Minimum height of mirror required to view image I wanted to know the minimum height of mirror required to be able to view a complete image of a person. I considered the following setup:

$HF$ is the person in question. $H$ denotes the head, $F$ the feet, and $E$, the eyes. For the person to see his complete image, a ray each from $H$ and $F$ has to come and reflect into his 
eyes ($E$). Let $HE = 0.16m$ and $HF = 1.84m$. $KG$ is the minimum height if mirror required. 
Now, since $HI = IE = \frac{HE}{2} = 0.08m$ and $FC = CE = \frac{EF}{2} = 0.92m$, $KG = 1m$. 
But this doesn't make any sense. This calculation doesn't take into account the distance of the person from the mirror. It is clear that the distance matters. If I have a really small mirror, and I go far away from it, I can see my whole body; which is not the case if I'm really close to it. 
 A: The minimum mirror height required to see your whole body in a mirror oriented in parallel to your body is half the height of your body. The top edge of the mirror should be half in-between the level of your eyes and the top of your head, and the bottom edge of the mirror should be at a level half way between your eyes and your feet. 
This is independent of your distance to the mirror. Such can easily be inferred by drawing a picture that shows straight lines from your eyes going to the top of the head and to the bottom of the feet of your mirror image.
A: To see the complete size of image of a person we must use the mirror of length is equals to half  of the length of  the person.
If  ab is the object then a ray of light moves from 'a' and reflected at p ,then reach at the eye which is at O ,then another ray moves from 'b' reflected at q , then reach at eye O if the image is  a'b'  then p & q be the mid point of  Oa' & Ob' ,so in triangle Oa'b'  pq is equals to half of a'b' (proved).
A: 
In image ,a person of height H is standing against a plane mirror of length AB.We will need the part AF of mirror so that the person will be able to see his full reflection .
In triangle CED we can say $\frac{CD}{CA}=\frac{DE}{AF}$
Or 
$\frac{2\not{x}}{\not{x}}=\frac{H}{minimumlength}$
Minimum length =$\frac{H}{2}$
We can see there is no x here ,so distance of person wil not matter
