An angular magnification (magnifying power) of `30 X` is desired using an objective of focal length `1.25 cm` and an eye piece of focal length `5 cm`. How will you set up the compound microscope ?

Explanation

In normal adjustment, image is formed at least distance of distinct vision, `d = 25 cm`.
Angular magnification of eye piece `= (1 + (d)/(f_(e))) = (1 + (25)/(5)) = 6`
As total magnification is `30`, magnification of objective lens,
`m = (30)/(6) = 5 :. m = (v_(0))/(-u_(0)) = 5 cm, or v_(0) = - 5 u_(0)`
As `(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0)) :. (1)/(- 5 u_(0)) - (1)/(u_(0)) = (1)/(1.25)`
`-(6)/(5 u_(0)) = (1)/(1.25)`
`u_(0) = (-6 xx 1.25)/(5) = - 1.5 cm`
i.e., object should be held at `1.5 cm` in front of objective lens.
As `v_(0) = - 5 u_(0) :. v_(0) = - 5(- 1.5) = 7.5 cm`
From `(1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e))`
`(1)/(u_(e)) = (1)/(u_(e)) = (1)/(v_(e)) - (1)/(f_(e)) = (1)/(-25) - (1)/(5) = -(6)/(25)`
`u_(e) = (-25)/(6) = -4.17 cm`
`:.` Separation between the objective lens and eye piece `= |u_(e)| + |v_(0)| = 4.17 + 7.5 = 11.67 cm`.