Magnifying power of an astronomical telescope for normal adjustments is 10 and length of the telescope is 110 cm. Magnifying power of the same telescope, when the image is formed at the near point and the object at $\infty$ is

Magnifying power in normal adjustment, where ${\text{f}}_{\text{o}}$ and ${\text{f}}_{\text{e}}$ are focal lengths of objective and eyepiece

$\text{m}=\frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}}$

${\text{f}}_{\text{o}}=10{\text{f}}_{\text{e}}$

Given, tube length, ${\text{f}}_{\text{o}}+{\text{f}}_{\text{e}}=110$

$10{\text{f}}_{\text{e}}+{\text{f}}_{\text{e}}=110$

${\text{f}}_{\text{e}}=10\text{cm}$ and ${\text{f}}_{\text{o}}=100\text{cm}$

Now, magnifying power, when image formed at near point

$\text{m}=\frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}}\left(1+\frac{{\text{f}}_{\text{e}}}{\text{D}}\right)$

$\text{m}=\frac{100}{10}\left(1+\frac{10}{25}\right)=10\times \frac{35}{25}$

$\Rightarrow \text{m}=14$