Q. In an astronomical telescope in normal adjustment a straight black line of length 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 I. The magnification of the telescope is :
A) $\frac{L}{I}$
B) $\frac{L}{I} + 1$
C) $\frac{L}{I} - 1$
D) $\frac{L+I}{L-I}$
Explanation
[Solution] Object distance for the eyepiece is $f_o + f_e$, giving image size ratio $\frac{I}{L} = \frac{f_e}{f_o}$; magnification $\frac{f_o}{f_e} = \frac{L}{I}$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.