Let fo and fe be the focal length of the objective and eyepiece respectively. For normal adjustment the distance from objective to eyepiece is fo + fe. Taking the line on the objective as object and eyepiece as lens
u = –(fo + fe ) and f = fe
\(\frac{1}{v}-\frac{1}{[-{f_0\ +\ f_e}]}\) = \(\frac{1}{f_e}\)
⇒ \( v = (\frac{f_0\ +\ f_e}{f_0})\)
Linear magnification (eyepiece) = \(\frac{v}{u}=\frac{image\ size}{object\ size}\)
= \(\frac{f_e}{f_0}=\frac{l}{L}\)
∴ Angular magnification of telescope
M = \(\frac{f_e}{f_0}=\frac{l}{L}\)