Question

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

Answer 1

Given: Radius of curvature, $R=20cm$

So, focal length, $f=R/2=-10cm$

For part $AB,PB=30+10=40cm$

So, $u=-40cm\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{-1}{10}-\left(\frac{1}{-40}\right)=\frac{-3}{40}$

$\Rightarrow v=\frac{-40}{3}=-13.3cm.$

So, $P{B}^{\prime }=13.3cm$

$m=\frac{A^{\prime }{B}^{\prime }}{AB}=-\left(\frac{v}{u}\right)=-\left(\frac{-13.3}{-40}\right)=-\frac{1}{3}.$

$\Rightarrow A^{\prime }{B}^{\prime }=-10/3=-3.33cm$

For part $CD,PC=30,$ So $u=-30cm$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{10}-\left(-\frac{1}{30}\right)=-\frac{1}{15}\Rightarrow v=-15cm=P{C}^{\prime }$

So, $m=\frac{C^{\prime }{D}^{\prime }}{CD}=-\frac{v}{u}=-\left(\frac{-15}{-30}\right)=-\frac{1}{2}.$

$\Rightarrow C^{\prime }{D}^{\prime }=5cm$

$B^{\prime }{C}^{\prime }=P{C}^{\prime }-P{B}^{\prime }=15-13.3=17cm$

So, total length $A^{\prime }{B}^{\prime }+B^{\prime }{C}^{\prime }+C^{\prime }{D}^{\prime }=3.3+1.7+5=10cm$.