Why does reflected light leave a smooth surface at the same angle as incident light?

[ValenceE]

Hello to all,

while I’m waiting for my reservation of Richard Feynman’s QED book to be available at the local library, here’s a question about light reflection …

Could someone please explain the process by which an incident ray of light is reflected in exactly the same angle after its interaction with a smooth reflective surface ?


Regards,


VE

 

[Meir Achuz]

The phase of the incident plane wave is given by exp(ik.r).
The phase of the reflected plane wave is given by exp(ik'.r).
In order for the boundary conditions at the interface to hold on the entire surface, k'.r must equal k.r, where r is in the plane of the surface.
The vector dot product k.r=kr\cos\theta, where \theta is the angle between k and r. k and the reflected wave number k' are equal in magnitude, so cos\theta=cos\theta'.
[Note: this angle \theta is (90 degrees - the "incident angle").]