Q. An object and its real image are located at distances $25\, cm$ and $40 \,cm$ respectively from the two principal foci of a convex lens. The linear magnification of the image is near to
AMUAMU 2013Ray Optics and Optical Instruments
Solution:
In the figure
$u=-(25+f)$
$ v=(40+f) $
using $ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} $
$\frac{1}{f+40} - \frac{1}{25+f}=\frac{1}{f} $
$65 f+2 f=1000+65 f+f^{2}$
$f^{2}=1000$
$ f=31.62 \,cm $
$u=25+31.62=56.62\, cm $
$v=40+31.62=71.62\, cm$
$ m=\frac{v}{u}=\frac{71.62}{-56.62}=-1.3$
