Q. A convex lens of focal length $f$ is placed some where in between an object and a screen. The distance between object and screen is $x$. If numerical value of magnification produced by lens is $m$, focal length of lens is
A) $\frac{mx}{(m + 1)^{2}}$
B) $\frac{mx}{(m - 1)^{2}}$
C) $\frac{(m + 1)^{2}}{m}x$
D) $\frac{(m - 1)^{2}}{m}x$
Explanation
(a) Here, $\ x = u + v$
As $m = \frac{f}{f + u} = \frac{f - v}{f}$
and image is real, magnification is negative
$\therefore - m = \frac{f}{f + u},u = \frac{- (m + 1)f}{m}$
From $–m = \frac{f - v}{f} \Rightarrow v = (m + 1)f$
Put in Eq.(i)
$x = \frac{- (m + 1)}{m}f + (m + 1)f$
Solving ,we get, $f = \frac{mx}{(m + 1)^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.