If F_o and F_e are the focal lengths of the o | vedclass

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(C) The magnifying power $(M)$ of an astronomical telescope in normal adjustment is defined as the ratio of the angle subtended by the image at the ey

Vedclass · Accepted Answer

$F_o / F_e$

Vedclass · Answer

(C) The magnifying power $(M)$ of an astronomical telescope in normal adjustment is defined as the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the objective.For a telescope in normal adjustment,the final image is formed at infinity.The formula for the magnifying power is given by $M = \frac{f_o}{f_e}$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.Therefore,the correct option is $C$.

If $F_o$ and $F_e$ are the focal lengths of the objective and eyepiece respectively of a telescope,then its magnifying power will be

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