Q. An infinitely long rod lies along the axis of concave mirror of focal length $f.$ The near end of the rod is at a distance $x\ > \ f$ from the mirror. Then the length of the image of the rod is
A) $\frac{f^{2}}{x + f}$
B) $\frac{f^{2}}{x}$
C) $\frac{xf}{x - f}$
D) $\frac{f^{2}}{x - f}$
Explanation
(d) $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ and $m = \frac{v}{u} = \ - \frac{I}{O}$
Using above relation, the length of image
$I = \frac{f^{2}}{x - f\ }$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.