A compound microscope has an objective of focal length `2 cm` and an eye-piece of focal length `3 cm`. An object is placed at a distance of `2.4cm` in front of the field lens. Find the magnifying power of the instrument and length of the tube is (a) final image is at infinity and (b) final image is at least distance of distinct vision.

Explanation

`f_(0) = 2cm, f_(e) = 5cm, u_(0) =- 2.4 cm`
(a) `(1)/(v_(0))-(1)/(u_(0)) = (1)/(f_(0))`
`(1)/(v_(0)) - (1)/(-2.4) = (1)/(2)`
`(1)/(v_(0)) = (1)/(2) - (1)/(2.4) = (2.4-2)/(4.8) = (1)/(12) rArr v_(0) = 12 cm`
Tuber length: `L = v_(0) +f_(e) = 12 +5 = 17cm`
`m = (v_(0))/(u_(0)) .(D)/(f_(e)) = (12)/(-2.4) xx (25)/(5) =- 25`
(b) `(1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e))`
`(1)/(-25) - (1)/(u_(e)) = (1)/(5) rArr -(1)/(u_(e)) = (1)/(5) +(1)/(25) = (6)/(25)`
`u_(e)=- (25)/(6)cm`
Tube length: `L = v_(0)+|u_(e)| = 12+(25)/(6) = (97)/(6) = 16.2cm`
`m =(v_(0))/(u_(0))(1+(D)/(f_(e))) = (12)/((-2.4)) (1+(25)/(5)) =- 30`