A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:

Solution

Physics sir · 3 steps

The lens shape stays the same, so only the relative index changes, and in this liquid it goes negative.

1. Lens maker formula. The surfaces of the lens do not change when it is put in the liquid. So the curvature part (1/R₁ - 1/R₂) is the same in both cases. Only the relative refractive index changes.

   1/f = (μₗₑₙₛ/μ_(medium) - 1)(1/R₁ - 1/R₂)

2. Take the ratio. In air the factor is 1.5 - 1 = 0.5. In the liquid it is 1.5/1.6 - 1 = -0.1/1.6 = -1/16. Dividing the two equations gives fₗ/fₐ = 0.5/(-1/16) = -8.
3. So. fₗ = -8 × 20 cm = -160 cm. The lens is now optically less dense than its surroundings, so it acts as a diverging lens. The answer is (d).

The idea to keep. A lens placed in a denser medium than itself flips its nature: a convex lens becomes diverging. Always use μₗₑₙₛ/μ_(medium) and keep the sign.

Answer from the NTA final answer key. Solution by PrepWiser.