An astronomical telescope is in normal adjustment position has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance

${\mathrm{f}}_0=40\mathrm{cm}$
${\mathrm{f}}_{\mathrm{e}}=4\mathrm{cm}$

$\begin{array}{l}{\mathrm{u}}_0=-200\mathrm{cm}\\ \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{-200}=\frac{1}{40}\\ \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{40}-\frac{1}{200}=\frac{5-1}{200}=\frac{4}{200}\\ \Rightarrow \mathrm{v}=50\mathrm{cm}\end{array}$
Image will be formed at first focus of  eyepiece lens. So, for normal adjustment distance between objective and eyepiece lens (length of tube) will be
$\mathrm{v}+{\mathrm{f}}_{\mathrm{e}}=50+4=54\mathrm{cm}$