Question

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$\frac{1}{4}$ 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 $\frac{1}{2}$ in the oil than it was in the air.

Answer 1

Answer: (C)

Resolving power will be four times in the oil than it was in the air.

Answer 2

The correct answer is (C) : Resolving power will be four times in the oil than it was in the air.
∵Resolving Power $=\frac{2µ\sinθ}{1.22λ}$
$\frac{P_1}{P_2}=\frac{µ_1}{µ_2}×\frac{µ_1}{µ_2}$
$=(\frac{µ_1}{µ_2})^2$
$⇒\frac{P_1}{P_2}=\frac{1}{4} ⇒P_2=4P_1$