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Question

Physics Question on Optical Instruments

The magnifying power of a telescope is 9.9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20cm.20\, cm. The focal length of lenses are

A

10 cm, 10 cm

B

15 cm, 5 cm

C

18 cm, 2 cm

D

11 cm, 9 cm

Answer

18 cm, 2 cm

Explanation

Solution

Magnifying power, m = fofe=9\frac{f_{o}}{f_{e}} = 9...(i)
where and are the focal lengths of the objective
and eyepiece respectively
Also, fo+fe=20cmf_{o} +f_{e} = 20\, cm...(ii)
On solving (i) and (ii), we get
fo=18cm,fe=2cmf_{o} = 18\, cm,\, f_{e} = 2\, cm