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" because (1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e)) therefore (1)/(u_(e)) = (1)/(v_(e)) - (1)/(f_(e)) = (1)/(-25) - (1)/(6.25) = (-1-4)/(25) = (-5)/(25) implies u_(e) = -5"cm"`
As distance between objective and eye piece = `15"cm" , v_(0) = 15-5 = 10"cm"`
`because (1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0)) implies (1)/(u_(0)) = (1)/(v_(0)) - (1)/(f_(0)) = (1)/(8.75) - (1)/(20) = (2-8.75)/(17.5) implies 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`