The limit of resolution of an oil immersion objective microscope of numerical aperture $0.8$ for light of wavelength $0.6 \mu m$ is

The resolving power of a microscope depends upon:

Wavelengths of light used in an optical instrument are $\lambda_1 = 4000 \; \mathring{A}$ and $\lambda_2 = 5000 \; \mathring{A}$. The ratio of their respective resolving powers (corresponding to $\lambda_1$ and $\lambda_2$) is:

The ratio of resolving powers of an optical microscope for two wavelengths $\lambda_1 = 4000\,\mathring{A}$ and $\lambda_2 = 6000\,\mathring{A}$ is

We use a simple microscope to magnify an object. The microscope has a numerical aperture of $\sin \alpha = 0.24$. The object is so small that the resolving power of the microscope is fully utilized. If the diameter of the eye's pupil is $d = 4.0 \ mm$ and the least distance of distinct vision is $D = 25 \ cm$,what is the minimum magnifying power of the microscope?

The aperture diameter of a telescope is $5\; m$. The separation between the moon and the earth is $4 \times 10^{5} \; km$. With light of wavelength $5500\; \mathring{A}$, the minimum separation between objects on the surface of the moon, so that they are just resolved, is close to......$m$.