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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
9.6 cm from P
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2.4 cm from Q
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
3.6 cm from Q
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C 2.4 cm from Q
As we can see △ TSP \& Δ TS'Q
are similar,
So, SPS′Q=TPTQ⇒82=TPTQ⇒TP=4TQTP+TQ=12Cm⇒4TQ+TQ=12Cm⇒TQ=2.4cm