A girl with two normal eyes wants to see the full width of her face using a plane mirror. The eye to eye and ear to ear distances of her face are 4 inch and 6 inch respectively. The minimum width of the required mirror is: 

We can view an item directly or indirectly if it emits light that reaches our eyes. The light might be self-produced or reflected (in case of luminous objects). If we wish to see a whole thing with dimensions, light from the bottom and top corners as well as the left and rightmost ones must fall on our eyes.

Here – A and D are ears; B and C are eyes.

In the figure, we can see that for viewing full face in mirror$M_1-M_2$, she need to have part ‘FE’ (Parallel sides) of the mirror for minimum part of the mirror for the rays to be reflected and hence all the face will be covered. Now, in the question, given ear separation = 6 inch and eye to eye distance = 4 inch.
So,$AD=6inchandBC=4inch$, where ‘o’ is the centre.
Minimum width needed  $GH=FE$
Now, $AG={\displaystyle \frac{AC}{2}}=DH={\displaystyle \frac{BD}{2}}$
Now, note that $AC=Ao+oC={\displaystyle \frac{AD}{2}}+{\displaystyle \frac{BC}{2}}={\displaystyle \frac{6}{2}}+{\displaystyle \frac{4}{2}}=5inch=BD$
Also $GH =AD - ( AG+HD)=AD-\left(\frac{AC}{2}+\frac{BD}{2}\right)=6-\left(\frac{5}{2}+\frac{5}{2}\right)=1 Inch$

So, the correct answer is “Option A”.