Question:
The magnifying power of a telescope with tube length 60 $\mathrm{cm}$ is 5 . What is the focal length of its eye piece?
Correct Option: 4
Solution:
(4) For telescope
Tube length $(\mathrm{L})=f_{o}+f_{e}=60$
and magnification $(m)=\frac{f_{o}}{f_{e}}=5 \Rightarrow f_{0}=5 f_{e}$
$\therefore f_{\mathrm{o}}=50 \mathrm{~cm}$ and $f_{\mathrm{e}}=10 \mathrm{~cm}$
Hence focal length of eye-piece, $f_{\mathrm{e}}=10 \mathrm{~cm}$
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