Question

In an astronomical telescope in normal adjustment, a straight black line of length $L$ is drawn on inside part of the objective lens. The eyepiece 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. $\frac{L+l}{L-l}$

Answer 1

The correct option is A $\frac{L}{l}$

We know that magnification of telescope:

$M=\frac{f_0}{f_e}$

The magnification of the eyepiece lens is given by:

$M_e=\frac{f_e}{f_e+u}=\frac{L_I}{L_O}$

Here, $L_I$ is the length of the image and $L_0$ is the length of the object.

Now for the eypiece lens, the distance of the object is $-(f_o+f_e)$

$\frac{f_e}{f_e-(f_0+f_e)}=\frac{-l}{L}$

$\Rightarrow \frac{f_e}{f_0}=\frac{l}{L}$

i.e $M=\frac{f_0}{f_e}=\frac{L}{l}$

Thus, option $\left(A\right)$ is correct.