Question

A $\text{U}-$ shaped wire $abcd$ of length $30\text{cm}$ is placed before a concave mirror having radius of curvature $20\text{cm}$. Find the total length of the image of wire.

• A. $11\text{cm}$
• B. $10\text{cm}$
• C. $12\text{cm}$
• D. $14\text{cm}$

Answer 1

The correct option is B $10\text{cm}$

For side $ab$,
$h_o=+10\text{cm}$
$u=-40\text{cm}$
$f=\frac{-20}{2}=-10\text{cm}$
So, $m=\frac{f}{f-u}$
$\Rightarrow m=\frac{-10}{-10-(-40)}=\frac{-1}{3}$
$\Rightarrow \frac{h_i}{h_o}=\left|\frac{-1}{3}\right|$
$\Rightarrow \frac{h_i}{10}=\frac{1}{3}$
$\Rightarrow h_i=\frac{10}{3}\text{cm}$ .........$\left(1\right)$
Similarly, for side $cd$,
$h_i=5\text{cm}$ .........$\left(2\right)$
For side $bc$,
End $b$,
$u=-40\text{cm}$
$f=-10\text{cm}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{40}=\frac{-1}{10}$
$\Rightarrow \frac{1}{v}=\frac{-3}{40}$
$\Rightarrow v=\frac{-40}{3}\text{cm}$
End $c$,
$u=-30\text{cm}$
$f=-10\text{cm}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{30}=\frac{-1}{10}$
$\Rightarrow \frac{1}{v}=\frac{-1}{15}$
$\Rightarrow v=-15\text{cm}$
So, length of image of $bc$ $=|-15|-\left|\frac{-40}{3}\right|=\frac{5}{3}\text{cm}$ .........$\left(3\right)$


From $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$
Total length of the image of wire $abcd=\frac{10}{3}+5+\frac{5}{3}=10\text{cm}$

$\begin{array}{c}\text{Why this question ?}\\ \text{In order to avoid mistakes, divide the whole object in segments}\\ \text{and then apply mirror formula and magnification formula for}\\ \text{boundary segments.}\end{array}$