A compound microsope of an objective lens of focal `2.0cm` and an eye-piece of focal length `6.25cm` separeated by a distance of `15cm`. How far form the objective shold an object be placed in order to obtain the final image at (a) least distance of disnict vision `(25 cm)`, (b) infinity? What is the magnifying power of the microscope in each case?

Explanation

Here `f_(0) =2.0 cm f_(e ) = 6.25 cm u_(0) =?`
(a) `v_(e ) = -25 cm`
`:' , (1)/( v_(e )) - (1)/(u_(e)) = (1)/(f_(e))`
`:. , (1)/(u_(e))= (1)/(v_(e)) - (1)/(f_(e)) = (1)/(-25)- (1)/(6.25) = (-1-4)/(25) = (-5)/(25)`
`rArr u_(e) = - 5 cm`
As distance between objective and eye piece = 15 cm
`v_(0) =15 - 5 = 10 cm`
`:. , (1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0))`
`:. , (1)/(u_(0)) = (1)/(v_(0)) - (1)/(f_(0)) = (1)/(10) - (1)/(2) = (1-5)/(10)`
`rArr u_(0) = (-10)/(4) = - 2.5 cm `
Magnifying power
`= (v_(0))/(|u_(0)|)[1+ (D )/(f_(e))] = (10)/(2.5) [1+(2.5)/(6.25)] =20`
(b) `:. , v_(e ) = prop , u_(e) = f_(e) = 6.25 cm `
`:. v_(0) = 15 - 6.25 = 8.75 cm `
`:. , (1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0))`
`:. (1)/(u_(0)) = (1)/(v_(0)) - (1)/(f_(0)) = (1)/(8.75) - (1)/( 20) = (2-8.75)/(17.5)`
`rArr u_(0) = (-17.5)/(6.75) = - 2.59 cm`
Magnifying power
`= (v_(0))/(|u_(0)|) xx [1+ (D )/(f_(e))] = (v_(0))/(|u_(0)|) xx (D )/(|u_(e)|) = (8.75)/(2.59) xx (25)/(6.25) = 13.51`