The resolving power of a compound microscope will be maximum when

The diameter of the objective of a telescope is $200 \text{ cm}$. What is the resolving power of the telescope? Take the wavelength of light $\lambda = 5000 \text{ \AA}$.

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 diameter of the objective of a telescope is $1 \ m$. Its resolving limit for light of wavelength $4538 \ \text{Å}$ will be:

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: An electron microscope can achieve better resolving power than an optical microscope.
Reason $R$: The de Broglie wavelength of the electrons emitted from an electron gun is much less than the wavelength of visible light.
In the light of the above statements, choose the correct answer from the options given below:

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