Question

A double convex lens $(\mu=3 / 2$ ) of focal length $20\, cm$ is totally immersed in water $(\mu=4 / 3)$. Its focal length now will be

• A. 20 cm
• B. 30 cm
• C. 40 cm
• D. 10 cm

Answer 1

Answer: (B)

30 cm

Answer 2

From lens formula $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ If focal length of lens in air is $f_{a}$ and in liquid is $f_{1},$ then $\frac{1}{f_{a}}=\left(_{a} \mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ $\frac{1}{f_{d}}=\left({ }_{1} \mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ $\frac{f_{l}}{f_{a}}=\frac{\left(a \mu_{g}-1\right)}{\left({ }_{l} \mu_{g}-1\right)}$ $\frac{f_{l}}{20} =\left[\frac{\frac{3}{2}-1}{\frac{4}{3}-1}\right]$ $=\frac{3}{2}$ $f_{l} =30\, cm$