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Q. A concave lens forms the image of an object such that the distance between the object and image is $10 cm$ and the magnification produced is 1 / 4 The focal length of the lens will be

Ray Optics and Optical Instruments

Solution:

A concave lens forms virtual image $I$ of a point object $O$.
As, $m=\frac{1}{4}=\frac{v}{u} $
$\therefore u=4 v$
If $v=-x, u=-4 x$
Distance between object and image,
image
$|O I=4 x-x=3 x=10\, cm$
or $x=\frac{10}{3} \,cm$
$\therefore u=-\frac{40}{3}\, cm$ and $v=\frac{-10}{3} \,cm$
Using lens formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{f}=-\frac{3}{10}+\frac{3}{40}=-\frac{9}{40}$ ;
$\therefore f=-\frac{40}{9} cm =-4.4 \,cm$