question_answer

The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its eye piece? [JEE MAIN Held On 08-01-2020 Morning]

A) 30 cm  

B) 10 cm

C) 20 cm

D) 40 cm

Correct Answer: B

Solution :

[b] For telescope Tube length \[\left( L \right)={{f}_{o}}+{{f}_{e}}\] And magnification \[(m)=\frac{{{f}_{e}}}{{{f}_{o}}}\] Where \[{{f}_{o}}\] and \[{{f}_{e}}\]are focal length of objective And eyepiece \[\therefore {{f}_{o}}+{{f}_{e}}=60\] and \[{{f}_{e}}=5{{f}_{o}}\] \[\therefore {{f}_{o}}=50\text{ }cm\] \[{{f}_{e}}=10\text{ }cm\]