Interference effect on refractionThe wavelength of light determines it's refraction in a prism.

But, what if two beams of light destructively interfere as they pass through a small prism, so there is no apparent wave or wavelength at that point in space?

For example, suppose you have two otherwise identical infrared laser beams traveling in almost the same direction intersecting at a prism with one beam 180 degrees out of phase producing total destructive interference at that small area in space. The beams intersect at the prism at a slight angle like an elongated x.

The index of refraction is wavelength dependent but if the waves are totally cancelled in the prism, would the beams exit the prism at same angle as if the beams constructively interfered?

I assume there are no “waves” and thus no detectable wavelength per se while beams are in the glass if there is total destructive interference at that point in space.

  • $\begingroup$ The configuration is rather unclear - a diagram would be extremely helpful in establishing unambiguously what give got in mind. $\endgroup$ – Emilio Pisanty Mar 27 at 19:47
  • $\begingroup$ If there is no electric field at the air-side of the prism, there will be no transmission into the prism (continuity of relevant fields accross the interface), but there still will be reflection. So all you will by sending two coherent anti-phase beams is to strengethen reflection from the prism and cancel transmission. $\endgroup$ – Cryo Mar 28 at 1:15
  • $\begingroup$ I assume for Beams 1 and 2 intersecting in the air (not a prism) at the vertex of an X, the beams would interfere at the vertex but continue on their merry ways and not interfere as the right side of the X. $\endgroup$ – jim lewis Mar 28 at 12:56

Thinking about water waves; interference is only a temporary thing. If 2 waves travel cross a pond from opposite sides, but let's say they are opposite in phase, in the middle of the pond when they meet they cancel, BUT this is only temporary. After they pass each other, they are visible again and continue along to the opposites shores!. Waves cancelling or "interference" is one of the most poorly explained phenomenons in physics, because they really don't literally cancel, they just temporarily interact with each other. The same could be said for light waves, 2 photons never cancel, that's a violation of energy conservation. In fact the only way to ever observe photons it to have them interact with matter (i.e. an electron in an atom/molecule). 2 photons arriving at an atom that are out of phase are very difficult for the atom to observe and 2 photons in phase increase the probability of observation greatly, its all Quantum Mechanics or probability.

  • $\begingroup$ Yes, I agree, the cancellation is temporary, and only at the vertex of the X. So, the questions is, the index of refraction is wavelength dependent but if the waves are temporarily cancelled in the prism, would the beams exit the prism at same angle as if the beams constructively interfered? I assume there is no detectable wavelength per se while beams are in the glass if there is temporary destructive interference at that small point in space, i.e., the prism at the vertex of the X. $\endgroup$ – jim lewis Mar 28 at 13:10
  • $\begingroup$ I would say you have to treat the light in the prism as many many interactions which eventually produce the refracted beam. It's not the result of that one infinity small point whereto field is zero, the beams continue on to the next atom where they are not aligned and do their thing . $\endgroup$ – PhysicsDave Mar 28 at 17:26

If you set up a shearing interferometer using collimated beams, interference planes are formed which show up as fringes if a screen is placed in the interference volume.

If you explore that volume using a piece of paper, you're likely to conclude that the light is cancelled out in the plane corresponding to a fringe. However, it's easy to prove that light is present in those "dark" fringes: stick something into the dark region and it will cast a shadow in both directions the beams are traveling.

The same thing is true in the prism experiment you have proposed. I assume the two light beams are mutually coherent, so have the same wavelength. They will form an interference pattern inside the prism, essentially the same as the interference pattern of a shearing interferometer: planes stacked together like a deck of thick playing cards. On a screen those will appear as light and dark fringes where the planes intersect the screen. Just like the shearing interferometer, light is passing through those "dark" planes even though they are "dark".

So, interference does not affect the propagation of any given light ray. The rays do not interact with each other; they interact with the detector, such as your eye or a light meter or a camera.

  • $\begingroup$ So, the beams are refracted by the same amount with constructive interference as with destructive interference. Here's what I do not understand, if you can help. Refraction is wavelength dependent. If the prism is the detector, and thus cannot "detect" the cancelled wave, nor presumably the wavelength of same, what determines the degree of refraction? $\endgroup$ – jim lewis Mar 28 at 19:17
  • $\begingroup$ The angle of refraction simply depends on the wavelength of the light, its angle of incidence, and the refractive indices of the prism and the surrounding medium. That's Snell's law. The prism is not a detector, any more than the air is; it's just a medium, with its own refractive index, through which the light propagates. $\endgroup$ – S. McGrew Mar 28 at 20:06
  • $\begingroup$ Aha! I see the light :) $\endgroup$ – jim lewis Apr 1 at 19:51

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