When the wavelengths of light used in optical instruments $A$ and $B$ are $4500 \, \text{Å}$ and $6000 \, \text{Å}$ respectively, the ratio of the resolving power of $A$ to $B$ will be:

An electron microscope is operated at $40 \ kV$. The ratio of the resolving power of this microscope to another one which uses yellow light of wavelength $6 \times 10^{-7} \ m$ is:

For a telescope to have large resolving power,the

Two point white dots are $1 \ mm$ apart on a black paper. They are viewed by an eye with a pupil diameter of $3 \ mm$. Approximately,what is the maximum distance at which the dots can be resolved by the eye? (Take wavelength of light $= 500 \ nm$)

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

According to Abbe,in the formula for the resolving power of a microscope,the numerical aperture is represented by: