Q 12-09-236JEE MainJEE Main 2025 (7 Apr, Shift 1)Easy
A lens having refractive index $1.6$ has focal length of $12\ \text{cm}$ when it is in air. Find the focal length of the lens when it is placed in water. (Take refractive index of water as $1.28$)
Answer: (B) $288\ \text{mm}$
By the lens maker's formula, $\dfrac1f = \left(\dfrac{\mu_L}{\mu_m} - 1\right)K$, where $K = \dfrac1{R_1} - \dfrac1{R_2}$ depends only on the shape.
$$\frac{f_w}{f_a} = \frac{\mu_L - 1}{\mu_L/\mu_w - 1} = \frac{0.6}{1.6/1.28 - 1} = \frac{0.6}{0.25} = 2.4$$
$$f_w = 2.4\times12 = 28.8\ \text{cm} = 288\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics