MHTCET 15 Apr 2026 Shift 2 Paper

Concept:For a thin prism, the deviation depends on the refractive index of the prism material relative to the surrounding medium.Explanation:The formula for minimum deviation of a thin prism is:$$\delta = (\mu - 1)A$$where $A$ is the prism angle and $\mu$ is the refractive index of the prism w.r.t. the surrounding medium.In air, the refractive index of glass w.r.t. air is:$${}^{a}n_{g} = \frac{3}{2}$$So, the deviation in air is:$$\delta_{1} = \left(\frac{3}{2} - 1\right)A = \frac{A}{2}$$When the prism is immersed in liquid, the refractive index of glass w.r.t. the liquid is:$${}^{\ell}n_{g} = \frac{{}^{a}n_{g}}{{}^{a}n_{\ell}}$$Given ${}^{a}n_{\ell} = \frac{1}{3}$, we get:$${}^{\ell}n_{g} = \frac{\frac{3}{2}}{\frac{1}{3}} = \frac{9}{2}$$Hence, the deviation in liquid is:$$\delta_{2} = \left(\frac{9}{2} - 1\right)A = \frac{7A}{2}$$Since $\delta_{1} = \frac{A}{2}$, we have:$$\delta_{2} = 7\delta_{1}$$Answer:$$\delta_{2} = 7\delta_{1}$$Option C.