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)

$1\,cm^2$

Answer 2

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
$=10\times10\times1\times1=100\,mm^{2}$
$=1\,cm^{2}$