Question

A convex glass lens $({\mu }_g=1.5)$ has a focal length of 8 cm when placed in air. What is the focal length lens when it is immersed in water? $({\mu }_{\omega }=\frac{4}{3})$

• A. $32cm$
• B. $60cm$
• C. $16cm$
• D. $30cm$

Answer 1

The correct option is A $32cm$

$\mu g=1.5$

$\mu w=4/3$

$fa=$ focal length of lens in air $=8cm$

To find $\to fw\to$ focal length of lens in water

Sol

we know that

$\frac{1}{fa}=(\frac{\mu g}{\mu a}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{1}{8}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$(\frac{1}{R_1}-\frac{1}{R_2})=\frac{1}{4}$ ...(1)

When lens is immersed into water,

$\frac{1}{fw}=(\frac{\mu g}{\mu w}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{1}{fw}=(\frac{1.5}{4/3}-1)(\frac{1}{R_1}-\frac{1}{R_2})$ ...(2)

From eqn (1) & (2)

$\frac{1}{fw}=(\frac{3\times 3}{2\times 4}-1)\left(\frac{1}{4}\right)$

$=\frac{(9\times -8)}{8}\times \frac{1}{4}$

$fw=32cm$