Question

The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D, as shown in the figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.

Answer 1

Given: An object is placed at a point A, the image of the object is seen at the point B, an observer is at point D,
and LM is a plane mirror .
To Prove: The image is as far behind the mirror as the object is in front of the mirror, i.e. BT = AT.
Proof:
∵ LM is a plane mirror
∴ i = r
(Angle of incidence is always equal to angle of reflection) .....(1)
(Both AB and CN are perpendicular to LM)
Also, AB∥CN
⇒∠TAC=∠ACN=i
And, ∠CBT=∠NCD=r
(Alternate interior angles) .....(2)
(Corresponding angles) .....(3)
From (1), (2) and (3), we get

∠TAC = ∠CBT.....(4)

Now,

In ∆TAC and ∆CBT,

∠TAC=∠CBT

therefore, ∆TAC≅∆CBT

From (4)

∠ATC=∠BTC=90°

CT=CT

Hence, AT = BT (CPCT)