1. Resolving power of a microscope

   For a microscope, the limit of resolution is given by $d = \frac{\lambda}{2\mu \sin\theta}$ where:

   • $\lambda$ = wavelength of light in the medium,
   • $\mu$ = refractive index of the medium between object and objective,
   • $\theta$ = half-angle of collection.

   Resolving power $R$ is inversely proportional to $d$: $R = \frac{1}{d} = \frac{2\mu\sin\theta}{\lambda}$

2. Effect of immersing in oil

   Initially, microscope is in air: $\mu_{\text{air}} = 1$ and wavelength in air is given as $\lambda$.

   When immersed in oil of refractive index $2$, wavelength becomes $\lambda_{\text{oil}} = \frac{\lambda}{2}$ because wavelength in a medium is inversely proportional to refractive index.

3. Resolving power in air

   $R_{\text{air}} = \frac{2(1)\sin\theta}{\lambda} = \frac{2\sin\theta}{\lambda}$

4. Resolving power in oil

   Now $\mu = 2$ and $\lambda_{\text{oil}}=\lambda/2$: $R_{\text{oil}} = \frac{2\cdot 2\sin\theta}{\lambda/2}$ $R_{\text{oil}} = \frac{4\sin\theta}{\lambda/2} = \frac{8\sin\theta}{\lambda}$

5. Compare the two

   $\frac{R_{\text{oil}}}{R_{\text{air}}} = \frac{8\sin\theta/\lambda}{2\sin\theta/\lambda} = 4$

   Hence, the resolving power becomes four times in oil.

6. Option check

   • A: $\frac{1}{4}$ times ❌
   • B: twice ❌
   • C: four times ✅
   • D: $\frac{1}{2}$ times ❌

Therefore, the correct option is C.

7. Comparison with stored answer

   Stored correct answer is B, but the derived result is C.

   The likely reason for the discrepancy is that if one uses only $R \propto \frac{\mu}{\lambda}$ and forgets that wavelength itself changes in the medium, one may incorrectly get a factor of $2$. Including both effects correctly gives a factor of $4$.