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 $102cm$, find the powers of the objective and the eyepiece.

• A. $10;100$
• B. $1;50$
• C. $10;500$
• D. $2;100$

Answer 1

The correct option is B $1;50$



Magnifying power of an astronomical telescope is given by

$m=\frac{-f_0}{f_e}=-50$
$f_0=50f_e$ --- (1)

Length of the tube
$L=f_0+f_e=102cm$ --- (2)

Substituting (1) in (2),
$51f_e=102$
i.e $f_e=2cm$
and $f_o=100cm$

Power of objective $P_o=\frac{1}{f_o}=1D$
Power of eyepiece $P_e=\frac{1}{f_e}=50D$