Question

Magnifying power of an astronomical telescope for normal adjustment is $10$ and length of the telescope is $110cm$. 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.
$m=\frac{f_o}{f_e}=10$
$f_o=10f_e$
Given, tube length, $f_o+f_e=110$
$10f_e+f_e=110$
$f_e=10cm$ and $f_o=100cm$
Now, magnifying power, when image formed at near point
$m=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)$
$m=\frac{100}{10}\left(1+\frac{10}{25}\right)=10\times \frac{35}{25}$
$\Rightarrow m=14$