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# 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 Open in App
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## 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 [correcsponding angles] ...(ii) Also, ∠i=∠r [∵ incident angle = reflected angle] …(iii) From Eqs.(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. Alternate Method : In ΔOBC and ΔOAC,∠1=∠2 [each 90∘] Also , ∠i=∠r [ incident angle = reflected angle] ...(i) On multiplying both sides of Eq. (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.  Suggest Corrections  0      Similar questions
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