Question

Two plane mirrors $M_1$ and $M_2$ are inclined at ${30}^{\circ }$to the vertical. A point object $\left(\text{O}\right)$ is placed symmetrically between them at a distance of $4\text{cm}$ from each mirror. Find the distance of the object from the second image formed in mirror $M_1$.

• A. $8\sqrt{3}\text{cm}$
• B. $4\sqrt{3}\text{cm}$
• C. $8\sqrt{2}\text{cm}$
• D. $8\text{cm}$

Answer 1

The correct option is A $8\sqrt{3}\text{cm}$


From $\Delta OPC,\sin {30}^{\circ }=\frac{4}{\text{OC}}$
$\Rightarrow \text{OC}=8\text{cm}$

Here:

$I_{11}\to$ 1st image formed on mirror $M_1$ considering an object O.

$I_{21}\to$ 1st image formed on mirror $M_2$ considering an object O.

$I_{1_{21}}\to$ 2nd image formed on mirror $M_1$ considering an object $I_{21}$ (Image formed by $M_2$ of object O)

Also, we can see that $△I_{1_{21}}CO$ is isosceles.
$(\because \Delta OCQ\text{is congurent to}\Delta QC{I}_{21}\text{and}C{I}_{1_{21}}=C{I}_{21}=OC)$
$\Rightarrow \angle O{I}_{1_{21}}C={30}^{\circ }$

Using sine rule in $△I_{1_{21}}CO$,

$\frac{x}{\sin {120}^{\circ }}=\frac{OC}{\sin {30}^{\circ }}$

$\Rightarrow x=OC\times \sqrt{3}=8\sqrt{3}\text{cm}$