A thin prism of refractive index 1.5 when placed in air deviates a light ray by a minimum angle of 5°. The angle of minimum deviation when it is immersed in oil of refractive index 1.25 is:

2°

The correct answer is Option (1) → 2°

Given:

Prism refractive index in air, $\mu = 1.5$

Minimum deviation in air, $\delta_\text{min} = 5^\circ$

Prism placed in oil, refractive index of oil, $\mu_\text{oil} = 1.25$

Relation for thin prism:

For prism in air: $\delta_\text{min} \approx (\mu - 1)A$

where $A$ is prism angle.

So, $A = \frac{\delta_\text{min}}{\mu - 1} = \frac{5^\circ}{1.5 - 1} = \frac{5^\circ}{0.5} = 10^\circ$

When prism is immersed in a medium of refractive index $\mu_\text{m}$:

New deviation: $\delta_\text{min}' = (\frac{\mu}{\mu_\text{oil}} - 1)A$

Substitute values:

$\delta_\text{min}' = \left(\frac{1.5}{1.25} - 1\right) \cdot 10^\circ = (1.2 - 1) \cdot 10^\circ = 0.2 \cdot 10^\circ = 2^\circ$

∴ Minimum deviation in oil = 2°