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:

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 $500 \, nm$.

$A$ telescope is used to observe two objects at a distance of $z = 10 \ km$ which are $s = 0.12 \ m$ apart and illuminated by light of wavelength $\lambda = 600 \ nm$. Estimate the diameter of the objective lens of the telescope if it can just resolve the two objects. Assume diameter $D >> \lambda$ and separation between objects $s << z$. The answer is in $cm$.

The value of numerical aperture of the objective lens of a microscope is $1.25$. If light of wavelength $5000\,\mathring{A}$ is used, the minimum separation between two points, to be seen as distinct, will be....$\mu m$

If the numerical aperture $(NA)$ of a microscope is increased,then its:

$A$ man can clearly see two poles at a distance of $11 \, km$. What should be the minimum distance between the poles in $m$?