Ray Optics And Optical Instruments Ques 69
69. 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 :
[2012]
(a) $10 $ $cm, 10 $ $cm$
(b) $15 $ $cm, 5 $ $cm$
(c) $18$ $ cm, 2$ $ cm$
(d) $11 $ $cm, 9 $ $cm$
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Answer:
Correct Answer: 69.(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
BETA
