Assertion : A double convex lens (μ=1.5) has focal length 10 ~cm ....

As $\frac{1}{f}=\left(\frac{\mu_g}{\mu_a}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$\frac{1}{10}=\left(\frac{3}{2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \Rightarrow\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{1}{5}$If$f_w$is the focal length of the lens when immersed in water,$\frac{1}{f_w}=\left(\frac{\mu_g}{\mu_w}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\left(\frac{3 / 2}{4 / 3}-1\right)\left(\frac{1}{5}\right)=\frac{1}{40}$Hence,$f_w=40 \mathrm{~cm}