m and n are two plane mirrors placed parallel to each other as shown in the figure. An incident ray AB to the first mirror is reflected twice in the direction CD, Prove that AB || CD.
Draw CN ⊥ n and BM ⊥ m. Since BM ⊥ m, CN ⊥ n and m || n.
Therefore CN || BM. Now according to laws of reflection, we know that angle of incidence is equal to angle of reflection
... At B, ∠1 = ∠2 ….(i)
and at C, ∠3 = ∠4 ….(ii)
Since CN || BN with transversal BC, \\ ∠2 = ∠3 ….(Alternate interior angles) Þ 2 ∠2 = 2 ∠3
Þ ∠2 +∠2 = ∠3 + ∠3
Þ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠ABC = ∠BCD
... AB || CD
Therefore CN || BM. Now according to laws of reflection, we know that angle of incidence is equal to angle of reflection
... At B, ∠1 = ∠2 ….(i)
and at C, ∠3 = ∠4 ….(ii)
Since CN || BN with transversal BC, \\ ∠2 = ∠3 ….(Alternate interior angles) Þ 2 ∠2 = 2 ∠3
Þ ∠2 +∠2 = ∠3 + ∠3
Þ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠ABC = ∠BCD
... AB || CD
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