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The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are:

1. 10 cm, 10 cm
2. 15 cm, 5 cm
3. 18 cm, 2 cm
4. 11 cm, 9 cm

Correct answer: 15 cm, 5 cm

Solution

The magnifying power of a telescope for parallel rays is given by M = f_o / f_e, where f_o and f_e are the focal lengths of the objective and eyepiece, respectively. Also, the distance between the lenses is f_o + f_e. Solving the equations M = 9 and f_o + f_e = 20 cm, we get f_o = 15 cm and f_e = 5 cm.

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