Question

Magnifying power of an astronomical telescope for normal adjustment 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 is

• A. 14
• B. 18
• C. 23
• D. 26

Answer 1

Answer: (A)

Magnifying power in normal adjustment, where $f_{o}$ and $f_{e}$ are focal lengths of objective and eyepiece.
$m=\frac{f_{o}}{f_{e}}=10 $
$f_{o}=10 f_{e}$
Given, tube length, $f_{o}+f_{e}=110$
$10 f_{e}+f_{e}=110$
$f_{e}=10 \,cm$ and $f_{o}=100\, cm$
Now, magnifying power, when image formed at near point
$m=\frac{f_{o}}{f_{e}}\left(1+\frac{f_{e}}{D}\right) $
$m=\frac{100}{10}\left(1+\frac{10}{25}\right)=10 \times \frac{35}{25} $
$\Rightarrow m=14 $