Optics: Phase difference

  • Thread starter Niles
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Hi

The phase difference associated with a light wave going through a birefringent material is given in this link (the first equation)

http://en.wikipedia.org/wiki/Wave_plate

My question is: Why do they only treat the spatial difference? I mean, there should also be a phase difference occuring because of the time-difference it takes for the light going through those paths?
 

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Born2bwire
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The phase for time is generally ignored since the phase progression is the same for waves of the same frequency. If there is a phase offset at the beginning, that is absorbed into the amplitude coefficients.
 
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The phase for time is generally ignored since the phase progression is the same for waves of the same frequency. If there is a phase offset at the beginning, that is absorbed into the amplitude coefficients.
The don't understand the bold part. The waves travel at different velocities (since different IOR), hence they spend different time in the material.
 
  • #4
Born2bwire
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The don't understand the bold part. The waves travel at different velocities (since different IOR), hence they spend different time in the material.
The time phase is independent of the phase/group velocities of the wave. It is dependent solely on the frequency of the wave. The change in the group velocity would be reflected in the spatial phase dependence since it changes the wave number (or you can think of it as changing the wavelength). That is because the phase dependence, of say a plane wave, is
[tex] ~ e^{i\left(\mathbf{k}\cdot\mathbf{r} - \omega t\right)}[/tex]
The phase velocity is \omega/k, the group velocity is d\omega/dk. Since \omega is constant, the changes in velocity would be reflected in the wave number k.
 
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The time phase is independent of the phase/group velocities of the wave. It is dependent solely on the frequency of the wave. The change in the group velocity would be reflected in the spatial phase dependence since it changes the wave number (or you can think of it as changing the wavelength). That is because the phase dependence, of say a plane wave, is
[tex] ~ e^{i\left(\mathbf{k}\cdot\mathbf{r} - \omega t\right)}[/tex]
The phase velocity is \omega/k, the group velocity is d\omega/dk. Since \omega is constant, the changes in velocity would be reflected in the wave number k.
This I agree with; but I don't believe it explains what happens to that extra ∆t the wave with the smaller vgroup spends in the material. I mean, at t=0 the wave enters the material, and a time t0 after the first component exits the material, and t0+∆t after the last component exits.
 
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Born2bwire
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This I agree with; but I don't believe it explains what happens to that extra ∆t the wave with the smaller vgroup spends in the material. I mean, at t=0 the wave enters the material, and a time t0 after the first component exits the material, and t0+∆t after the last component exits.
But the phase progression is the same. The slower wave will have the exact same time induced phase as the faster one when they are measured. At t_0+\delta t, the faster wave will also have the extra \delta t worth of phase. It doesn't matter how long the wave spends in the material, the phase progression from time is independent of the material, it only depends on frequency.
 
  • #7
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Niles, you are correct, the effect is the same if you calculate it one way or the other. The reason why we use the spatial description is, because we can manipulate it directly by making the crystal thinner or thicker.

If you would try to calculate it in the time domain, you would have to find out how long it takes each beam to pass the crystal from the thickness and the respective light speeds, and how much phase difference that time difference translates into.

One more multiplication... that's all.
 

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