Question

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

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

Answer 1

The correct option is B $200$

Given,
$f_0=1.2\text{cm},f_e=3\text{cm}$
$u_0=-1.25\text{cm}$
When final image is formed at infinity, magnifying power is given by, $MP=\frac{v_0}{|u_0|}\left(\frac{D}{f_0}\right)$
Using lens formula for objective,
$\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0}$
$\Rightarrow \frac{1}{v_0}-\frac{1}{-1.25}=\frac{1}{1.2}\Rightarrow \frac{1}{v_0}=\frac{1}{1.2}-\frac{1}{1.25}$
$\therefore v_0=30\text{cm}$.
Now, $MP=\frac{v_0}{|u_0|}\left(\frac{D}{f_e}\right)$

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

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