An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm.

solution:

Given, object height, $\mathrm{h}_{1}=+7 \mathrm{~cm}$;

object distance, $\mathrm{u}=-27 \mathrm{~cm}$;

focal length, $\mathrm{f}=-18 \mathrm{~cm}$;

image distance, $\mathrm{v}=? ;$ image height, $\mathrm{h}_{2}=?$

Mirror formula,$\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \quad$ or $\quad \frac{1}{v}+\frac{1}{(-27)}=\frac{1}{(-18)}$

or $\frac{1}{v}=\frac{1}{27}-\frac{1}{18}=\frac{3-4}{54}=\frac{-1}{54}$ or $v=-54 \mathrm{~cm}$

Now, magnification, $m=-\frac{v}{u}=-\frac{(-54)}{(-27)}=-2$

Also, $m=\frac{h_{2}}{h_{1}}$ or $-2=\frac{h_{2}}{(+7)}$ or $h_{2}=-14 \mathrm{~cm}$

The image is located at a distance of $54 \mathrm{~cm}$ in front of the mirror ; it is real, inverted and magnified of size $14 \mathrm{~cm}$.