A lens having refractive index 1.6 has focal length of 12 cm , when it is in air. Find the focal length of the lens when it is placed in water.
(Take refractive index of water as 1.28)

As we know,
\(\frac{1}{\mathrm{f}}=\left[\frac{\mu_{\mathrm{L}}}{\mu_{\mathrm{m}}}-1\right]\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]\)
For air \(\mu_{\mathrm{m}}=1\)
\(\begin{aligned} & \frac{1}{12}=[1.6-1]\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\ & \frac{1}{12}=\frac{6}{10}\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\ & {\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]=\frac{10}{72}}\end{aligned}\)
For water
\(\frac{1}{\mathrm{f}}=\left[\frac{1.6}{1.28}-1\right]\left[\frac{10}{72}\right]=\frac{32}{128} \times \frac{10}{72}\)
\(\begin{aligned} & \frac{1}{\mathrm{f}}=\frac{1}{4} \times \frac{10}{72} \\ & \mathrm{f}=28.8 \mathrm{~cm} \\ & \mathrm{f}=288 \mathrm{~mm}\end{aligned}\)