Question

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

Answer 1

Given,
Radius of curvature of concave mirror, R = 20 cm
So its focal length will be f = $\frac{R}{2}$ = −10 cm
For part AB of the U shaped wire, PB = 30 + 10 = 40 cm



Therefore, u = −40 cm
Using mirror formula,
$\Rightarrow \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
${\displaystyle \frac{1}{v}}=-\frac{1}{10}-\frac{1}{-40}=-\frac{3}{40}$
$\Rightarrow v=\frac{-40}{3}=13.3\mathrm{cm}$
$\mathrm{So},\mathrm{PB}\text{'}=13.3\mathrm{cm}$
$m=\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=\frac{-\left(v\right)}{u}$
$$\begin{gathered} =\frac{-(-13.3)}{-40}=-\frac{1}{3} \\ \Rightarrow \mathrm{A}\text{'}\mathrm{B}\text{'}=-\frac{10}{3}=-3.33\mathrm{cm} \end{gathered}$$

For part CD of the U shaped wire, PC = 30 cm
Therefore, u = −30 cm
Again, using mirror equation:
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$$\begin{gathered} {\displaystyle \frac{1}{v}}=-\frac{1}{10}-\frac{1}{-30} \\ {\displaystyle \frac{1}{v}}=-\frac{1}{10}+\frac{1}{30}=-\frac{1}{15} \end{gathered}$$
⇒ v = −15 cm = PC'
$m=\frac{\mathrm{C}\text{'}\mathrm{D}\text{'}}{\mathrm{CD}}=-\frac{v}{u}$
$=-\frac{(-15)}{-30}=-\frac{1}{2}$
⇒ C'D' = 5 cm
⇒ B'C' = PC' − PB'
= 15 − 13.3 = 1.7 cm

Hence, the total length of the U- shaped wire is A'B' + B'C' + C'D'
= (3.3) + (1.7) + 5 = 10 cm