The image of the 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 figure.prove that the image is as far behind the mirror as the object is in front of the mirror.

Explanation

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_|_" and " AB_|_LM`
`rArr" "AB||CN`
`angleA=anglei" [alternate interior angles]...(i)"`
`angle B=angle r" [corresponding angles]...(ii)"`
Also `" "anglei=angler" "[ because "incident angle = reflected angle"`]...(iii)
From Eqs. (i), (ii) and (iii),`" "angle A=angle B`
In `DeltaCOB" and " Delta COA," "angleB=angleA" [Proved above]"`
`angle1=angle2" [each"90^(@)]`
`"and " CO=CO ["common side"]`
`:." "DeltaCOBcongDeltaOAC " [by AAS congruence rule]" `
`rArr" "OB=OA" [by CPCT]"`
Alternate Method
In`DeltaOBC " and "DeltaOAC," "angle1=angle2" [each "90^(@)]`
`"Also, " anglei=angler" "[ :'" incident angle =redlected angle]...(i)"`
On multiplying both sides of Eq. (i) by - 1 and than adding `90^(@)` both sides, we get
`90^(@)-anglei=90^(@)-angler`
`rArr " "angleACO=angle BCO`
`" and "OC=OC" [Common side]`
`:." "DeltaOBCcongDeltaOAC" [by ASA conguence rule]"`
`rArr" "OB=OA" [by CPCT]"`
Hence, the image as for behind the mirror as the object is in front of the mirror.