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 diameter of the objective lens of a telescope is $5.0\, m$ and the wavelength of light is $6000\ \mathring{A}$. The limit of resolution of this telescope will be......$sec$.

$A$ person is observing a bacteria through a compound microscope. For better analysis and to improve the resolving power,he should:

$A$ telescope uses light having a wavelength of $5000 \, \mathring{A}$ and lenses with focal lengths of $2.5 \, \text{cm}$ and $30 \, \text{cm}$. If the diameter of the aperture of the objective is $10 \, \text{cm}$,what are the resolving limit and the magnifying power of the telescope,respectively?

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 nearly: (in $\mu m$)

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