Let the image is formed at normal near point $25 \mathrm{~cm}$.
$\therefore$ Angular magnification
$$
=\left(1+\frac{\mathrm{d}}{\mathrm{f}_{\mathrm{e}}}\right)=\left(1+\frac{25}{5}\right)=6 \text {. }
$$
Magnification of the objective lens,
$$
\begin{aligned}
&\mathrm{m}=\frac{30}{6}=5 \\
&\text { As, } \mathrm{m}=\frac{\mathrm{v}_0}{-\mathrm{u}_0} \Rightarrow 5=\frac{\mathrm{v}_0}{-\mathrm{u}_0} \quad \therefore \mathrm{v}_0=-5 \mathrm{u}_0 \\
&1
\end{aligned}
$$
As, $\frac{1}{\mathrm{v}_0}-\frac{1}{\mathrm{u}_0}=\frac{1}{\mathrm{f}_0}$
$\therefore \frac{1}{-5 \mathrm{u}_0}-\frac{1}{\mathrm{u}_0}=\frac{1}{1.25}, \frac{-1-5}{5 \mathrm{u}_0}=\frac{1}{1.25}$,
$\frac{-6}{5 \mathrm{u}_0}=\frac{1}{1.25}$
$\therefore \mathrm{u}_0=\frac{-6 \times 1.25}{5}=-1.5 \mathrm{~cm}$.
$\therefore \mathrm{v}_0=-5 \mathrm{u}_0=-5 \times(-1.5)=7.5 \mathrm{~cm}$
As, $\frac{1}{\mathrm{v}_{\mathrm{e}}}-\frac{1}{\mathrm{u}_{\mathrm{e}}} \therefore \frac{1}{\mathrm{u}_{\mathrm{e}}}=\frac{1}{-25}-\frac{1}{5}=\frac{-1-5}{25}=\frac{-6}{25}$
$\therefore \mathrm{u}_{\mathrm{e}}=-4.17 \mathrm{~cm}$
The separation between the objective and the eye piece $=7.5+4.17=11.67 \mathrm{~cm}$.