Q. A square card of side length 1mm is being seen through a magnifying lens of focal length 10 cm. The card is placed at a distance of9 cm from the lens. The apparent area of the card through the lens is
A) $1\ {cm}^{2}$
B) $0.81\ {cm}^{2}$
C) $0.27\ {cm}^{2}$
D) $0.60\ {cm}^{2}$
Explanation
(a) Focal length of converging lens $f$ = +10 cm
$u\ $= -9 cm
From lens formula
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$or\ \ \ \ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{( - 9)}$
$\frac{1}{v} = \frac{1}{10} - \frac{1}{9}$
$Or\ \ \ \ \ v\ = \ - 90\ cm$
Magnification , $m = \frac{v}{u} = \ \frac{- 90}{- 9} =$10m
$\therefore$ Apparent area of card through lens
= 10 $\times \ $10 $\times$ 1 $\times$ 1 = 100 $mm^{2}\ = \ 1\ cm^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.