The aperture of the objective lens of a telescope is made large so as to

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 $\,m$)

There are two white dots on a black paper separated by a distance of $1 \, mm$. They are observed by the naked eye. If the diameter of the eye lens is $3 \, mm$,what is the maximum distance between the dots and the eye so that they can be seen as separate (in $, m$)? The wavelength of light is $500 \, nm$.

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 $= 500 \ nm$)

Assuming the human pupil to have a radius of $0.25 \ cm$ and a comfortable viewing distance of $25 \ cm$,the minimum separation between two objects that the human eye can resolve at $500 \ nm$ wavelength is nearly: (in $\mu m$)

Which of the following is the smallest size grain that can be seen using a normal microscope using visible light $(400 \ nm < \lambda < 700 \ nm)$ (in $nm$)?