Question

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

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

Answer 1

Answer: (D)

Here, $f_0=1.5cm,f_e=6.25cm$,
$u_0=-2cm{v}_e=-25cm$
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 $v_0=6cm$
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 = $|v_0|+|u_e|$
$=6cm+5cm=11cm$