A compound microscope has a magnifying power `30`. The focal length of its eye-piece is `5cm`. Assuming the final to be at the least distance of distinct vision `(25 cm)`, calculate the magnification produced by objective.

Explanation

In case of compound microscope,
`M=m_(0)xxm_(e)to 1`
And in case of final image at least distance of distinct vision,
`m_(e)[1+D.f_(e)]to 2`
so, from eqs,(1) and (2), `M=m_(0)[1 + D/f_(e)]`
Here M=-30,D=25cm and `f_(e)=5cm`
So, `-30=m[1 + 25/5] ie m_(0)=(-30)/6=-5`
Negative sign inplies that image formed by objective is inverted.