The diameter of the objective of a telescope is $a$,its magnifying power is $m$,and the wavelength of light is $\lambda$. The resolving power of the telescope is:

The average distance between the earth and moon is $38.6 \times 10^4 \ km$. The minimum separation between the two points on the surface of the moon that can be resolved by a telescope whose objective lens has a diameter of $5 \ m$ with $\lambda = 6000 \ \mathring{A}$ is ...... $m$.

Two luminous point sources separated by a certain distance are at $10 \,km$ from an observer. If the aperture of his eye is $2.5 \times 10^{-3} \,m$ and the wavelength of light used is $500 \,nm$, the distance of separation between the point sources just seen to be resolved is (in $\,m$)

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

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: