The diameter of a human eye lens is $2\,mm$. What will be the minimum distance between two points to resolve them,if they are situated at a distance of $50\,m$ from the eye? The wavelength of light used is $5000\,\mathring{A}$.

The human eye has an approximate angular resolution of $\phi = 5.8 \times 10^{-4} \, rad$ and a typical photoprinter prints a minimum of $300 \, dpi$ (dots per inch, $1 \, inch = 2.54 \, cm$). At what minimal distance $z$ should a printed page be held so that one does not see the individual dots?

$A$ telescope has a large diameter of the objective. Then,its resolving power is

The limit of resolution of a telescope is $2.5 \times 10^{-7} \text{ rad}$. If the telescope is used to detect light of wavelength $500 \text{ nm}$ coming from a star, the diameter of the objective lens used by the telescope is: (in $\text{ cm}$)

The correct relation between limit of resolution and resolving power is

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 $\,m$)