Question

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​

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Answer 1

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

Answer 2

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)✓

hope it helps