# Question #303a3

Apr 3, 2017

A)

#### Explanation: In the figure above let $A B$ is a person. $A \text{B}$ 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 ${0}^{\circ}$. We see that for part height $E B$ the required length of mirror is $E ' B '$

Now if we draw a normal $B ' N$ as shown,
length $E N = E ' B '$ .......(1)

In $\Delta E B ' N \mathmr{and} \Delta N B B '$

1. From laws of reflection we know that angle of incidence is equal to angle of reflection.
2. Angles at $N = {90}^{\circ}$
3. Side $B ' N$ is common.

$\implies$ Triangles are congruent (ASA).
$\implies$ sides $E N = N B$ ......(2)

Combining (1) and (2) we get
$E ' B ' = \frac{1}{2} E B$ .....(3)
Similarly it can be shown that
$A ' E ' = \frac{1}{2} A E$ ......(4)

Combining (3) and (4)
Total height of the mirror required $A ' B ' = \frac{1}{2} A B$.