Question

Two concave mirrors each of radius of curvature $40cm$ are placed such that their principal axes are parallel to each other and at a distance of $1cm$ to each other. Both the mirrors are at a distance of $100cm$ to each other. Consider first reflection at $M_1$ and then at $M_2$. If the y-coordinate of the final image thus formed is found to be $-\frac{x}{5}\text{cm}$. Then find the value of $x$. (Take the location of the object as the origin).

Answer 1

At $M_1$,
$f=-20\text{cm}$
$\frac{1}{v}+\frac{1}{(-60)}=\frac{1}{(-20)}$
$\frac{1}{v}=\frac{1}{60}-\frac{1}{20}=\frac{-40}{60\times 20}$
$\Rightarrow v=-30\text{cm}$
$\Rightarrow$ x-coordinate $=-30cm$

Magnification $m=-\frac{v}{u}=\frac{(-30\times -1)}{(-60)}=\frac{-1}{2}$
Image height $=-1\times \frac{-1}{2}=\frac{1}{2}cm$

$\Rightarrow$ y-coordinate $=1+\frac{1}{2}=\frac{3}{2}cm$

So image coordinates are $(-30,\frac{3}{2})$ after reflection from $M_1$

At $M_2$,
$\frac{1}{v}+\frac{1}{(-70)}=\frac{1}{-20}$ $\Rightarrow \frac{1}{v}=\frac{1}{70}-\frac{1}{20}=\frac{-50}{70\times 20}$

$\Rightarrow v=\frac{-140}{5}=-28\text{cm}$
Magnification $m=\frac{-v}{u}=\frac{-(-28)}{(-70)}=-\frac{2}{5}$
Object height $=\frac{3}{2}\text{cm}$

Image height $=\left(\frac{3}{2}\right)(-\frac{2}{5})=-\frac{3}{5}\text{cm}$
$\Rightarrow$ y-coordinate $=-\frac{3}{5}\text{cm}$
$\Rightarrow x=3$