MHTCET 18 Apr 2026 Shift 2 Paper

Concept:The power of a lens depends on the refractive index of the surrounding medium, as given by the lens maker's formula.Explanation:For a lens placed in a medium, the power is:$$P = \left(\frac{\mu_l}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$Here, $\mu_l = 1.5$ is the refractive index of glass.The radius of curvature is the same for both surfaces, so the factor $\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ cancels when taking the ratio.For the lens in air, $\mu_m = 1$:$$P_{\text{air}} \propto \frac{1.5}{1} - 1 = 0.5$$For the lens immersed in liquid, $\mu_m = 1.25$:$$P_{\text{liquid}} \propto \frac{1.5}{1.25} - 1 = 1.2 - 1 = 0.2$$Therefore,$$\frac{P_{\text{liquid}}}{P_{\text{air}}} = \frac{0.2}{0.5} = \frac{2}{5}$$Thus, the required ratio is $2:5$.Answer:$P_{\text{liquid}} : P_{\text{air}} = 2:5$Correct option: B.