MHTCET 15 May 2024 Shift 2 Paper

Concept:For a thin prism, the angle of minimum deviation depends on the refractive index of the prism material relative to the surrounding medium.Explanation:The formula for angle of minimum deviation of a thin prism is $\delta = (\mu - 1)A$, where $\mu$ is the refractive index of the prism with respect to the surrounding medium and $A$ is the prism angle.In air, the relative refractive index is $\mu_1 = \frac{3}{2}$.So, $\delta_1 = \left(\frac{3}{2} - 1\right)A = \frac{1}{2}A$.When the prism is immersed in water, the relative refractive index of glass with respect to water is:$$\mu_2 = \frac{\mu_{\text{glass}}}{\mu_{\text{water}}} = \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{9}{8}$$.Therefore, $\delta_2 = \left(\frac{9}{8} - 1\right)A = \frac{1}{8}A$.Taking the ratio of the two deviations:$$\frac{\delta_1}{\delta_2} = \frac{\frac{1}{2}A}{\frac{1}{8}A} = 4$$.So, $\delta_2 = \frac{\delta_1}{4}$.Answer:The angle of minimum deviation when the prism is immersed in water is $\frac{\delta_1}{4}$.Correct option: D.