Question

In a compound microscope the objective and eye-piece have focal lengths of $9.5cm$ and $5cm$ respectively, and are kept at a distance of $20cm$ . The last image is formed at a distance of $25cm$ from eye-piece. What is the total magnification of the microscope?

• A. $95$
• B. $95/6$
• C. $94/6$
• D. None of these

Answer 1

The correct option is D None of these

In the case of eye-piece lens

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{25}-\frac{1}{u}=\frac{1}{5}$

$\frac{1}{u}=\frac{1}{25}-\frac{1}{5}$

$u=\frac{-5}{4}$

$u=-1.25cm$

For objective the image distance

$v=20-1.25=18.75cm$

Again

$\frac{1}{18.75}-\frac{1}{u}=\frac{1}{9.5}$

$\frac{1}{u}=\frac{4}{75}-\frac{2}{19}$

$\frac{1}{u}=\frac{76-150}{75\times 19}$

$u=\frac{75\times 19}{-74}$

$u=-19.25cm$

Total magnification $m=m_1{m}_2$

$m=\frac{18.75}{19.25}\times \frac{25}{1.25}$

$m=19.6$