Q 12-09-109JEE MainJEE Main 2023 (24 Jan, Shift 2)Medium
A convex lens of refractive index $1.5$ and focal length $18\ \text{cm}$ in air is immersed in water. The change in focal length of the lens will be ______ cm. (Given refractive index of water $=\frac43$)
Numerical value type. Enter your answer.
Answer: 54
$\dfrac{f_w}{f_a}=\dfrac{\mu-1}{\dfrac{\mu}{\mu_w}-1}=\dfrac{0.5}{\dfrac{1.5\times3}{4}-1}=\dfrac{0.5}{0.125}=4$.
$f_w=72\ \text{cm}$, so the change is $72-18=54\ \text{cm}$.
Solution by Sreeraj P, M.Sc Physics