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Imagine the following setup:

I have a coherent single frequency electromagnetic wave (laser beam) that is imaged on a surface. It is reflected onto a detector (photodiode). I can easily take measurements there.

The question is: What happens when there is a second wave with similar, but not identical frequency, traveling the same path? There will be constructive and destructive interference along the propagating path. If I place the reflecting surface at a position, where it happens to interfere destructive, the electric field strength is zero on the surface.

I imagine, there is still an electric field composed of both beams. However, the physical interactions that lead to reflection or scattering are based on electric field which is zero at the interface.

Can I still measure the reflected light as both beams independently will have energy? What happens if I filter one of the frequencies after the reflection happened (assuming there is reflection)? Is there a difference between a reflecting and scattering surface?

Many thanks!

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  • $\begingroup$ If the beams don't have identical frequency, the interference will alternate between constructive and destructive at a given position. $\endgroup$
    – John Doty
    Jul 11, 2022 at 14:48
  • $\begingroup$ Thanks for this insight. Is this also true for multiple modes of a laser cavity? The waves inside the cavity are standing waves that have their nodes at the mirrors so I'd assume that the distance where the destructive interference occurs is constant. Also this is what one can observe with HeNe-Lasers that have two modes. The nodes are static and spaced equally to the resonator length. $\endgroup$
    – kpa
    Jul 12, 2022 at 14:58
  • $\begingroup$ If you place a mirror in a resonant cavity, it makes a new cavity with different mode frequencies. If the mirror is placed at a node of a mode, the specific frequency of that mode will be a resonance of both the original cavity and the modified cavity. $\endgroup$
    – John Doty
    Jul 12, 2022 at 15:45

1 Answer 1

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Interference doesn't destroy energy. The energy in the incoming like will go somewhere. This means that if destructive interference produces no or less light in one place, it produces more light through constructive interference in another.

Suppose you have monochromatic light at normal incidence on a plate of glass with a single layer antireflection coating. Without the coating a small amount is reflected.

enter image description here

With the coating, some light reflects from the top surface of the coating and some from the bottom surface. The thickness of the coating has been arranged so that the extra distance traveled to the bottom surface and back is $1/2$ wavelength. This means destructive interference will complete cancel the reflected wave. It also means all the light will be transmitted into the glass.

In reality it is more complex. Multiple reflections must be accounted for, there can be different phase shifts on reflection at the bottom and top, and there can be multiple layers.

enter image description here


You could take two coherent beams of identical wavelength from two sources that are $180^o$ out of phase. Total destructive interference means there would be no energy in the beam.

You would find that to combine two beams like that, you would wind up with the energy in the light going someplace else. For example, you can combine two beams with a partially reflective mirror. Not shown is the reflected input beam and transmitted pointer beam. If there is total destructive interference with one, there is total constructive interference with the other.

Image from AliExpress

You can ask what happens to the beam with $0$ intensity. Of course, it is "totally transmitted". But there is nothing to transmit.


If you have two different frequencies, you get beats. The incoming wave changes intensity with time. When the incident light is intense, the transmitted light is intense. When the incident light has no intensity, neither does the transmitted beam.

enter image description here


This has talked about two possibilities for light when it hits a surface. It can be transmitted or reflected. There is a third possibility. It can be absorbed. In this case, light energy is converted to another form, perhaps heat.

Again, energy is conserved. All three things can happen together. All the energy of the incident beam will go into a reflected beam, transmitted beam, and/or absorption.

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    $\begingroup$ The question refers to the last figure's blue wave as "interference", when it is just ordinary superposition; moreover, it asked to place a mirror at the node (ii), but that node moves so the question is ill-posed. $\endgroup$
    – JEB
    Jul 11, 2022 at 15:01
  • $\begingroup$ @JEB Yes. And, since the node moves at the speed of light, this is a variation on sites.pitt.edu/~jdnorton/Goodies/Chasing_the_light $\endgroup$
    – John Doty
    Jul 11, 2022 at 15:39
  • $\begingroup$ As commented above, one can observe these nodes with a HeNe Laser that has two modes (two adjacent frequencies). But there they semm to be static over time. If that is another phenomenon I appologise. Regardless, there is at least an instant with destructive interference at the surface as @JEB wrote. Can I consider the red and violet waves seperately (which would both reflect ordinary) or rather the combined blue wave (which would just transmit/absorbe)? $\endgroup$
    – kpa
    Jul 12, 2022 at 15:07

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