Question

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

Magnification of telescope: $M=\frac{f_0}{f_e}$
$\frac{f_e}{f_e+u}=-\frac{I}{L}$
$\therefore \frac{f_e}{f_e-(f_0+f_e)}=-\frac{I}{L}$
Solving, we get
$\frac{f_0}{f_e}=\frac{L}{I}=M$