The magnifying power of a telescope with tube length 60 $\mathrm{cm}$ is 5 . Then the focal length of its…
The magnifying power of a telescope with tube length 60 $\mathrm{cm}$ is 5 . Then the focal length of its eye piece is
$20 \mathrm{~cm}$
$40 \mathrm{~cm}$
$30 \mathrm{~cm}$
$10 \mathrm{~cm}$
Solution
The magnifying power of a telescope is given by
$\begin{aligned}
& m=\frac{f_0}{f_e}=5 \\
& f_0=5 f_e \\
& f_0+f_e=60 \mathrm{~cm} \Rightarrow 5 f_e+f_e=60 \\
& 6 f_e=60 \Rightarrow f_e=10 \mathrm{~cm}
\end{aligned}$
The focal length of its eye piece, $f_e=10 \mathrm{~cm}$