A lens having refractive index 1.6 has focal length of 12 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)

Solution

Physics sir · 3 steps

Write the lensmaker's equation in air and in water, then divide so the curvature terms cancel.

1. Lensmaker's equation. The power of a lens depends on how much its refractive index exceeds that of the surroundings: 1/f = (μₗₑₙₛ/μ_(medium) - 1)(1/R₁ - 1/R₂). The curvature bracket stays the same in both media.

   f_w/fₐ = (μ - 1)/(μ/μ_w - 1)

2. Put in numbers. In air, μ - 1 = 1.6 - 1 = 0.6. In water, 1.6/1.28 - 1 = 1.25 - 1 = 0.25. So f_w/fₐ = 0.6/0.25 = 2.4.
3. So. f_w = 2.4 × 12 cm = 28.8 cm = 288 mm, which is option (c).

The idea to keep. When a lens goes into a new medium, only the factor (μₗₑₙₛ/μ_(medium) - 1) changes. Take the ratio of the two cases and the radii drop out.

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