1. What determines resolving power of a microscope?

For a microscope, the resolving power increases when the limit of resolution decreases.

The minimum resolvable distance is approximately $d_{\min} = \frac{1.22\,\lambda}{2\mu \sin\theta}$ or equivalently, $d_{\min} \propto \frac{\lambda}{\mu \sin\theta}$ where:

• $\lambda$ = wavelength of light used,
• $\mu$ = refractive index of the medium between object and objective,
• $\theta$ = نصف-angle subtended by the objective lens,
• $\mu \sin\theta$ is the numerical aperture.

So, to improve resolving power, we need to decrease $d_{\min}$, i.e.:

• decrease $\lambda$,
• increase $\mu$,
• increase aperture.

2. Check each option

Option A: Decrease the focal length of the eyepiece

This mainly increases magnifying power, not resolving power.

So this is not the best option.

Option B: Increase the wavelength of the light

Since $d_{\min} \propto \lambda,$ increasing $\lambda$ makes resolution worse.

So this is incorrect.

Option C: Increase the refractive index of the medium between the object and objective lens

Since $d_{\min} \propto \frac{1}{\mu},$ increasing $\mu$ decreases the minimum resolvable distance and hence improves resolving power.

So this is correct.

Option D: Decrease the diameter of the objective lens

A smaller objective diameter means smaller aperture, hence smaller numerical aperture, and poorer resolving power.

So this is incorrect.

3. Best option

Therefore, the scientist should: $\boxed{\text{Increase the refractive index of the medium between the object and objective lens}}$

So the correct option is $\boxed{\text{C}}$