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 following figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.

[Hint: CN is normal to the mirror. Also, angle of incidence = angle of reflection].

Answer 1

Given: An object OA placed at a point A, LM be a plane mirror, D be an observer and OB is the image.

To prove: The image is as far behind the mirror as the object is in front of the mirror i.e., OB = OA.

Proof: CN ⊥ LM and AB ⊥ LM

⇒ AB || CN

∠A = ∠i  [Alternate interior angles] ...(i)

∠B = ∠r  [Corresponding angles] ...(ii)

Also, ∠i = ∠r  [∵ incident angle = reflected angle]  ...(iii)

From equations (i), (ii) and (iii),

∠A = ∠B

In ΔCOB and ΔCOA,

∠B = ∠A  ...[Proved above]

∠1 = ∠2   ...[Each 90°]

And CO = CO   ...[Common side]

∴ ΔCOB ≅ ΔCOA   ...[By AAS congruence rule]

⇒ OB = OA   ...[By CPCT]

Hence proved.

Answer 2

In ΔOBC and ΔOAC,

∠1 = ∠2   ...[Each 90°]

Also, ∠i = ∠r  [∵ incident angle = reflected angle]  ...(i)

On multiplying both sides of equation (i) by –1 and then adding 90° both sides, we get

90° – ∠i = 90° – ∠r

⇒ ∠ACO = ∠BCO

And OC = OC  ...[Common side]

∴ ΔOBC ≅ ΔOAC   ...[By ASA congruence rule]

⇒ OB = OA   ...[By CPCT]

Hence, the image is as far behind the mirror as the object is in front of the mirror.