We know that in a homogeneous medium light propagates rectilinearly, i.e. a speedy way. But the light chooses the quickest way even when it goes from one point to another directly and reaches its pre reflected from the mirror.
Follow it through. Let the letter A in figure oboznachaet the light source, the line MN - mirror, and the line ABC - the path of the beam from the spark to the eyes C. Direct KB perpendicular to MN.
Fig. 1. The reflection angle 2 equals the angle of incidence 1.
According to the laws of optics, the reflection angle 2 equals the angle of incidence 1. Knowing this, it is easy to prove that all possible paths from A to C, associated with the achievement of the mirror MN, the path ABC is the most imminent. To do this, compare the path of the beam ABC with some other, for example with the ADC (Fig. 2). Lower the perpendicular AE from point A to MN and continue it next to the intersection with the continuation of the ray BC at point F. also connect points F and D. make Sure, first of all, the equality of the triangles ABE and EBF. They are rectangular, and they have a common leg EV; in addition, the corners of the EFB and ESS are equal, as are respectively equal to the angles 2 and 1. Therefore, AE = EF. Hence the equality of right-angled triangles AED and EDF on two legs and, consequently, the equality AD and DF.
The light reflecting, chooses the shortest path.
Because of this we can path ABC to replace equal by CBF (since AB = FB), a path ADC - by CDF. Comparing between a length CBF and CDF, we can see that a straight line CBF shorter broken CDF. From here the path ABC shorter ADC, what we wanted to prove!
Wherever the point D, the path ABC is always shorter path ADC, if only the angle of reflection equals the angle of incidence. Therefore, the light really chooses the shortest and the most speedy way possible between the source, the mirror and the eye. This fact was first pointed out another Heron of Alexandria, a wonderful Greek engineer and mathematician of the second century.