Optics Question 392
Question: 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 length of the two lenses are
[MP PMT 1986]
Options:
A) 18 cm, 2 cm
B) 11 cm, 9 cm
C) 10 cm, 10 cm
D) 15 cm, 5 cm
Show Answer
Answer:
Correct Answer: A
Solution:
In this case $ |m|\ =\frac{f _{o}}{f _{e}}=9 $ …. (i)
and length of telescope $ =f _{o}+f _{e}=20 $ …. (ii)
Solving (i) and (ii), we get fe = 2 cm, $ f _{o}=18cm. $
