The separation between two microscopic particles is measured as $P_A$ and $P_B$ by two different lights of wavelength $2000 \; \mathring{A}$ and $3000 \; \mathring{A}$ respectively. Then:

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?

An astronomical telescope has a large aperture to

Wavelengths of light used in an optical instrument are $\lambda_1 = 4000 \; \mathring{A}$ and $\lambda_2 = 5000 \; \mathring{A}$. The ratio of their respective resolving powers (corresponding to $\lambda_1$ and $\lambda_2$) 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 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?