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 $$\lambda$$, calculate the change of microscope's resolving power due to oil

A microscope is moved from air (refractive index 1) to oil (refractive index 2). We need to find the change in resolving power.

The resolving power of a microscope is:

$$RP = \frac{2n \sin\theta}{1.22\lambda}$$

where $$n$$ is the refractive index of the medium, $$\theta$$ is the half-angle of the cone of light, and $$\lambda$$ is the wavelength of light in vacuum (or air).

In air: $$RP_{\text{air}} = \frac{2 \times 1 \times \sin\theta}{1.22\lambda}$$

In oil: $$RP_{\text{oil}} = \frac{2 \times 2 \times \sin\theta}{1.22\lambda}$$

$$\frac{RP_{\text{oil}}}{RP_{\text{air}}} = \frac{2 \times 2}{2 \times 1} = 2$$

The resolving power in oil is twice that in air.

The correct answer is Option B: Resolving power will be twice in the oil than it was in the air.