# Do reflected and incident rays interfere?

I have a very simple question about incident an reflected waves. Consider for instance an incident light ray on an interface between two different materials. We know that there will be a reflected wave (with same angle of incidence) and a transmitted wave.

Let's focus on incident and reflected waves. My question is: do these waves interfere and become one single wave, or will they be separate?

I have inserted the following picture to explain better my doubt: Consider a plane object (in red). If the reflected ray is independent from the incident ray, that surface will see that ray arriving orthogonally at it. If incident and reflected ray interfere and become one single wave, I'd say that it would propagate on a direction which is different from both directions of incident and reflected waves.

From what I know from electromagnetic waves, I'd say they would interfere (because they have the same frequencies). But if it is true, I do not understand why we usually draw two rays with specific directions for incident and reflected waves as if they were separate entities.

• The rays in that diagram can be taken to mean a narrow pencil of light, or the propagation vector of a plane wave, or the mean propagation vector of a beam of light with a non-zero, non-infinite (such quantities are sometimes called ponderable) cross section. Also note that the wave equation is linear, which means the interference does not change the directions of the two beams. They simply overlap. – garyp May 28 at 1:45
• there is a series of MIT video on laser beam interferences, example youtube.com/watch?v=J4Ecq7hIzYU . Note that interference is not interaction, (the two beams do not scatter off each other) – anna v May 28 at 7:10

In a more realistic way you would draw wave beams instead of rays, like in the image below. And here is the same situation as an animation.

So you are right. The reflected wave and the incident wave indeed interfere with each other. Where the incident and reflected beams overlap, we get kind of a standing wave pattern. Where there is only one beam, we have just a simple wave.

Drawing rays instead of waves is just a convenient simplification. This is valid because waves propagate in a straight direction (like rays), and waves get reflected at surfaces (also like rays).

• Why would the interference pattern go away on the exiting beam on the right? Think about a beam splitter in an Interferometer once the two beams interfere the patterns persist long after the b/s. Once destructive interference occurs that wave is gone. Yet, in your diagram the waves “regenerate”. – Lambda May 28 at 21:38
• @Lambda I tried to improve my explanation and added an animation. – Thomas Fritsch May 29 at 11:01

If the incident beam had nonzero diameter, or were a plane wave, then they would interfere to form a standing wave.

But rays are an idealization, which assumes a zero-diameter beam. In this approximation, the incident and reflected waves don't overlap (except at the exact point where they meet the surface) so there is no standing wave.

• Why if the incident wave is plane, it does not interfere with the reflected one? – Kinka-Byo May 27 at 21:12
• @Kinka-Byo, My answer says If the incident beam ... were a plane wave, then they would interfere. – The Photon May 27 at 21:14