Question

The width of man's face is $D$. The distance between the eyes of the man is $d$. Then the minimum width of plane mirror to see his full face is

• A. $\frac{D-d}{4}$
• B. $\frac{D-d}{2}$
• C. $D-d$
• D. $\frac{D+d}{2}$

Answer 1

The correct option is B $\frac{D-d}{2}$

Let us draw the asked situation as shown below
Here, the length $GA$ represents the width of man's face, the length $E_1{E}_2$ represents the distance between eyes.
Assuming the face to be presented is symmetric as shown in the figure.
Also, the part of the face seen by eye $E_1$ will be toward $A$ and part of the face seen by eye $E_2$ will be toward $G.$
Here, we need to find the length $CB$ which the width of mirror required to see the face.
Now, from geometry $\Delta E_{1}BD\simeq \Delta ABD$
$\Rightarrow E_{1}A=E_{1}O+OA=\frac{d}{2}+\frac{D}{2}=\frac{D+d}{2}$
and, $E_{1}D=DA=\frac{E_{1}A}{2}=\frac{D+d}{4}$
Now we have
$OD=O^{\text{'}}B=OA-DA$
$\Rightarrow O^{\text{'}}B=\frac{D}{2}-\frac{D+d}{4}=\frac{D-d}{4}$
Also from symmetry $O^{\text{'}}B=O^{\text{'}}C=\frac{D-d}{4}$
Thus, width of mirror is $CB=O^{\text{'}}C+O^{\text{'}}B=\frac{D-d}{4}+\frac{D-d}{4}=\frac{D-d}{2}$

Key point: Use the concept of lateral inversion in such problem while drawing the ray diagram.