Q 12-09-182JEE MainJEE Main 2020 (8 Jan, Shift 1)Easy
The magnifying power of a telescope with tube length $60$ cm is $5$. What is the focal length of its eye piece?
Answer: (D) $10$ cm
In normal adjustment: $f_o + f_e = 60$ cm and $\dfrac{f_o}{f_e} = 5$.
So $6f_e = 60 \Rightarrow f_e = 10$ cm.
Solution by Sreeraj P, M.Sc Physics