Why angle of incidence equal to angle of reflection

[prashant13b]

I want to why in ray optics. Angle between incident ray and reflecting surface normal is equal to angle between reflected ray and reflecting surface normal. Simply why angle of incidence is equal to angle of reflection?

 

[A.T.]

I want to why in ray optics. Angle between incident ray and reflecting surface normal is equal to angle between reflected ray and reflecting surface normal. Simply why angle of incidence is equal to angle of reflection?

http://www.physics.ucdavis.edu/Classes/Physics9B_Animations/ReflRefr.html

 

[prashant13b]

http://www.physics.ucdavis.edu/Classes/Physics9B_Animations/ReflRefr.html

This law was made before the evolution of wave optics. I want answer on basis of classical physics

 

[PaulDirac]

Write Maxwell equations and apply them to the inner and outer medium. One out of three consequences you get would be the equality of incident and reflected angles. Let me know if it is not clear.

 

[ehild]

I want to why in ray optics. Angle between incident ray and reflecting surface normal is equal to angle between reflected ray and reflecting surface normal. Simply why angle of incidence is equal to angle of reflection?

Ray Optics is based on Fermat Principle. http://en.wikipedia.org/wiki/Fermat's_principle
In the simplest form it states that the light ray travels along a path between two points which needs the shortest time.

In optics, Fermat's principle or the principle of least time is the principle that the path taken between two points by a ray of light is the path that can be traversed in the least time. This principle is sometimes taken as the definition of a ray of light.[1] However, this version of the principle is not general; a more modern statement of the principle is that rays of light traverse the path of stationary optical length with respect to variations of the path.[2] In other words, a ray of light prefers the path such that there are other paths, arbitrarily nearby on either side, along which the ray would take almost exactly the same time to traverse.[3]

See picture. You want the light reach from A to B in the way that it is reflected from the horizontal mirror. How can you draw the the light ray so it reaches B in the shortest time? It must travel along the shortest path.
Draw the mirror image of B: it is B'. Connect A and B' with a straight line. The line segment AB' is the shortest of all possible paths between A and B'.
AB' intersects the mirror at C. Connect C with B. The length of CB is equal to CB'. So the length of the path ACB is the shortest possible between A and B which touches the mirror. The three alpha angles are congruent, so the beta-s are also equal ($\beta=90-\alpha$

)ehild