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

Answer

(a) `v_(0) = -2.5cm` and `f_(e) = 6.25cm` give `mu_(e) = -5cm; v_(0) = (15-5)cm = 10cm`
`f_(0) = u_(0) = -2.5cm;`
Magniflying power `= (10)/(2.5) xx (25)/(5) = 20`
(b) `mu_(e) = -6.25cm, v_(0) = (15-6.25)cm = 8.75, f = 2.0cm`. Therefore
`u_(0) = -(70//27) = -2.59cm`
Magnifying power
`= (v_(0))/(|u_(0)|) xx (25//6.25) = (27)/(8) xx 4 = 13.5`

Explanation

(a) `v_(e) = -2.5cm` and `f_(e) = 6.25cm` give `u_(e) = -5cm, v_(0) = (15-5)cm = 10cm`.
`f_(0) = u_(00 = -2.5cm,` Magnifying power `= (10)/(2.5) xx (25)/(5) = 20`
(b) `u_(theta) = -6.25cm, v_(o) = (15-6.25)cm = 8.75, f_(0) = 2.0cm`. Therefore
`mu_(0) = -(70//27) = -2.59cm`
Magnifiying power `= (v_(0))/(|mu_(0)|) xx (25//6.25) = (27)/(8) xx 4 = 13.5`