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Magnifying power of an astronomical telescope for normal adjustment is 10 and length of the telescope is $110 \mathrm{~cm}$. Magnifying power of the same telescope, when the image is formed at the near point is
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Verified Answer
The correct answer is:
14
Magnifying power in normal adjustment, where $f_{\mathcal{o}}$ and $f_e$ are focal lengths of objective and eyepiece.
$$
\begin{aligned}
& \quad m=\frac{f_o}{f_e}=10 \\
& f_o=10 f_e
\end{aligned}
$$
Given, tube length, $f_o+f_e=110$
$$
\begin{aligned}
10 f_e+f_e & =110 \\
f_e & =10 \mathrm{~cm} \text { and } f_o=100 \mathrm{~cm}
\end{aligned}
$$
Now, magnifying power, when image formed at near point
$$
\begin{aligned}
& \quad m=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right) \\
& \quad m=\frac{100}{10}\left(1+\frac{10}{25}\right)=10 \times \frac{35}{25} \\
& \Rightarrow \quad m=14
\end{aligned}
$$
$$
\begin{aligned}
& \quad m=\frac{f_o}{f_e}=10 \\
& f_o=10 f_e
\end{aligned}
$$
Given, tube length, $f_o+f_e=110$
$$
\begin{aligned}
10 f_e+f_e & =110 \\
f_e & =10 \mathrm{~cm} \text { and } f_o=100 \mathrm{~cm}
\end{aligned}
$$
Now, magnifying power, when image formed at near point
$$
\begin{aligned}
& \quad m=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right) \\
& \quad m=\frac{100}{10}\left(1+\frac{10}{25}\right)=10 \times \frac{35}{25} \\
& \Rightarrow \quad m=14
\end{aligned}
$$
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