A converging lens made of glass (μ=1 · 5) has its spherical faces of radii of curvature 10 cm and 20 cm. Find its focal length (a) in air, and (b)…

Finding the focal length

In Air

In Liquid

$$\frac{1}{\mathrm{f}_{\mathrm{a}}}=\left({ }_{\mathrm{a}} \mu^{\mathrm{g}}-$\displaystyle 1$\right)\left(\frac{1}{\mathrm{R}_{$\displaystyle 1$}}-\frac{1}{\mathrm{R}_{$\displaystyle 2$}}\right)

$$$$ \mathrm{R}_{$\displaystyle 1$}=$\displaystyle 10$ \mathrm{~cm} & \mathrm{R}_{$\displaystyle 2$}=-$\displaystyle 20$ \mathrm{~cm},{ }_{\mathrm{a}} \mu^{\mathrm{g}}=$\displaystyle 1.5$

$$$\frac{1}{\mathrm{f}_{\mathrm{a}}}=(1.5-1)\left(\frac{1}{10}-\frac{1}{-20}\right)$

$$\mathrm{f}_{\mathrm{a}}=\frac{40}{3}=$\displaystyle 13.3$ \mathrm{~cm}

$$b) $\displaystyle \frac{1}{f_{l}}=\left({ }_{l} \mu^{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

$$=\left(\frac{1.5}{1.25}-$\displaystyle 1$\right)\left(\frac{1}{10}-\frac{1}{-20}\right)

$$$f_{l}=\frac{100}{3}=33.3 \mathrm{~cm}$