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#.