If $f_{0} = 5 \, cm$,$\lambda = 6000 \, \mathring{A}$,and $a = 1 \, cm$ for a microscope,what will be its resolving power?

The limit of resolution of an oil immersion objective microscope of numerical aperture $0.8$ for light of wavelength $0.6 \mu m$ is

The wavelengths of light used in an optical instrument are ${\lambda _1 = 4000 \ \text{\AA}}$ and ${\lambda _2 = 5000 \ \text{\AA}}$. The ratio of their respective resolving powers (corresponding to ${\lambda _1}$ and ${\lambda _2}$) is:

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..... $\mu m$.

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 $5000\,\mathring{A}$.

Resolving power of a microscope can $NOT$ be increased by