Question

In a compound microscope, the focal lengths of two lenses are $1.5\,cm$ and $6.25\,cm.$ An object is placed at $2\,cm$ from objective so what is the distance ( in $cm$ ) between two lenses if the final image is formed at $25\,cm$ from eye lens?

Answer 1

From lens maker's equation,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
For objective,
$\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$
$ \therefore v_0=\frac{u_0 f_0}{\left(u_0+f_0\right)} \ldots(i)$
For eye-piece,
$\frac{1}{f_e}=\frac{1}{V_e}-\frac{1}{u_e}$
Here, $v_e=D=25 \,cm$
$\therefore \frac{1}{u_e}=\frac{1}{D}-\frac{1}{f_c} $
$\therefore u_e=\frac{f_e D}{f_e-D} \ldots$ (ii)
From equations (i) and (ii),
$v_0=\frac{(-2)(1.5)}{(-2)+(1.5)}=6$
$u_e=\frac{(6.25)(-25)}{(6.25)-(-25)}=-5$
As negative sign denotes direction neglecting it,
$\therefore L-v_0+u_e-6+5-11 \,cm$