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The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is $20 \mathrm{~cm}$. The focal length of lenses are
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Verified Answer
The correct answer is:
$18 \mathrm{~cm}, 2 \mathrm{~cm}$
Given, $M=\frac{f_0}{f_e}=9$ and $f_0+f_e=20$
$f_o=9 f_e$
So, $9 f_e+f_e=20$
$\begin{aligned}
& \therefore f_e=2 \mathrm{~cm} \\
& f_o=9 \times 2 \\
& f_0=18 \mathrm{~cm}
\end{aligned}$
$f_o=9 f_e$
So, $9 f_e+f_e=20$
$\begin{aligned}
& \therefore f_e=2 \mathrm{~cm} \\
& f_o=9 \times 2 \\
& f_0=18 \mathrm{~cm}
\end{aligned}$
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