# 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.

• "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. – Johannes Jul 28 '13 at 5:00
• I'm sorry. I meant $KG$ instead of $KB$ in the last statement. Edited. Thanks for pointing out the mistake. – Gerard Jul 28 '13 at 5:13

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.

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).

• 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 ;) – Jim Sep 28 '16 at 13:30

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