The magnifying power of an astronomical telescope in the normal adjustment position is 100. The distance between the objective and the eyepiece is 101 cm. Calculate the focal lengths of the objective and of the eyepiece.

Solution

Here we are given an astronomical telescope.
Distance between objective and eyepiece, l = 101 cm

Magnification, $m = \frac{f_{0}}{f_{e}} = 100$

$\Rightarrow$    $f_{0} = 100 f_{e}$

Since, $f_{0} + f_{e} = 101$

Then,    $100 f_{e} + f_{e} = 101$

$∴$    $f_{e} = 1 cm and, f_{0} = 100 cm .$

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Wave Optics

The magnifying power of an astronomical telescope in the normal adjustment position is 100. The distance between the objective and the eyepiece is 101 cm. Calculate the focal lengths of the objective and of the eyepiece.

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