Q. The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye-piece is found to be 20 cm. The focal lengths of the lenses are
A) 18 cm, 2 cm
B) 11 cm, 9 cm
C) 10 cm, 10 cm
D) 15 cm, 5 cm
Explanation
(a) $M = \frac{f_{0}}{f_{e}}$
$9 = \frac{f_{0}}{f_{e}}\ \ $or$\ f_{0} = 9f_{e}$
Also, $L = f_{o} + f_{e}\ $or$\ 20 = f_{o} + f_{e}$
Or $20 = 9f_{e} + f_{e}\ $or$\ 20 = 10\ f_{e}$
Or $f_{e} = 2\ cm$
$\therefore f_{o} = 9 \times 2\ cm = 18\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.