Question

What is the resolving power of a telescope, for light of wavelength $5000\,A^o $.If the aperture of the objective of the telescope is $1.22m$?

Answer 1

Hint:To solve this question we should know that the resolving power of the telescope is depends on the aperture and the wavelength of light and it is given as $\dfrac{aperture}{1.22\times \lambda }$ where, $\lambda $ is the wavelength of the light.To get the answer of this question we simply put the value of the diameter of the aperture and the wavelength.

Complete step by step answer:
Astronomical telescope: it is used to increase visual angle of distant large objects.
Its objective lens is of large focal length and aperture while eye lens of short focal length and aperture and both are convergent.Final image is inverted, virtual and enlarged at a distance D to infinity from the eye.

Magnifying power(MP) of this telescope becomes $\dfrac{1}{{{m}^{2}}}$ times of its initial values if the objective and eye lense are interchanged as MP$\sim [\dfrac{{{f}_{o}}}{{{f}_{e}}}]$. Magnifying power is increased by increasing the focal length of the objective lens and by decreasing the focal length of the eyepiece.

Resolving power(RP) is increased by increasing the aperture of the object. $(\because RP=\dfrac{D}{1.22\times \lambda })$. In normal adjustment, to have large magnifying power, ${{f}_{o}}$ must be as large as practically possible while ${{f}_{e}}$ has to be kept small.

Given: $\lambda=5000\,A^o$ and aperture(D) = $1.22\,m$
So the resolving power (RP) = $\dfrac{aperture}{1.22\times \lambda }$
$RP=\dfrac{1.22}{1.22\times 5000\times {{10}^{-10}}}$
$\therefore RP= 2\times {{10}^{6}}\,m$

Hence, resolving power of telescope is $2\times {{10}^{6}}\,m$.

Note:In a telescope, the aperture and focal length of objective are greater than that of eyepiece.Reflecting telescope in which an objective lens is replaced by a concave mirror.The reflecting telescope is cheap, light, and portable as compared to the astronomical telescope.