Question

A source of light S is located at a distance of 8 cm from a plane mirror. The reflected ray S' is detected 2 cm from the mirror at a vertical displacement of 12 cm. The point of reflection is at a distance of

• A. 6.0 cm from P
• B. 9.6 cm from P
• C. 2.4 cm from Q
• D. 3.6 cm from Q

Answer 1

The correct option is C 2.4 cm from Q

As we can see $△$ TSP \& $\Delta$ TS'Q

are similar,

$\begin{array}{c}\text{So,}\frac{SP}{S^{\prime }Q}=\frac{TP}{TQ}\\ \Rightarrow \frac{8}{2}=\frac{TP}{TQ}\\ \Rightarrow TP=4TQ\\ TP+TQ=12Cm\\ \Rightarrow 4TQ+TQ=12Cm\\ \Rightarrow TQ=2.4cm\end{array}$

$\therefore$ Reflection point is $2.4cm$

from $Q$.