A U - shaped wire is placed before a concave mirror having radius of curvature $20\text{cm}$ as shown in the figure. Find the total length of the image.

The correct option is B $10\text{cm}$
For part MN,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-10}-\frac{1}{-30}=-\frac{1}{10}+\frac{1}{30}$
$\frac{1}{v}=\frac{-3+1}{30}=\frac{-2}{30}$
$v=-15\text{cm}$
Magnification
$m=\frac{I}{O}=-\frac{v}{u}$
$\Rightarrow \frac{I}{10}=-\frac{(-15)}{(-30)}=-\frac{1}{2}\Rightarrow I=5\text{cm}$


For image of $ST$,
$u=-40\text{cm}$
$f=-10\text{cm}$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}+\frac{1}{(-40)}=\frac{1}{(-10)}$
$\frac{1}{v}=-\frac{1}{10}+\frac{1}{40}$
$\frac{1}{v}=-\frac{-4+1}{40}=\frac{-3}{40}$
$v=-\frac{40}{3}\text{cm}$

For length of image of $ST$,
$m=\frac{I}{O}=-\frac{v}{u}$
$\frac{I}{10}=-\left(\frac{-\frac{40}{3}}{-40}\right)\Rightarrow I=-\frac{10}{3}\text{cm}$

Length of image of $NT$,
$T^{\prime }{N}^{\prime }=15-\frac{40}{3}=\frac{5}{3}\text{cm}$


Length of wire in image $=5+\frac{10}{3}+\frac{5}{3}=10\text{cm}$

Why this question? Applications of mirror formula are important for competitive examinations.