Law of reflection?

Question

Suppose a man stands in front of a mirror. His eyes are 1.65 m above the floor, and the top of his head is 0.13 m higher. What is the height above the floor of the top and bottom of the smallest mirror in which he can see both the top of his head and his feet. How is this distance related to the man's height?

Answer 1

Let:

A be the top of the man's head with reflection A',

B be the man's eyes with reflection B',

C be the man's feet with reflection C'.

MN be the wall (M at the top) on which the mirror is fixed.

Join A'B meeting MN at P (top of the required mirror),

and C'B meeting MN at Q (bottom of the required mirror).

As the image is as far behind the mirror as the man is in front,

tringles PQB and A'CB are similar, with the dimensions of  the former being half those of the latter.

Therefore:

QN = BC / 2 = 0.825 m.

PQ = A'C' / 2 = AC / 2 = 1.78 / 2 = 0.89 m.

PN = PQ + QN = 1.715 m.

The mirror is half the man's height, the top is half way vertically between his eyes and the top of his head, and the bottom is half way vertically between his eyes and his feet.