A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of f₁ in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of f₂ when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5 , the ratio of f₁ and f₂ will be

Solution

Physics sir · 4 steps

In a liquid, the lens bends light only by the glass-to-liquid ratio, so use 1.5/1.2, not 1.5.

1. Lens maker formula. For a plano-convex lens the flat side has R = ∞, so only the curved surface counts. The index that matters is the glass index divided by the index of the surrounding medium.

   1/f = (μ_g/μₘ - 1)1/R

2. First lens in air. Here μₘ = 1, so 1/f₁ = (1.5 - 1) × 1/2 = 1/4. This gives f₁ = 4 cm.
3. Second lens in liquid. The relative index is 1.5/1.2 = 1.25, so 1/f₂ = (1.25 - 1) × 1/3 = 1/12. This gives f₂ = 12 cm.
4. So. f₁ : f₂ = 4 : 12 = 1 : 3, which is option (b).

The idea to keep. A lens dipped in a liquid gets weaker because μ_g/μₘ - 1 shrinks. Always use the glass index relative to whatever surrounds the lens.

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