Question

A square wire of side $2.0cm$ is placed $20cm$ in front of a concave mirror of focal length $10cm$ with its center on the axis of the mirror and its plane normal to the axis. The area enclosed by the image of wire is

• A. $7.5{cm}^2$
• B. $6{cm}^2$
• C. $2{cm}^2$
• D. $4{cm}^2$

Answer 1

The correct option is D $4{cm}^2$

Given, square wire of side $=2cm$, $u=-20cm$, $f=-10cm$.
Area of second wire $=$?
$\therefore \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}+\frac{1}{(-20)}=\frac{1}{-10}\Rightarrow \frac{1}{v}=-\frac{1}{10}+\frac{1}{20}$
$\Rightarrow v=-20cm$
We know that, $m=-\frac{v}{u}=\frac{-(-20)}{-20}=-1$
$\because \frac{Areaofimage}{Areaofobject}=m^2\Rightarrow \frac{A_i}{A_o}={(-1)}^2$
$\frac{A_i}{2\times 2}=1$
$\Rightarrow A_i=4{cm}^2$