A convex lens made of glass (refractive index $=1.5$) has focal length 24 cm in air. When it is totally immersed in water (refractive index $=1.33$), its focal length changes to

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Aicharya · Accepted Answer

$\begin{aligned}
& \frac{1}{8}=\left(\frac{\mu_{\ell}}{\mu_{\mathrm{s}}}-1\right)\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\ & \frac{1}{24}=(1.5-1)\left[\frac{2}{\mathrm{R}}\right]...(i)\\ & \frac{1}{\mathrm{f}^{\prime}}=\left(\frac{1.5}{1.33}-1\right)\left(\frac{2}{\mathrm{R}}\right) \\ & \frac{1}{\mathrm{f}^{\prime}}=\left(\frac{1.5 \times 3}{4}-1\right) \frac{2}{\mathrm{R}}...(ii)
\end{aligned}$
(i) divided by (ii)
$\begin{aligned}
& \frac{\mathrm{f}^{\prime}}{24}=4 \\ & \mathrm{f}^{\prime}=96 \mathrm{~cm}
\end{aligned}$

A convex lens made of glass (refractive index $=1.5$) has focal length 24 cm in air. When it is totally…

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