Question

A compound microscope consists of an objective of focal length `1 cm` and an eye-piece of focal length `5cm`. An object is placed at a distance of `0.5cm` from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on the screen `30 cm` behind the eye-piece ?

Answer 1

`f_(0) = 1cm, f_(e) = 5cm, u_(0) =- 0.5cm`
For objective:
`(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0))`
`(1)/(v_(0)) - (1)/(-0.5) = (1)/(1)`
`(1)/(v_(0)) +2=1 rArr v_(0) =- 1cm`

For eye-piece:
`u_(e) =- (L+1)cm, v_(e) = 30cm, f_(e) = 5cm`
`(1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e)) rArr (1)/(30) - (1)/(-(L+1)) = (1)/(5)`
`(1)/(L+1) = (1)/(5)-(1)/(30) = (1)/(6)`
`L+1 =6 rArr L =5cm`