Question

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

Answer 1

Answer: 11

From lens maker's equation,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
For objective,
$\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$
$\therefore v_0=\frac{u_0{f}_0}{\left(u_0+f_0\right)}\dots (i)$
For eye-piece,
$\frac{1}{f_e}=\frac{1}{V_e}-\frac{1}{u_e}$
Here, $v_e=D=25cm$
$\therefore \frac{1}{u_e}=\frac{1}{D}-\frac{1}{f_c}$
$\therefore u_e=\frac{f_{e}D}{f_e-D}\dots$ (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,
$\therefore L-v_0+u_e-6+5-11cm$