Question

The eyepiece of an astronomical telescope has a focal length of 10 cm. The telescope is focussed for normal vision of distant objects when the tube length is 1.0. m. Find the focal length of the objective and the magnifying power of the telescope.

Answer 1

Given :

Focal length of the eyepiece, fe = 10 cm

Length of the tube, L = 1 m = 100 cm

Focal length of the objective, fo = ?

We know:

fo + fe = L

∴fo = L – fe = 100 – 10 = 90 cm

Magnifying power  in normal adjustment:

m = fo/fe=90/10=9

Answer 2

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