Home
Class 12
PHYSICS
The magnifying power of a telescope is 9...

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

A

10 cm, 10 cm

B

15 cm, 5 cm

C

18 cm, 2 cm

D

11 cm, 9 cm

Text Solution

Verified by Experts

The correct Answer is:
C
`m = (f_(o))/(f_(e)) = 9 `
`and f_(o) + f_(e) = 20 rArr 9f_(e) + f_(e) = 20 rArr f_(e) = 2 cm `.
`f_(0) = 18 ` cm.
Promotional Banner

Similar Questions

Explore conceptually related problems

The magnifying power of an astronomical telescope is 5 . When it is set for normal adjustment, the distance between the two lenses is 24 cm . Calculate the focal lengths of eye piece and objective lens.

The magnifying power of an astronomical telescope for relaxed vision is 16. On adjusting, the distance between the objective and eye lens is 34 cm . Then the focal length of objective and eye lens will be respectively

The magnifying power of an astronomical telescope in the normal adjustment position is 100 . The distance between the objective and eye piece is 101 cm . Calculate the focal lengths of objective and eye piece.

An astronomical telescope has a magnifying power of 10 . In normal adjustment, distance between the objective and eye piece is 22 cm . The focal length of objective lens is.

The magnifying power of terrestrial telescope is 25 when it is in normal adjustment and the length of the telescope is 124 cm. If the focal length of the erecting lens is 5 cm, the focal lengths of the objective and the eye-piece are respectively.

An astronomical telescope has a magnifying power of 10 . In normal adjustment, distance between the objective and eye piece is 22 cm calculate focal length of objective lens.