Question

A convex lens, $({\mu }_{lens}=1.5)$ initially placed in air, is now immersed in water $({\mu }_w=\frac{4}{3})$. The focal length of the lens will,

• A. increase
• B. decrease
• C. remain same
• D. can't say

Answer 1

The correct option is A increase

According to Lens-maker's equation,
$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

Where, $\mu$ is the refractive index of the material of the lens, w. r. t. the surrounding medium.

Here, ${}_{air}{\mu }_{lens}=\frac{{\mu }_{lens}}{{\mu }_{air}}\text{and}{}_w{\mu }_{lens}=\frac{{\mu }_{lens}}{{\mu }_w}$
Now, ${\mu }_{lens}>{\mu }_w>{\mu }_{air}$

${\Rightarrow }_{air}{\mu }_{lens}{>}_w{\mu }_{lens}$
Applying the above condition to Lens-maker's equation, we can say that,
$f_w>f_{air}$

Hence, $\left(A\right)$ is the correct answer.