What is the effect on the focal length of a convex lens when it is immersed in a liquid whose refractive index is equal to refractive index of the lens?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Focal length will become infinity
A lens bends light only because its refractive index differs from that of the surrounding medium. The lens maker's formula in a medium of refractive index \( n_m \) is
\( \displaystyle \frac{1}{f} = \left(\frac{n_l}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
With \( n_l = n_m \):
\( \displaystyle \frac{1}{f} = (1 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0 \;\Rightarrow\; f \rightarrow \infty \)
Light crosses the surfaces without refraction, so the lens behaves like a plane sheet of the liquid; its power is zero and its focal length is infinite.
Option (1) Focal length will become infinity — Correct. With equal refractive indices \( 1/f = 0 \), so the focal length becomes infinite and the lens becomes invisible to the light.
Option (2) Focal length will become zero — Incorrect. A zero focal length would need infinite power; here the power falls to zero, the opposite case.
Option (3) Focal length will reduce but not become zero — Incorrect. The focal length increases in a denser liquid, it does not reduce; when the indices are equal it does not stay finite at all.
Option (4) Focal length remains unchanged — Incorrect. \( f \) depends on \( n_l/n_m \), so changing the surrounding medium changes the focal length.