If the numerical aperture $(NA)$ of a microscope is increased,then its:

An electron microscope is operated at $40 \ kV$. The ratio of the resolving power of this microscope to another one which uses yellow light of wavelength $6 \times 10^{-7} \ m$ 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?

If $f_{0} = 5 \, cm$,$\lambda = 6000 \, \mathring{A}$,and $a = 1 \, cm$ for a microscope,what will be its resolving power?

Some distant star is to be observed by a telescope with an objective lens diameter of $a$, at an angular resolution of $3.0 \times 10^{-7}$ radian. If the wavelength of light from the star reaching the telescope is $500$ nm, the minimum diameter of the objective lens of the telescope is . . . . . . cm. (nearest integer)

When the object is self-luminous,the resolving power of a microscope is given by the expression