Find the maximum magnifying power of a compound microscope having a `25 D` lens as the objetcive, a `5D` lens as the eye-piece and the separation `30 cm` between the two lenses. The least distance for clear vision is `25 cm`.

Explanation

`P_(0) = 25D,P_(e) = 5D, L = 30cm, D = 25cm`
`f_(0) = (100)/(P_(0)) = (100)/(25) = 4cm`
`f_(e) = (100)/(P_(e)) = (100)/(5) = 20cm`
For eye-piece: `v_(e) = -D=- 25cm, f_(e) = 20cm, u_(e) =- x`
`(1)/(v_(e)) - (1)/(u_(2)) = (1)/(f_(e))`
`(1)/(-25) -(1)/(-x) = (1)/(20)`
`(1)/(x) = (1)/(20) +(1)/(25) = (9)/(100)`
`x = (100)/(9)cm,u_(e) =-x =(-100)/(9)cm`
`L =v_(0)+|u_(e)|`
`30 = v_(0)+(100)/(9) rArr v_(0) = 30-(100)/(9) = (170)/(9)cm`
For objective: `u_(0) =- x', v_(0) = (170)/(9)cm, f_(0) = 4cm`
`(1)/(v_(0))-(1)/(u_(0)) = (1)/(f_(0))`
`(1)/(170//9) - (1)/(-x') = (1)/(4)`
`(1)/(x') = (1)/(4)-(9)/(170) = (170-36)/(4xx170) = (134)/(4xx170)`
`x' = (4xx170)/(134)cm, u_(0) =- x' =- (4xx170)/(134)`
`m=(v_(0))/(u_(0))(1+(D)/(f_(e))) = (170//9)/(-(4xx170)/(134))(1+(25)/(20))`
`=- (134)/(36) (1+(5)/(4)) =- (134)/(36)xx(9)/(4) =- (134)/(16)`
`=- 8.4`