A telescope has an objective of focal length `50 cm` and eye piece of focal length `5 cm`. The least distance of distinct vision is `25 cm`. The telescope is focussed for distinct vision on a scale `200 cm` away from the objective. Calculate
(i) the separation between objective and eye piece
(ii) the magnification produced.

Options

A. 75 cm
B. 60 cm
C. 71 cm
D. 74 cm

Answer

C

Explanation

(c ) For objective,
`1/v_(o)-1/(-200)=1/50`
`therefore v_(o)=200/3cm`
For eye-pieces,
`1/(-25)-1/((-u_(e)))=1/5`
`therefore u_(e)=25/6`
Therefore, seperation between objecctive and eye-piece,
`L=v_(o)+u_(e)=200/3+25/6=425/6=71 cm`