A convex lens made of glass has focal length \(0.15 \mathrm{~m}\) in air. If the refractive index of glass is \(\frac{3}{2}\) and that of water is \(\frac{4}{3}\), the focal length of lens when immersed in water is

Given, \(\mathrm{f}_{\mathrm{a}}=0.15 \mathrm{~m}, \mu_{\mathrm{g}}=\frac{3}{2}, \mu_{\mathrm{w}}=\frac{4}{3}\)
According to Lens maker's formula
\(\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right) \quad\) where \(\mu=\frac{\mu_{\mathrm{L}}}{\mu_{\mathrm{M}}}\)
\(\frac{1}{\mathrm{f}_{\mathrm{a}}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
\(=\left(\frac{(3 / 2)}{1}-1\right) C\) where \(C=\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
or \(\quad \frac{1}{f_{a}}=\frac{C}{2} \quad \text{...(i)}\)
Also, \(\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\left(\frac{(3 / 2)}{(4 / 3)}-1\right) \mathrm{C}\)
or \(\quad \frac{1}{f_{\mathrm{W}}}=\frac{C}{8} \quad \text{...(ii)}\)
From Eqs. (i) and (ii), we get
\[
\frac{\mathrm{f}_{\mathrm{w}}}{\mathrm{f}_{\mathrm{a}}}=\frac{\mathrm{C}}{2} \times \frac{8}{\mathrm{C}}=4
\]
or
\[
\begin{aligned}
\mathrm{f}_{\mathrm{w}} &=4 \mathrm{f}_{\mathrm{a}} \\
&=4 \times 0.15=0.6 \mathrm{~m}
\end{aligned}
\]