A convex lens of refractive index 1.5 and focal length 18cm in air is immersed in water. The change in focal length of the lens will be ___________ cm.
(Given refractive index of water =4/3)

Solution

Physics sir · 3 steps

Same lens, same curvatures: only (μᵣₑₗ - 1) changes, so divide the two lensmaker equations.

1. Curvatures cancel. The lensmaker formula gives 1/f = (μᵣₑₗ - 1)(1/R₁ - 1/R₂), where μᵣₑₗ is the lens index relative to the surrounding medium. The lens shape does not change in water, so the bracket cancels when you divide air by water.

   f_w/fₐ = (μ_g - 1)/(μ_g/μ_w - 1) = (1.5 - 1)/(1.5/(4/3) - 1) = 0.5/(9/8 - 1) = 0.5/0.125 = 4

2. Focal length in water. f_w = 4 × 18 = 72 cm. The lens still converges, but much more weakly, because glass and water have close refractive indices.
3. So. The change in focal length is 72 - 18 = 54 cm, so the answer is 54.

The common slip: 72. Stopping at the new focal length gives 72, but the question asks for the change, which is 72 - 18 = 54.

The idea to keep. In a new medium, only the relative index changes and the lens shape stays the same. Divide the lensmaker equations so the radii cancel, then use (μ_g/μ_(medium) - 1).

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