Question

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.

Answer 1

Here, `f_(0) = 50 cm, f_(e) = 5 cm, u_(0) = - 200 cm`
If `v_(0)` is distance of image from the objective lens, then
`(1)/(v_(0)) = (1)/(f_(0)) + (1)/(u_(0)) = (1)/(50) + (1)/(-200) = (3)/(200)`
`v_(0) = (200)/(3) cm`
This image formed by objective acts as object for eye lens, whose final image is formed at least distance of distinct vision.
From `(1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e))`
`(1)/(u_(e)) = (1)/(v_(e)) - (1)/(f_(e)) = (1)/(-25) - (1)/(5) = (-6)/(25)`
`u_(e) = (-25)/(6) cm`
`:.` Separation between objective lens and eye piece
`L = |v_(0)| + u_(e)|`
`L = (200)/(3) + (25)/(6) = (425)/(6) = 70.83 cm`
Now, `m_(0) = (v_(0))/(u_(0)) = (200//3)/(-200) = -(1)/(3)`
and `m_(e) = (v_(e))/(u_(e)) = (-25)/(-25//6) = 6`
`:.` magnification produced by telescope,
`m = m_(0) xx m_(e) = -(1)/(3) xx 6 = -2`
Negative sign shows that final image is real and inverted.