In the figure above let #AB# is a person. #A"B"# is his image as seen by him in the mirror.
Let a ray of light start from #B#, hit the mirror at #B'#, get reflected and enter eye at #E#. The image of #B# is formed at #B#". Similarly image of eye #E# is formed at #E#", where #E E'# hits the mirror making an angle of incidence of #0^@#. We see that for part height #EB# the required length of mirror is #E'B'#
Now if we draw a normal #B'N# as shown,
length #EN=E'B'# .......(1)
In #Delta EB'N and Delta NBB'#
- From laws of reflection we know that angle of incidence is equal to angle of reflection.
- Angles at #N=90^@#
- Side #B'N# is common.
#=># Triangles are congruent (ASA).
#=># sides #EN=NB# ......(2)
Combining (1) and (2) we get
#E'B'=1/2EB# .....(3)
Similarly it can be shown that
#A'E'=1/2AE# ......(4)
Combining (3) and (4)
Total height of the mirror required #A'B'=1/2AB#.
