The length of the compound microscope is $18\text{cm}$. The magnifying power for the relaxed eye is $28$, find the object distance from the objective lens, if the focal length of the eyepiece is $4\text{cm}$.

The correct option is A $3.125\text{cm}$
Given,
$L=18\text{cm}$ and $m=28$ (for relaxed eye), $f_e=4\text{cm}$

We know for relaxed eye, the image is formed at infinity. So, for eyepiece object distance will be equal to $f_e$.

$\therefore$ Length of microscope,

$L=v_0+f_e$

$\Rightarrow 18=v_0+4$

$\therefore v_0=14\text{cm}$

Now, $m=\frac{v_0}{u_0}\times \frac{D}{f_e}$

$\Rightarrow 28=\frac{14}{u_0}\times \frac{25}{4}$

$\therefore u_0=3.125\text{cm}$

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