Question

A square card of side length $ 1 \,mm $ is being seen through a magnifying lens of focal length $ 10 \,cm $ . The card is placed at a distance of $ 9 \,cm $ from the lens. The apparent area of the card through the lens is

• A. $ 1\,cm^2 $
• B. $ 0.8\,cm^2 $
• C. $ 0.27\,cm^2 $
• D. $ 0.60\,cm^2 $

Answer 1

Answer: (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, $m=\frac{v}{u}=\frac{-90}{-9}=10$
$\therefore $ Apparent area of card through lens
$=10\times10\times1\times1=100\,mm^{2}$
$=1\,cm^{2}$