Question

A square wire of side $3.0$cm is placed $25$cm away from a concave mirror of focal length $10$cm. Find area enclosed by the image of the wire in $c{m}^2$. The centre of the wire is on the axis of the mirror, with the all sides normal to the axis.

Answer 1

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{v}=\frac{1}{-10}-\frac{1}{-25}=-\frac{3}{50}$

$v=-\frac{50}{3}$

$m=\frac{h^{{}^{\prime }}}{h}=-\frac{v}{u}$

h is the height of the object

$\frac{h^{{}^{\prime }}}{3}=-\frac{\left(\frac{-50}{3}\right)}{-25}$ $h=-2cm$

$={(-2cm)}^2=4{cm}^2$ so the area enclosed by image of wire.