A compound microscope has an objective of focal length `2.0 cm` and an eye-piece of focal length `6.25cm` and distance between the objective and eye-piece is `15cm`. If the final image is formed at the least distance vision `(25 cm)`, the distance of the object form the objective is

Explanation

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