Magnifying power of an astronomical telescope for normal adjustment is 10 and length of the telescope is…
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
14
18
23
26
Solution
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}
$