The radius of curvature of each surface of a convex lens having refractive index 1.8 is 20 cm. The lens is now immersed in a liquid of refractive index 1.5 . The ratio of power of lens in air to its power in the liquid will be x: 1. The value of x is _________.

Solution

Physics sir · 4 steps

The shape factor stays the same in any medium, so only the (μ - 1) part changes when you dip the lens.

1. Lensmaker formula. For a biconvex lens with R₁ = +20 cm and R₂ = -20 cm, the curvature bracket is 1/20 + 1/20 = 0.1 cm⁻¹. This bracket depends only on the glass shape, so it is the same in air and in the liquid.

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

2. In air. The relative index is 1.8/1 = 1.8, so 1/fₐᵢᵣ = 0.8 × 0.1 = 0.08 cm⁻¹, a power of 8 D.
3. In the liquid. The relative index is 1.8/1.5 = 1.2, so 1/f_(liq) = 0.2 × 0.1 = 0.02 cm⁻¹, a power of 2 D.
4. So. Pₐᵢᵣ/P_(liq) = 0.8/0.2 = 4, so the ratio is 4 : 1 and x = 4.

The common slip: 0.25. Dividing the liquid power by the air power instead of the other way round gives 0.25, not 4. The question asks for air to liquid.

The idea to keep. When a lens changes medium, the curvature bracket cancels in the ratio. Just compare (μₗₑₙₛ/μ_(medium) - 1) in the two media.

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