Question

In the Fig. 11.12, m and n are two plane mirrors perpendicular to each other.Show that the incident ray CA is parallel to the reflected ray BD.​​

Answer 1

Explanation:

In order to prove that CA || BD. It is sufficient to show that

→∠CAB + ∠ABD=180°

Since mirrors m and n are perpendicular and OB and OA are perpendicular to m and n respectively.

OA ⊥ AB → ∠ BOA = 90°

In ∆BOA, we have

→∠2+∠3+∠BOA=180°

→∠2+∠3+90°=180° [∠BOA = 90°]

→∠2+∠3=90°

→2(∠2+∠3)=180° [Multiplying 2 on both sides]

→2(∠2)+2(∠3)=180

→∠CAB+∠ABD=180° [ Angle incidence = Angle of reflection ∴ ∠1 = ∠2 and ∠3 = ∠4. 2∠2 = ∠CAB and 2∠3 = ∠BAD]

Thus, CA and BD are two lines intersected by a transversal AB such that ∠CAB + ∠ABD = 180° i.e, the sum of the interior angles on the same side of AB is 180°. Hence,CA || BD.

Answer 2

Given:

Two plane mirrors m and n, perpendicular to each other. CA is incident ray and BD is reflected ray.

To Prove:

CA∥DB

Construction: OA and OB are perpendiculars to m and n respectively.

Proof:

∵m⊥n,OA⊥m and OB⊥n

∴∠AOB=90°

(Lines perpendicular to two perpendicular lines are also perpendicular.)

In ΔAOB,

∠AOB+∠OAB+∠OBA=180°

⇒90°+∠2+∠3=180°⇒∠2+∠3=90°

⇒2(∠2+∠3)=180°

(Multiplying both sides by 2)

⇒2(∠2)+2(∠3)=180°

⇒∠CAB+∠ABD=180°

(Angle of incidence = Angle of reflection)

∴∠1=∠2 and ∠3=∠4)

⇒CA∥BD (∠CAB & ∠ABD form a pair of consecutive interior angles and are supplementary)