Question

An astronomical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tube is 102 cm, find the powers of the objective and the eyepiece.

Answer 1

For the astronomical telescope,
Magnifying power, m = 50
Length of the tube, L = 102 cm
Let the focal length of objective and eye piece be f₀ and f_e, respectively.
Now, using $m=\frac{f_o}{f_e}$, we get:
f_o = 50f_e …(1)
And,
L = f_o + f_e = 102 cm …(2)
On substituting the value of f_o from (1) in (2), we get:
50fe + fe = 102
⇒ 51fe = 102
⇒ fe = 2 cm = 0.02 m
And,
f_o = 50$\times$0.02 = 1 m
Power of the objective lens = $\frac{1}{f_o}$ = 1 D
And,
Power of the eye piece lens = $\frac{1}{f_e}=\frac{1}{0.02}$ = 50 D