Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If the focal length of objective and eye lens are $1.2\, cm$ and $3\, cm$ respectively and the object is put $1.25\, cm$ away from the objective lens and the final image is formed at infinity. The magnifying power of the microscope is :

Solution:

$m _{\infty} - \frac{ v _{0}}{ u _{0}} \times \frac{ D }{ f _{ e }}$
From $\frac{1}{ f _{0}}=\frac{1}{ v _{0}}-\frac{1}{ u _{0}}$
$\Rightarrow \frac{1}{(+1.2)}=\frac{1}{ v _{0}}-\frac{1}{(-1.25)}$
$\Rightarrow v _{0}=30\, cm$
$\therefore \left| m _{\infty}\right|=\frac{30}{1.25} \times \frac{25}{3}=200$