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?

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:

To increase both the resolving power and magnifying power of a telescope:

Resolving power of a telescope increases when

Two point white dots are $1 \ mm$ apart on a black paper. They are viewed by an eye with a pupil diameter of $3 \ mm$. Approximately,what is the maximum distance at which the dots can be resolved by the eye? (Take wavelength of light $= 500 \ nm$)

$A$ telescope has an objective lens of $10\, cm$ diameter and is situated at a distance of $1\, km$ from two objects. The minimum distance between these two objects which can be resolved by the telescope,when the mean wavelength of light is $5000\,\mathring{A}$,is the order of