A source of light O is located at a distance of 8 cm from a plane mirror. The reflected ray O' 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 X (B) 7.2 cm from X
(C) 2.4 cm from Y (D) 3.6 cm from Y
Answers
Answer:
Step-by-step explanation:
Answer:
According to laws of reflection (I think you presumably know it) that the angle of reflection is always is equal to angle of incidence .
∠i = ∠r which implies that :
⇒ 90° - ∠i = 90° - ∠r
This means that we can confidently say that ∠OPX = ∠O'PY [ See the figure once , don't be too lazy ] .
Let's see what we have got in ΔOPX and ΔO'PY . At the blink of an eye I can see that the two triangles are similar .
∠X = ∠Y [ 90° each ]
∠OPX = ∠O'PY [ as said above ]
Δ OPX ≈ Δ O'PY [ A.A criterion ]
Assume the distance is x cm from Y .
Then we have :
x / (12 - x) = 2/8
⇒ 8 x = 24 - 2 x
⇒ 8 x + 2 x = 24
⇒ 10 x = 24
⇒ x = 24/10
⇒ x = 2.4 cm
The distance is 2.4 cm from Y
SOLUTION
sini = (12-x)/8
=) sin r= x/2
by the law of reflection
angle of incidence = angle of reflection
=) sin i = sin r
=) (12-x)/8= x/2
=) 24- 2x= 8x
=) 24= 10x
=) x= 2.4cm
OPTION (c)✓