A square wire of side 3.0 cm is placed 25 cm away from a concave mirror of focal length 10 cm. What the area enclosed by the image of the wire? (The centre of the wire is on the axis of the mirror, with its two sides normal to the axis)

Here , u= -25cm,       f= -10cm
As  $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
$$\begin{gathered} \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-10}+\frac{1}{25} \\ =\frac{-5+2}{50}=-\frac{3}{50} \end{gathered}$$

or  $v=-\frac{50}{3}$

Now,  $m=-\frac{v}{u}=-\frac{50}{3\times 25}=-\frac{2}{3}$

$=\frac{side{\text{ of image of wire (h}}_2)}{sideofsquare{\text{ wire (h}}_1)}$

Side of image of wire, $h_2$

$=-\frac{2}{3}\times h_1=-\frac{2}{3}\times 3=-2cm$

Area enclosed by the image of wire= ${\left(2\right)}^2=4c{m}^2$