In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is I. The magnification of the telescope is

For normal adjustment, the fi nal image is formed at infinity.
$\therefore$ Magnifying power, $M=\frac{-f_0}{f_e}\dots \left(1\right)$
Length of telescope, $f_0+f_e\dots \left(2\right)$
Consider the line on objective as an object and eye piece as lens.
$\therefore$ For eye piece
$\begin{array}{l}{u}_e=-\left(f_0+f_e\right)\dots \left(3\right)\\ f=f_e\\ \therefore \frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}\\ \frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{-\left(f_0+f_e\right)}\\ \Rightarrow v_e=\frac{\left(f_0+f_e\right)f_e}{f_0}\dots \left(4\right)\end{array}$
Magnification of eye piece,
$M_e=\frac{v_e}{u_e}=\frac{I_i}{h_0}\dots \left(5\right)$
Put equation (3) and (4) in (5),
$\begin{array}{l}\frac{\left(f_0+f_e\right)f_e}{f_0\times \left(f_0+f_e\right)}=\frac{h_1}{h_0}\\ \frac{f_e}{f_0}=\frac{\ell }{L}\end{array}$
$\therefore$Magnification of telescope,
$M=\frac{f_0}{f_e}=\frac{L}{\ell }.$