The resolving power of an astronomical telescope is $0.2$ seconds. If the central half portion of the objective lens is covered,the resolving power will be......$sec$

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 $\mu m$)

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?

The resolving power of a telescope whose lens has a diameter of $1.22 \ m$ for a wavelength of $5000 \ \mathring{A}$ is

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

The aperture of the objective of a telescope is $24.4 \, cm$. What is the resolving power of this telescope if light of wavelength $2440 \, \mathring{A}$ is used to view the object?