Question

Find the new focal length of a convex lens of focal length $20\text{cm}$ and refractive index $1.5$ when it is immersed in water.
[refractive index of the water is $\frac{4}{3}$]

• A. $40\text{cm}$
• B. $40\text{cm}$
• C. $60\text{cm}$
• D. $80\text{cm}$

Answer 1

The correct option is D $80\text{cm}$

Given,

${\mu }_l=1.5=\frac{3}{2};{\mu }_w=\frac{4}{3};{\mu }_a=1$

$f_a=20\text{cm}$

Where,

${\mu }_l-\text{Refractive index of the lens}{\mu }_a-\text{Refractive index of air}{\mu }_w-\text{Refractive index of water}{f}_a-\text{Focal length of the lens in air}{f}_w-\text{Focal length of the lens in water}$

The focal length of the lens in air is,

$\frac{1}{f_a}=(\frac{{\mu }_l}{{\mu }_a}-1)[\frac{1}{R_1}-\frac{1}{R_2}]......\left(1\right)$

The focal length of the lens when immersed in water,

$\frac{1}{f_w}=(\frac{{\mu }_l}{{\mu }_w}-1)[\frac{1}{R_1}-\frac{1}{R_2}]......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{f_w}{f_a}=\frac{(\frac{{\mu }_l}{{\mu }_a}-1)}{(\frac{{\mu }_l}{{\mu }_w}-1)}$

$\frac{f_w}{20}=\frac{(\frac{3}{2}-1)}{(\frac{\frac{3}{2}}{\frac{4}{3}}-1)}=\frac{8}{2}$

$\Rightarrow f_w=4\times 20=80\text{cm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.