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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:

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:

Option: 1

10 cm, 10 cm


Option: 2

15 cm, 5 cm


Option: 3

18 cm, 2 cm


Option: 4

11 cm, 9 cm


Answers (1)

best_answer

As we learnt in

Astronomical Telescope -

m= frac{-f_{o}}{f_{e}}left ( 1+frac{f_{e}}{D} ight )
 

- wherein

f_{o} = focal length of objective

f_{e}= focal length of eyepiece

 

 

 

magnifying power = \frac{f_{0}}{f_{e}}=9

f_{0}=9f_{e}----------(1)

Also f_{0}+f_{e}=20 cm

\Rightarrow 10 f_{e}=20cm

f_{e}=2cm

f_{0}=18cm

Posted by

Gaurav

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