A thin lens made of glass (refractive index = 1.5) of focal length f = 16 cm is immersed in a liquid of refractive index 1.42. If its focal length in liquid is f₁ , then the ratio f₁/f is closest to the integer :

Solution

Physics sir · 4 steps

The lens shape stays the same in the liquid, so divide the two lensmaker's equations.

1. Lensmaker in each medium. The power of a thin lens is (μₗₑₙₛ/μ_(medium) - 1) times the curvature factor (1/R₁ - 1/R₂). The curvature factor is the same in air and in the liquid, so it cancels in the ratio.

   f₁/f = (μ_g - 1)/(μ_g/μₗ - 1)

2. Put in numbers. In air the factor is 1.5 - 1 = 0.5. In the liquid it is 1.5/1.42 - 1 = 0.08/1.42, which is about 0.0563.
3. Take the ratio. f₁/f = (0.5 × 1.42)/0.08 = 0.71/0.08 = 8.875. The liquid's index is so close to the glass's that the lens barely bends light, so its focal length grows to about 142 cm.
4. So. The ratio 8.875 is closest to the integer 9, which is option (c).

The idea to keep. When a lens is moved into a new medium, only the (μₗₑₙₛ/μ_(medium) - 1) factor changes. Divide the two lensmaker's equations and the radii drop out.

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