The image of an object placed at a point A before a plane mirror LM is

1.	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» According to the figure we need to prove that AT = BT We know that Angle of incidence = Angle of reflection So we get ∠ ACN = ∠ DCN ….. (1) We know that AB \|\| CN and AC is the transversal From the figure we know that ∠ TAC and ∠ ACN are alternate angles ∠ TAC = ∠ CAN …… (2) We know that AB \|\| CN and BD is the transversal From the figure we know that ∠ TBC and ∠ DCN are corresponding angles ∠ TBC = ∠ DCN ….. (3) By considering the equation (1), (2) and (3) We get ∠ TAC = ∠ TBC …… (4) Now in △ ACT and △ BCT ∠ ATC = ∠ BTC = 90^o CT is common i.e. CT = CT By AAS congruence criterion △ ACT ≅ △ BCT AT = BT (c. p. c. t) Therefore, it is proved that the image is as far behind the mirror as the object is in front of the mirror.

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»

According to the figure we need to prove that AT = BT

We know that

Angle of incidence = Angle of reflection

So we get

∠ ACN = ∠ DCN ….. (1)

We know that AB || CN and AC is the transversal

From the figure we know that ∠ TAC and ∠ ACN are alternate angles

∠ TAC = ∠ CAN …… (2)

We know that AB || CN and BD is the transversal

From the figure we know that ∠ TBC and ∠ DCN are corresponding angles

∠ TBC = ∠ DCN ….. (3)

By considering the equation (1), (2) and (3)

We get

∠ TAC = ∠ TBC …… (4)

Now in △ ACT and △ BCT

∠ ATC = ∠ BTC = 90^o

CT is common i.e. CT = CT

By AAS congruence criterion

△ ACT ≅ △ BCT

AT = BT (c. p. c. t)

Therefore, it is proved that the image is as far behind the mirror as the object is in front of the mirror.