Question

A wire is bent into square shape of side $3\text{cm}$ and placed at $25\text{cm}$ in front of a concave mirror of focal length $10\text{cm}$ with one of its edge on the principal axis of the mirror and its plane normal to the axis. The area enclosed by the image of the wire will be

• A. $6{\text{cm}}^2$
• B. $5{\text{cm}}^2$
• C. $4{\text{cm}}^2$
• D. $7{\text{cm}}^2$

Answer 1

The correct option is C $4{\text{cm}}^2$

Given:
Distance of object, $u=-25\text{cm}$
Focal length, $f=-10\text{cm}$
Object height, $l_o=3\text{cm}$
On applying mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{25}=\frac{-1}{10}$
$\Rightarrow \frac{1}{v}=\frac{-3}{50}$
$\Rightarrow v=\frac{-50}{3}$
Linear magnification, $m=\frac{-v}{u}=\frac{\frac{50}{3}}{-25}=\frac{-2}{3}$
Side length of the image of the square,
$l^{{}^{\prime }}=l_i=l_o\times |m|=3\times \frac{2}{3}=2\text{cm}$
So, the area of the image of the square,
$A={\left(l^{{}^{\prime }}\right)}^2=2^2=4{\text{cm}}^2$