Find the focal length of plano-convex lens of refractive index $\displaystyle 1.5$ and radius of curvature $\displaystyle 10$ cm when it is immersed in a liquid of refractive index 1.25.

$\displaystyle \frac{1}{f}=\left(\frac{n_{2}}{n_{1}}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$ For plano-convex lens $\displaystyle R_{1}=R$ and $\displaystyle R_{2}=\infty$: $\displaystyle \frac{1}{f}=\left(\frac{n_{2}}{n_{1}}-1\right) \frac{1}{R}$ $\displaystyle =\left(\frac{1.5}{1.25}-1\right) \times \frac{1}{10}$ $\displaystyle \frac{1}{f}=\frac{1}{50}$ $\displaystyle \therefore f=50 \mathrm{~cm}$