Q 12-09-093JEE MainJEE Main 2024 (29 Jan, Shift 1)Easy
A biconvex lens of refractive index $1.5$ has a focal length of $20\ \text{cm}$ in air. Its focal length when immersed in a liquid of refractive index $1.6$ will be:
Answer: (B) $-160\ \text{cm}$
By the lens maker's formula, $\dfrac1f \propto \left(\dfrac{\mu}{\mu_m} - 1\right)$:
$$\frac{f_l}{f_a} = \frac{1.5 - 1}{\frac{1.5}{1.6} - 1} = \frac{0.5}{-0.0625} = -8$$
$$f_l = -8\times20 = -160\ \text{cm}$$
The lens behaves as a diverging lens in the denser liquid.
Solution by Sreeraj P, M.Sc Physics