From what I have gathered, complete constructive or destructive interference results in all light or no light traveling a given path of an interferometer (correct me if this is incorrect). However, when waves are partially constructive (the waves are mostly, but not completely in phase with each other) or partially destructive (the waves are mostly, but not completely out of phase with each other), I do not know what happens. Since light is made up of photons, does partial constructive/destructive interference simply result in more/fewer photons going down a given path? This is what I think, but I am looking for clarification.
2 Answers
Largely correct. Photons are sort of like classical particles and sort of like classical waves. Thinking of them as such can help or lead your thoughts into confusion.
In this case, photons do not go down a path. Where photons go is wave like. What happens when they are detected is particle like. So complete destructive interference means few photons are detected at a point. Constructive interference means lots of photons are detected there. Partial cases are in between.
That said, there are cases where you can use rays to approximate waves. These are cases where you don't need the phase of waves to figure out where they are headed. This means you do not have multiple waves interfering. You do not have small apertures where diffraction matters. (For diffraction, you treat each point across the opening as a point source of waves. At a distant point, you add up all the waves, taking phase into account.)
In this case, you draw rays perpendicular to wave fronts. Waves travel in the direction of the rays. If rays spread out, the intensity of the wave gets smaller. Rays are really tools to tell you what waves are doing. Waves are still the real story.
What you're taught in high school/1st year college is based on 1920s science/textbooks, it hasn't changed. And this explanation sort of works until you dig a little deeper ..... like you have. Two famous scientists Dirac and Feynman said every photon "interferes with itself" and this is a basis for the modern interpretation that "every photon determines it own path". The dark areas in the double slit experiment or in an interferometer have no photons .... the bright areas get them all. The Feynman path integral provides an explanation ..... you can find out more on this site.