Optics Question 389
Question: The magnifying power of an astronomical telescope for relaxed vision is 16. On adjusting, the distance between the objective and eye lens is 34 cm. Then the focal length of objective and eye lens will be respectively
[MP PMT 1989]
Options:
A) 17 cm, 17 cm
B) 20 cm, 14 cm
C) 32 cm, 2 cm
D) 30 cm, 4 cm
Show Answer
Answer:
Correct Answer: B
Solution:
In this case $ |m|\ =\frac{f _{o}}{f _{e}}=16 $ …. (i)
and length of telescope $ =f _{o}+f _{e}=34 $ …. (ii)
Solving (i) and (ii), we get fe = 20 cm, $ f_{o}=14\text{cm} $.
BETA
