In the given figure, m and n are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.
In the given figure, 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:
AP is normal to the plane mirror OA and BP is normal to the plane mirror OB.
It is given that the two plane mirrors are perpendicular to each other.
Therefore, BP || OA and AP || OB.
So, BP ⊥ AP
(OA ⊥ OB)
⇒ ∠APB = 90° .....(1)
In ∆APB,
∠ 2 + ∠3 + ∠APB = 180° (Angle sum property)
∴ ∠2 + ∠3 + 90° = 180° [Using (1)]
⇒ ∠2 + ∠3 = 180° − 90° = 90°
⇒ 2∠2 + 2∠3 = 2 × 90° = 180°
By law of reflection, we have.....(2)
∠1 = ∠2 and ∠3 = ∠4.....(3)
(Angle of incidence = Angle of reflection)
From (2) and (3), we have
∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠BAC + ∠ABD = 180°
(∠1 + ∠2 = ∠BAC and ∠3 + ∠4 = ∠ABD)
Thus, the lines CA and BD are intersected by a transversal AB such that the interior angles on the same side of
the transversal are supplementary.
∴ CA || BD
ANSWER:
AP is normal to the plane mirror OA and BP is normal to the plane mirror OB.
It is given that the two plane mirrors are perpendicular to each other.
Therefore, BP || OA and AP || OB.
So, BP ⊥ AP
(OA ⊥ OB)
⇒ ∠APB = 90° .....(1)
In ∆APB,
∠ 2 + ∠3 + ∠APB = 180° (Angle sum property)
∴ ∠2 + ∠3 + 90° = 180° [Using (1)]
⇒ ∠2 + ∠3 = 180° − 90° = 90°
⇒ 2∠2 + 2∠3 = 2 × 90° = 180°
By law of reflection, we have.....(2)
∠1 = ∠2 and ∠3 = ∠4.....(3)
(Angle of incidence = Angle of reflection)
From (2) and (3), we have
∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠BAC + ∠ABD = 180°
(∠1 + ∠2 = ∠BAC and ∠3 + ∠4 = ∠ABD)
Thus, the lines CA and BD are intersected by a transversal AB such that the interior angles on the same side of
the transversal are supplementary.
∴ CA || BD
