According to Abbe,in the formula for the resolving power of a microscope,the numerical aperture is represented by:

The numerical aperture of a microscope is $0.12$, and the wavelength of light used is $600 \, nm$. Then its limit of resolution will be nearly $............. \, \mu m$.

Two luminous point sources separated by a certain distance are at $10 \,km$ from an observer. If the aperture of his eye is $2.5 \times 10^{-3} \,m$ and the wavelength of light used is $500 \,nm$, the distance of separation between the point sources just seen to be resolved is (in $\,m$)

If the wavelengths of light used in an optical instrument are $\lambda_1 = 4000 \, \mathring A$ and $\lambda_2 = 5000 \, \mathring A$,what will be the ratio of their resolving powers?

By increasing the aperture of the objective lens,how do the wavelength of light,the focal length of the objective lens,and the resolving power of an astronomical telescope change,respectively?

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?