A convex lens (n = $\displaystyle 1.52$) has a focal length of $\displaystyle 15.0$ cm in air. Find its focal length when it is immersed in liquid of refractive index 1.65. What will be the nature of the lens?

For convex lens in air: $\displaystyle \frac{1}{f_{a}}=\left(\frac{n_{g}}{n_{a}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ For convex lens in liquid: $\displaystyle \frac{1}{f_{l}}=\left(\frac{n_{g}}{n_{l}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ $\displaystyle \frac{f_{l}}{f_{a}}=\frac{\dfrac{1.52-1}{1}}{\dfrac{1.52-1.65}{1.65}}$ $\displaystyle =-6.6$ $\displaystyle f_{l}=-6.6 f_{a}$ $\displaystyle =-99 \mathrm{~cm}$ Nature of the lens: diverging / behaves like a concave lens.