Question:
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
Updated On: Jul 13, 2024
- 10 cm, 10 cm
- 15 cm, 5 cm
- 18 cm, 2 cm
- 11 cm, 9 cm
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The Correct Option is C
Solution and Explanation
Magnifying power, m = $\frac{f_{o}}{f_{e}} = 9$...(i)
where and are the focal lengths of the objective
and eyepiece respectively
Also, $f_{o} +f_{e} = 20\, cm$...(ii)
On solving (i) and (ii), we get
$f_{o} = 18\, cm,\, f_{e} = 2\, cm$
where and are the focal lengths of the objective
and eyepiece respectively
Also, $f_{o} +f_{e} = 20\, cm$...(ii)
On solving (i) and (ii), we get
$f_{o} = 18\, cm,\, f_{e} = 2\, cm$
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