If the focal length of the eyepiece of a tele | vedclass

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(C) The magnifying power $(m)$ of an astronomical telescope in normal adjustment is given by the formula: $m = -\frac{f_o}{f_e}$, where $f_o$ is the f

Vedclass · Accepted Answer

$\frac{m}{2}$

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(C) The magnifying power $(m)$ of an astronomical telescope in normal adjustment is given by the formula: $m = -\frac{f_o}{f_e}$, where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.If the focal length of the eyepiece is doubled, the new focal length becomes $f_e' = 2f_e$.The new magnifying power $(m')$ will be: $m' = -\frac{f_o}{f_e'} = -\frac{f_o}{2f_e} = \frac{1}{2} \left( -\frac{f_o}{f_e} \right) = \frac{m}{2}$.Therefore, the magnifying power becomes half of its original value.

If the focal length of the eyepiece of a telescope is doubled, its magnifying power $(m)$ will be

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