# Why angle of incidence equal to angle of reflection

1. Oct 11, 2014

### 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?

2. Oct 11, 2014

### A.T.

Last edited by a moderator: Apr 18, 2017
3. Oct 11, 2014

### prashant13b

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

Last edited by a moderator: Apr 19, 2017
4. Oct 11, 2014

### 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.

5. Oct 11, 2014

### ehild

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.
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