Question

In a compound microscope the focal length of an objective lens is $1.2\text{cm}$ and the focal length of the eye piece is $3\text{cm}$. When the object is kept at $1.25\text{cm}$ in front of the objective lens, the final image is formed at infinity. Magnifying power of the compound microscope should be:

• A. $400$
• B. $200$
• C. $150$
• D. $100$

Answer 1

The correct option is B $200$

Given:
$f_o=1.2\text{cm},f_e=3\text{cm}$ and $u_o=-1.25\text{cm}$

When the final image is formed at infinity, the magnifying power is given as :

$m=\frac{v_o}{\left|\begin{array}{c}{u}_o\end{array}\right|}\left(\frac{D}{f_e}\right)$

Using lens formula:

$\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$

$\Rightarrow \frac{1}{v_o}-\frac{1}{-1.25}=\frac{1}{1.2}$

$\Rightarrow \frac{1}{v_o}=\frac{1}{1.2}-\frac{1}{1.25}$

$\therefore v_o=30\text{cm}$

Now,

$m=\frac{v_o}{\left|\begin{array}{c}{u}_0\end{array}\right|}\left(\frac{D}{f_e}\right)$

$=\frac{30}{1.25}\times \left(\frac{25}{3}\right)=200$

Hence, $\left(B\right)$ is the correct answer.