A compound microscope has a magnifying power of `100` when the image is formed at infinity. The objective has focal length `0.5cm` and the tube length is `6.5cm`. Find the focal length of the eye-piece.

Explanation

`m=- 100, f_(0) = 0.5cm, L = 6.5cm`
Final image is at infinite.
`m = (v_(0))/(u_(0))(D)/(f_(e))`
`-100=(v_(0))/(u_(0))xx(25)/(f_(e)) rArr (v_(0))/(u_(0)f_(e)) =- 4 ….(a)`
`L = v_(0) +f_(e) rArr 6.5 -f_(e) = v_(0) …(b)`
For objective
`(1)/(v_(0))-(1)/(u_(0))=(1)/(f_(0))`
`(v_(0))/(v_(0)) - (v_(0))/(u_(0)) = (v_(0))/(f_(0))`
`1-(v_(0))/(u_(0)) = (v_(0))/(f_(0)) rArr (v_(0))/(u_(0)) =1-(v_(0))/(f_(0)) =1-(v_(0))/(0.5) =1- 2v_(0) ...(c )`
`(v_(0))/(u_(0)f_(e)) =- 4rArr (1-2v_(0))/(f_(e)) =- 4`
`1-2(6.5-f_(e)) =- 4f_(e)`
`1-13+2f_(e) =- 4f_(e)`
`-12 =- 6f_(e) rArr f_(e) =2cm`