In a compound microscope, the focal length of the objective lens is $1.2\text{cm}$ and the focal length of the eyepiece 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. The magnifying power of the compound microscope is -

The correct option is B $200$
Given:
$f_o=1.2\text{cm}$
$f_e=3\text{cm}$
$u_o=-1.25\text{cm}$

Using lens formula for objective lens,

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

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

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

Now, magnifying power of the compound microscope when the final image is formed at infinity,

$m=\frac{v_o}{|u_o|}\left(\frac{D}{f_e}\right)$

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

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