\circ}{2} = 3.5^\circ$.
This direct ratio method is extremely fast.

Answer 1

The Correct Option is C

Step 1: Notice the prism angle A is common to both cases.
For a thin prism, $\delta = (\mu_{\text{rel}}-1)A$. Since $A$ does not change when the prism is dipped in the liquid, we can skip solving for it and work directly with a ratio.
Step 2: Write the ratio of the two deviations. \[ \frac{\delta_2}{\delta_1} = \frac{\mu_{\text{rel},2}-1}{\mu_{\text{rel},1}-1} \] In air, $\mu_{\text{rel},1} = 1.5$. In the liquid, $\mu_{\text{rel},2} = \frac{1.5}{1.2} = 1.25$.
Step 3: Substitute and simplify. \[ \frac{\delta_2}{7^\circ} = \frac{1.25-1}{1.5-1} = \frac{0.25}{0.5} = 0.5 \]
Step 4: Solve for the new minimum deviation. \[ \delta_2 = 0.5 \times 7^\circ \] \[ \boxed{\delta_2 = 3.5^\circ} \]