A compound microscope has lenses of focal length `10 mm and 30 mm`. An object placed at `1.2 cm` from the first lens is seen through the second lens at `0.25 m` from the eye lens. Calculate
(i) magnifying power
(ii) distance between the two lenses.

Answer

`-46.7 ; 8.7 cm`

Explanation

Here, `f_(0) = 10 mm = 1 cm, f_(e) = 30 mm = 3 cm`
`u_(0) = - 1.2 cm, v_(e) = - 0.25 m = - 25 cm`
`M = ? L = ?`
From `(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0))`
`(1)/(v_(0)) = (1)/(f_(0)) + (1)/(u_(0)) = (1)/(1) - (1)/(1.2) = (0.2)/(1.2) = (1)/(6)`
`v_(0) = 6 cm`
From `(1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e))`
`(1)/(u_(e)) - (1)/(v_(e)) - (1)/(f_(e)) = (1)/(-25) - (1)/(3) = (-28)/(75)`
`u_(e) = (- 75)/(28) = - 2.7 cm`
`M = (v_(0))/(-u_(0))(1 + (d)/(f_(e))) = (6)/(-1.2)(1 + (25)/(3)) = -46.7`
`L = v_(0) + |u_(0)| = 6 + 2.7 = 8.7 cm`