Question

Find the distance between object $\left(O\right)$ and final image formed after two successive reflections. Take the first reflection on mirror $M_1$ having radius of curvature $20\text{cm}$

• A. $20\text{cm}$
• B. $15\text{cm}$
• C. $40\text{cm}$
• D. $35\text{cm}$

Answer 1

The correct option is D $35\text{cm}$

For 1st reflection on $M_1$
Sign convention: Taking positive along the incident ray.
$f=\frac{-R}{2}=\frac{-20}{2}=-10\text{cm}$
$u=-15\text{cm}$
Let first image by the mirror $M_1$ is formed at distance $v$
Using mirror formula,
$\Rightarrow \frac{1}{v}+\frac{1}{(-15)}=\frac{1}{(-10)}$
$\Rightarrow \frac{1}{v}=\frac{1}{15}-\frac{1}{10}=\frac{2-3}{30}$
$\therefore v=-30\text{cm}$
$\Rightarrow$ First image formed by concave mirror lie at $I_1$ and $I_1$ will form another image at $I_2$ by plane mirror.
Thus, object distance for plane mirror, $|N{I}_1|=10\text{cm}$

Now the distance between object and image formed by plane mirror at $I_2$ is
$O{I}_2=|NO|+|N{I}_2|$
$\because |NO|=40-15=25\text{cm}$
$\therefore O{I}_2=25+10=35\text{cm}$