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 \[{{f}_{e}}=5{{f}_{o}}\]

\[\therefore {{f}_{o}}=50\text{ }cm\]

\[{{f}_{e}}=10\text{ }cm\]