A U-shaped wire is placed before a concave mirror having radius of curvature 20cm as shown in Fig. Find the total length of the image 

For part MN 
u = – 30cm, f = – 10cm
$\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$
$\frac{1}{\mathrm{V}}=\frac{1}{\mathrm{u}}-\frac{1}{\mathrm{f}}=\frac{1}{-10}-\frac{1}{-30}=-\frac{1}{10}+\frac{1}{30}$

$\frac{1}{\mathrm{V}}=\frac{-3+1}{30}=\frac{-2}{30}\mathrm{V}=-15\mathrm{cm}$

$\mathrm{m}=\frac{\mathrm{I}}{\mathrm{O}}=-\frac{\mathrm{v}}{\mathrm{u}}\Rightarrow \frac{\mathrm{I}}{10}=-\frac{(-15)}{(-30)}=-\frac{1}{2}$

So length of Image of MN $\Rightarrow \mathrm{I}=5\mathrm{cm}$

for image of ST

u = – 40 cm ; f = – 10cm 
$\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\Rightarrow \frac{1}{\mathrm{v}}+\frac{1}{(-40)}=\frac{1}{(-10)}$
$\frac{1}{\mathrm{V}}=-\frac{1}{10}+\frac{1}{40}$
$\frac{1}{\mathrm{V}}=-\frac{-4+1}{40}=\frac{-3}{40}\Rightarrow \mathrm{V}=\frac{-40}{3}\mathrm{cm}$
For length of image of ST,

$\mathrm{m}=\frac{\mathrm{I}}{\mathrm{O}}=-\frac{\mathrm{v}}{\mathrm{u}};\frac{\mathrm{Iength} \mathrm{of} \mathrm{image}}{10}=-\left(\frac{-40/3}{-40}\right)\Rightarrow \mathrm{Iength} \mathrm{of} \mathrm{image}=\frac{10}{3}\mathrm{cm}$ 
Length of image of NT, 
${\mathrm{T}}^{\mathrm{\prime }}{\mathrm{N}}^{\mathrm{\prime }}=15-\frac{40}{3}=\frac{5}{3}\mathrm{cm}$
Length of wire in image 
$=5+\frac{10}{3}+\frac{5}{3}=10\mathrm{cm}$