Ray Optics and Optical Instruments - Result Question 68
72. The magnifying power of a telescope is 9 . When it is adjusted for parallel rays the distance between the objective and eyepiece is $20 cm$. The focal length of lenses are :
(a) $10 cm, 10 cm$
(b) $15 cm, 5 cm$
(c) $18 cm, 2 cm$
(d) $11 cm, 9 cm$
[2012]
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Answer:
Correct Answer: 72. (c)
Solution:
(c) M.P. $=9=\frac{f_0}{f_e}$
$\Rightarrow f_0=9 f_e$
$f_0+f_e=20$
on solving
$f_0=18 cm=$ focal length of the objective
$f_e=2 cm=$ focal length of the eyepiece
