Q. A thin rod of length $f/3$ lies along the axis of a concave mirror of focal length $f$. One end of its magnified image touches an end of the rod. The length of the image is
A) $f$
B) $\frac{1}{2}f$
C) $2f$
D) $\frac{1}{4}f$
Explanation
(b) If end $A$ of rod acts an object for mirror then it’s image will be $A'$ and if
$u = 2f - \frac{f}{3} = \frac{5f}{3}$
So by using $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\Rightarrow \frac{1}{- f} = \frac{1}{v} + \frac{1}{\frac{- 5f}{3}} \Rightarrow v = - \frac{5}{2}f$
$\therefore\ $Length of image $= \frac{5}{2}f - 2f = \frac{f}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.