# All mirrors always shrink to 50% scale?

I have this geometric optics exercise here, in which a man is looking at himself in a mirror.

1. Determine the minimum height at which the bottom of the mirror must be placed so the man can see his feet (his eyes are at 1.7m).
2. How does this height vary depending on the distance from the mirror to the man's eyes?
3. What is the minimum length of the mirror so that the man can see his entire body (assume he is 1.8m tall).

From the principle that the reflected angle is equal to the incident angle, we show that the minimum height of the bottom of the mirror is directly between the feet and the eyes (1), and that this does not depend on the distance to the mirror (2). The top of the mirror must be directly between the man's crown and eyes at the lowest, in order to reflect the top of his head, therefore the mirror must be half the height of the man, at minimum.

Thus, the mirror reflects the man's image at a scale of 50%. But this doesn't depend on the distance from the mirror, and from what I can tell, doesn't even depend on the size of the mirror. These results, to me, indicate that no matter what kind of mirror you use or where you stand, your image should appear half way up the mirror, and at 50% scale.

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for the condition that the mirror should be half the height of the man the mirror should be placed I don't remember exactly but it was 18 or 25 cm from the man – Dimensionless Mar 8 '13 at 21:16
@Akash No. It just has be be (1) flat and (2) vertical. This is the result of $\theta_{inc} = \theta_{ref}$ and simple geometry. Jack, not all mirrors have that property, only those that are flat: go to a carnival funhouse to see counter examples, or look at a shaving mirror. – dmckee Mar 8 '13 at 21:23