JEE Main 2022 · 27 July · morning · Single correct · +4 / −1
A microscope was initially placed in air (refractive index 1). It is then immersed in oil (refractive index 2). For a light whose wavelength in air is λ, calculate the change of microscope's resolving power due to oil and choose the correct option.
- (a) Resolving power will be 1/4 in the oil than it was in the air.
- (b) Resolving power will be twice in the oil than it was in the air.
- (c) Resolving power will be four times in the oil than it was in the air.
- (d) Resolving power will be 1/2 in the oil than it was in the air.
Answer and solution
Answer (b) Resolving power will be twice in the oil than it was in the air.
Solution
Physics sir · 4 steps
Oil shortens the wavelength inside it, and a microscope resolves finer detail with shorter waves.
- Resolving power. The smallest separation a microscope can resolve is d = (1.22 λ)/(2n sin β), where n is the refractive index of the medium between object and objective. Resolving power is 1/d.
RP = (2n sin β)/(1.22 λ)
- Why n helps. In a medium of index n, the wavelength becomes λ/n. That is where the factor n in the numerator comes from. Shorter waves mean finer detail can be separated.
- Compare the media. With λ and β unchanged, RP ∝ n. So RPₒᵢₗ/RPₐᵢᵣ = 2/1 = 2.
- So. The resolving power in oil is twice that in air, which is option (b).
The idea to keep. A microscope's resolving power is proportional to the refractive index n of the medium in front of the objective. That is the whole point of oil-immersion objectives.
Answer from the NTA final answer key. Solution by PrepWiser.