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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:

image $HF$ 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.

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  • $\begingroup$ "Now, since $HI=IE=HE2=0.08m$ and $FC=CE=EF2=0.92m$, $KD=1m$" - this is incorrect. $KD$ should be $KG$ and $KG = HF - HI - FC = IE + CE = 0.92$, exactly half of the persons length. $\endgroup$ – Johannes Jul 28 '13 at 5:00
  • $\begingroup$ I'm sorry. I meant $KG$ instead of $KB$ in the last statement. Edited. Thanks for pointing out the mistake. $\endgroup$ – Gerard Jul 28 '13 at 5:13
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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.

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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).

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  • $\begingroup$ you've used your own set of labels (a, b, p, and q) without showing us where they are. This post could really do with a diagram to help the readers. I get the sense that in general it is a sound answer, but it could be presented in a much better way. Who knows, it might even get you a populist badge ;) $\endgroup$ – Jim Sep 28 '16 at 13:30
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enter image description here

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

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