Question

The focal length of a convex lens in air is $10\text{cm}$. Find its focal length in water.
Given : Refractive index of lens and water are ${\mu }_g=\frac{3}{2}$ and ${\mu }_w=\frac{4}{3}$ respectively.

• A. $60\text{cm}$
• B. $40\text{cm}$
• C. $30\text{cm}$
• D. $25\text{cm}$

Answer 1

The correct option is B $40\text{cm}$

Given that,
Focal length of lens in air, $f_a=10\text{cm}$
Case I: When lens is in air.
$\frac{1}{f_a}=(\frac{{\mu }_g}{{\mu }_a}-1)(\frac{1}{R_1}-\frac{1}{R_2})$....(i)
Case II: When lens is placed in water,
$\frac{1}{f_w}=(\frac{{\mu }_g}{{\mu }_w}-1)(\frac{1}{R_1}-\frac{1}{R_2})$......(ii)
On dividing equation (i) with equation (ii).
$\frac{f_w}{f_a}=\frac{(\frac{3}{2}-1)}{\left(\frac{\frac{3}{2}}{\frac{4}{3}}\right)-1}=\frac{\left(\frac{1}{2}\right)}{\frac{9}{8}-1}$
or, $\frac{f_w}{f_a}=\frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{8}\right)}=4$
$\Rightarrow f_w=4\times f_a=40\text{cm}$
$$\begin{array}{c}\text{Why this question ?}\\ \text{Bottom line : When lens is immersed in water the}\text{focal length of lens increases as}\text{compared to air.}\end{array}$$