Question

Question 2
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 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 :
$\because CN\perp LMandAB\perp LM$
$\Rightarrow AB||CN$
$\angle A=\angle I$ [alternate interior angles]…(i)
$\angle B=\angle r$ [correcsponding angles] ...(ii)
Also, $\angle i=\angle r$ $[\because$ incident angle = reflected angle] …(iii)
From Eqs.(i), (ii) and (iii),
$\angle A=\angle B$
$In\Delta COBand\Delta COA$,
$\angle B=A$ [proved above]
$\angle 1=\angle 2$ [each ${90}^{\circ }]$
And CO=CO [common side]
$\therefore \Delta COB\cong \Delta COA$ [ by AAS congruence rule]
$\Rightarrow OB=OA$ [by CPCT]
Hence proved.

Alternate Method :
$In\Delta OBCand\Delta OAC,\angle 1=\angle 2$ [each ${90}^{\circ }]$
Also , $\angle i=\angle r$ [ incident angle = reflected angle] ...(i)
On multiplying both sides of Eq. (i) by-1 and then adding ${90}^{\circ }$ both sides, we get ${90}^{\circ }-\angle i={90}^{\circ }-\angle r$
$\Rightarrow \angle ACO=\angle BCO$
And OC=OC [common side]
$\therefore \Delta OBC\cong \Delta OAC$ [by ASA congruence rule]
$\Rightarrow OB=OA$ [by CPCT]
Hence, the image is as far behind the mirror as the object is in front of the mirror.