Question

In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the inside part of the 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. $\frac{L}{l}$
B. $\frac{L}{l}+1$
C. $\frac{L}{l}-1$
D. $\frac{L+1}{l-1}$

Answer 1

Use the formula for magnification of the telescope and eye-piece lens.

Formula used: ${{M}_{t}}=\frac{{{f}_{o}}}{{{f}_{e}}}$
Here
Mt is the magnification of the telescope
fo is the focal length of the objective lens
fe is the focal length of the eye-piece lens
Also
${{M}_{e}}=\frac{{{f}_{e}}}{{{f}_{e}}+u}=\frac{{{L}_{i}}}{{{L}_{o}}}$
Here
Me is the magnification of the eye-piece lens
u is the object distance
Li is the length of the image
L0 is the length of the object

Complete step by step solution:
The object distance is
$u=-({{f}_{o}}+{{f}_{e}})$
Substituting in the formula for magnification of eye-piece lens
$\begin{aligned} & \frac{{{f}_{e}}}{{{f}_{e}}+(-{{f}_{o}}-{{f}_{e}})}=\frac{{{L}_{i}}}{{{L}_{o}}} \\\ & ~\frac{{{f}_{e}}}{-{{f}_{o}}}=\frac{-l}{L} \\\ & \frac{{{f}_{e}}}{{{f}_{o}}}=\frac{l}{L} \\\ \end{aligned}$

Hence
${{M}_{t}}=\frac{{{f}_{o}}}{{{f}_{e}}}=\frac{L}{l}$

The correct answer is option A.

Note: According to the sign convention used, distance is taken as negative when it is on the left side of the lens and positive when it is on the right side of the lens. Also, negative sign of image length is taken for real image as it is inverted. Consequently, positive image length would imply a virtual and erect image.