Question

In a compound microscope, the focal lengths of two lenses are $1.5\, cm$ and $6.25\, cm$. If an object is placed at $2 \,cm$ from objective and the final image is formed at $25 \,cm$ from eye lens, the distance between the two lenses is

• A. $6.00\, cm$
• B. $7.75\, cm$
• C. $9.25\, cm$
• D. $11.0\, cm$

Answer 1

Answer: (D)

Here, $f_0 = 1.5 \,cm, f_e = 6.25\, cm$,
$u_0 = -2 \,cm\,\, v_e = -25 \,cm$
For objective,
$\frac{1}{v_0} -\frac{1}{u_0}=\frac{1}{f_0}$
$\therefore \frac{1}{v_0} - \frac{1}{-2} = \frac{1}{1.5}$ or $v_0 = 6\,cm$
For eye piece,
$\therefore \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$;
$\frac{1}{-25}-\frac{1}{u_e} = \frac{1}{6.25}$
$-\frac{1}{u_e} = \frac{1}{6.25} + \frac{1}{25}$ or
$u_e = -5cm$
Distance between two lenses = $|v_0|+|u_e|$
$ = 6\,cm + 5\,cm = 11\,cm$