Question

A convex lens of focal length 0.2 m and made of glass ${(}^a{\mu }_g=1.5)$ is immersed in water ${(}^a{\mu }_w=1.33)$. find the change in the focal length of the lens.

• A. $6.8cm$
• B. $5.8cm$
• C. $0.58m$
• D. $7.8cm$

Answer 1

Answer: (C)

$0.58m$

Answer : $0.58m$

Solution : For glass lens in air, ${.}^a{\mu }_g=1.5$ and $f_a=0.2m$

$\frac{1}{f_a}=({.}^a{\mu }_g-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{1}{0.2}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\therefore \frac{1}{R_1}-\frac{1}{R_2}=10cm$

For the same lens in water,

$\frac{1}{f_a}=(\frac{{.}^a{\mu }_g}{{.}^a{\mu }_w}-1)(\frac{1}{R_1}-\frac{1}{R_2})=(\frac{1.5}{1.33}-1)\times 10$

$f_w=\frac{1.33}{1.71}=0.78m$

Change in focal length $=f_w-f_a=0.78-0.20=0.58m$