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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.
Solution
In the astronomical telescope (in normal adjustment),
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:
f0 +fe =L
∴f0 = L - fe = 100 - 10 = 90 cm
Magnifying power (m) in normal adjustment:
m =`f_0/f_e = 90/10 = 9`
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Useful Constants & Relations:
| 1 | Charge of a proton | e | 1.6 × 10-19 C |
| 2 | Speed of light in vacuum | c | 3 × 108 ms-1 |
| 1 u = 931 MeV | |||
