A  square $\mathrm{ABCD}$ of side 1mm is kept at distance 15cm in front of the concave mirror as shown in fig. The focal length of the mirror is 10cm. The length of the perimeter of its image will be __________mm

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

For the heights AB and CD :

$\frac{1}{v}+\frac{1}{-15}=\frac{1}{-10}$

$\mathrm{v}=-30\mathrm{cm}$

$\mathrm{m}=\frac{{\mathrm{h}}^1}{\mathrm{h}}=\frac{-\mathrm{v}}{\mathrm{u}}$

$\frac{{\mathrm{h}}^1}{0.1}=\frac{30}{-15}\Rightarrow {\mathrm{h}}^1=-0.2\mathrm{cm}$

For the lengths BC and AD

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

$\Rightarrow \frac{∆\mathrm{v}}{{\mathrm{v}}^2}+\frac{∆\mathrm{u}}{{\mathrm{u}}^2}=0$

$\Rightarrow \frac{∆\mathrm{v}}{∆\mathrm{u}}=-\frac{{\mathrm{v}}^2}{{\mathrm{u}}^2}$

$\Rightarrow \frac{∆\mathrm{v}}{∆\mathrm{u}}=-\frac{{30}^2}{{15}^2}=-4$

Therefore, the imagewill have a length of $0.1\times 4\mathrm{cm}$

$\text{ Perimeter }\left(p\right)=2\times 0.2+2\times 0.4=0.12\mathrm{cm}$