In an astronomical telescope in normal adjustment a straight black line of the length $\text{L}$ is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is $l$ . The magnification of the telescope is:

• A

  $\frac{\mathrm{L}}{l}-1$
• B

  $\frac{\mathrm{L}+l}{\mathrm{L}-l}$
• Correct answer
  C

  $\frac{\mathrm{L}}{l}$
• D

  $\frac{\mathrm{L}}{l}+1$

Solution

Magnification by eyepiece
$\mathrm{m}=\frac{\mathrm{f}}{\mathrm{f}+\mathrm{u}}$
$-\frac{\mathrm{l}}{\mathrm{L}}=\mathrm{ }\frac{{\mathrm{f}}_{\mathrm{e}}}{{\mathrm{f}}_{\mathrm{e}}+(-({\mathrm{f}}_0+{\mathrm{f}}_{\mathrm{e}})}$
$\Rightarrow \mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\frac{\mathrm{l}}{\mathrm{L}}=\frac{{\mathrm{f}}_{\mathrm{e}}}{{\mathrm{f}}_0}$
$\mathrm{m}.\mathrm{p}.\mathrm{ }=\frac{{\mathrm{f}}_0}{{\mathrm{f}}_{\mathrm{e}}}=\frac{\mathrm{L}}{\mathrm{l}}$

Hence, the correct answer is option: 'C'.