Question

In fig 6.33 PQ and RS are two mirrors placed parallel to each other.Anincident ray AB strike the mirror PQ at B,the reflected ray moves along a path BC and striks the mirror RS at C and again reflects back long CD. prove that AB // CD.​

Answer 1

Answer:

1. construction: draw Ray BL

Answer 2

Solutions:

Draw BE and CF normals to the mirrors PQ and RS at B and C respectively.

Then, BE ⊥ PQ and CF ⊥ RS.

Since, BE and CF are perpendicular to parallel lines PQ and RS respectively. Therefore, BE || CF.

Since, BE || CF and transversal BC intersects BE and CF at B and C respectively.

Hence, ∠3 = ∠2 .............. [Alternate angles]..... (i)

But, ∠3 = ∠4 and ∠1 = ∠2 ........... [Since, angle of incidence = angle of reflection] ........ (ii)

=> ∠4 = ∠1

=> ∠3 + ∠4 = ∠2 + ∠1 .......... [Adding corresponding sides of (i) and (ii)]

=> ∠ABC = ∠BCD

Thus, transversal BC intersects lines AB and CD such that alternate interior angles ∠ABC and ∠BCD are equal. Hence, AB || CD