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.

Explanation

Given `u=-200cm, f=50cm`
For image `I_(1)` object formed by objective lens,
`(1)/(f)=(1)/(v)-(1)/(u)`
we have
`(1)/(v)=(1)/(f)+(1)/(u)=(1)/(50)+(1)/(200)=(4-1)/(200)=(3)/(200)`
`rArr v=+(200)/(3) cm`
Also, magnification produced by objective lens
`m_(0)=(v)/(u)=-(200//3)/(200)=(1)/(3)`
Image `I_(1)` acts as an object for eye lens.
Here, `v=-25cm, f=5 cm`
`:. (1)/(f)=(1)/(v)-(1)/(u)`
`rArr (1)/(u)=(1)/(v)-(1)/(f)=-(1)/(25)-(1)/(5)=-(1+5)/(25)`
`:. u=-(25)/(6)cm`
And magnification produced by eye lens,
`m_(e)=(v)/(u)=(-25)/((-25//6))=6`
a. The separation between objective and eyepiece
`=|V|+|u|=(200)/(3)+(25)/(6)=(425)/(6)=70.73cm`
b. Magnification produced, `m=m_(0)xxm_(e)=-(1)/(3)xx6=-2`
The negative sign show that the final image is inverted.