Q. The distance between an object and a divergent lens is $m$ times the focal length of the lens. The linear magnification produced by the lens is
A) $m$
B) $1/m$
C) $m + 1$
D) $\frac{1}{m + 1}$
Explanation
(d) Here $u = mf$
For divergent lens (concave)
$\therefore v = \frac{mf}{m + 1}$
Now magnification $= \frac{v}{u} = \frac{mf}{m + 1} \times \frac{1}{mf} = \frac{1}{m + 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.