A lens having refractive index 1.6 has focal length of 12 cm , when it is in air. Find the focal length of…
(Take refractive index of water as 1.28)
- 355 mm
- 288 mm
- 555 mm
- 655 mm
Solution
$\frac{1}{\mathrm{f}}=\left[\frac{\mu_{\mathrm{L}}}{\mu_{\mathrm{m}}}-1\right]\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]$
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
$\frac{1}{\mathrm{f}}=\left[\frac{1.6}{1.28}-1\right]\left[\frac{10}{72}\right]=\frac{32}{128} \times \frac{10}{72}$
$\begin{aligned} & \frac{1}{\mathrm{f}}=\frac{1}{4} \times \frac{10}{72} \\ & \mathrm{f}=28.8 \mathrm{~cm} \\ & \mathrm{f}=288 \mathrm{~mm}\end{aligned}$
Asked in: JEE Main 2025 (07 Apr Shift 1)