There are two mirrors PQ and RS placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflect back along CD. Show that AB parallel CD.
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The two mirrors are placed parallel to each other and the angle of the incident ray = angle of the reflected ray.
Hence we can conclude that AB // CD
you will understand better by this-
Hence we can conclude that AB // CD
you will understand better by this-
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shivpatil2005:
And 1 more thing.....
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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
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