The diameter of the objective of a telescope | vedclass

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(D) The resolving power of a telescope is defined as the reciprocal of the minimum angular separation $(	heta)$ between two distant objects that can

Vedclass · Accepted Answer

$a / (1.22 \lambda)$

Vedclass · Answer

(D) The resolving power of a telescope is defined as the reciprocal of the minimum angular separation $(	heta)$ between two distant objects that can be just distinguished by the telescope.The formula for the angular resolution (limit of resolution) is given by:$	heta = 1.22 \lambda / a$Therefore,the resolving power $(R)$ is:$R = 1 / 	heta = a / (1.22 \lambda)$Where:$a$ is the diameter of the objective lens.$\lambda$ is the wavelength of light used.Thus,the correct option is $D$.

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:

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