Why is the refractive index for red lower than for blue?

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SUMMARY

The discussion centers on the refractive index of water, specifically why it is lower for red light than for blue light. The Drude dispersion model is highlighted as a key framework for understanding this phenomenon, where the refractive index (n) is derived from the electric susceptibility (χ) through the equation n² = 1 + χ. The conversation also delves into the implications of phase delay (Δφ) in relation to light propagation and the complex refractive index, particularly in media with attenuation. Participants express confusion over the relationship between phase delay and the speed of light, emphasizing the need for a clearer visual representation of these concepts.

PREREQUISITES
  • Understanding of the Drude dispersion model
  • Familiarity with electric susceptibility (χ) and its role in optics
  • Knowledge of complex refractive index and its application in media with attenuation
  • Basic concepts of phase delay (Δφ) in wave propagation
NEXT STEPS
  • Study the Drude model in detail, focusing on its application to light propagation
  • Explore the relationship between electric susceptibility and refractive index in various materials
  • Investigate the concept of complex refractive index and its implications in optical physics
  • Review Feynman Lectures on Physics, specifically chapter 31, for insights on wave behavior in different media
USEFUL FOR

Physicists, optical engineers, and students of optics seeking to deepen their understanding of light propagation, refractive indices, and the underlying physics of the Drude model.

spareine
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By observing that red is the outermost color of the rainbow it is possible to conclude the refractive index of water, n, is lower for red than for blue. However, why is n lower for red? This seems to be answered by the Drude dispersion model, with a resonance in the ultraviolet region, but I don't get it because by definition n is related to the speed of light, v, whereas the Drude model discusses the electric susceptibility, χ, as a function of the frequency f, without mentioning the speed of light. In the Drude model n is introduced indirectly by the equation n2 = 1+χ. Should the phase lag of the Drude model, Δφ, be used to explain the variation of v? (Like: v(f) is monotonic increasing because Δφ(f) is monotonic increasing, at the left side of the graph)

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spareine said:
by definition n is related to the speed of light, v, whereas the Drude model discusses the electric susceptibility, χ, as a function of the frequency f, without mentioning the speed of light
Isn’t the speed of light related to the susceptibility?
 
Dale said:
Isn’t the speed of light related to the susceptibility?

Apparently my intention is unclear, so let me rephrase. Yes, n2 = 1+χ, but when possible I am interested in a visual image of the mechanics of how the propagating wave is delayed, instead of equations that do not contain the propagating wave. Somehow the Drude model seems to facilitate such a visual image, but the only time parameter provided by the model seems to be the phase delay. I am simply unsure about how to use the phase delay.
 
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Ah, I found an old post on physics forum, "Microscopic reason why refractive index typically increases with frequency?", that uses the Drude model and its phase delay to explain that the refractive index increases with frequency. Still not entirely clear to me, unfortunately (because in that post, 180° should be 90°, I think, and because it says the amplitude has an effect on the speed as well)

Lord Jestocost said:
When regarding the propagation of light in a medium with attenuation, one has to exchange the simple refractive index by a complex refractive index that is directly related to the complex dielectric function. See, for example: …

Thanks! The Drude model uses a complex refractive index as well (n2 = 1+χ, where χ is complex)
 
spareine said:
Ah, I found an old post on physics forum, "Microscopic reason why refractive index typically increases with frequency?", that uses the Drude model and its phase delay to explain that the refractive index increases with frequency. Still not entirely clear to me, unfortunately (because in that post, 180° should be 90°, I think, and because it says the amplitude has an effect on the speed as well)

No, the phase delay is only 90° exactly at resonance. Above resonance it quickly approaches 180° as stated correctly in that old post. Actually, the physics behind the behavior at resonance is not different from a driven antenna or any other periodically driven harmonic oscillator. If you drive it below resonance, the oscillating charge follows the external force exactly and almost instantaneously. At resonance, there is a 90° phase shift, which yields the typical resonance behavior and above resonance you get the 180° phase shift, which is the typical "force that points in the opposite direction of the displacement" scenario.
 
Thanks, in the way you rephrased "slightly above the resonance frequency, the oscillator is 180 degree out of phase", it corresponds better to the Δφ-f graph that I made in post #1 on basis of the exact equations of the Drude model.

I still do not understand the remainder of the old post:
In the case of light, the phase velocity of light will be higher than in vacuum as the emitted wave appears to have "jumped forward" in comparison with the driving light wave. However, the amplitude decreases when the frequency increases so that the phase velocity of the light will decrease, too.
When using a continuous wave, I don't understand why the phase velocity would be higher than in vacuum.
 
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Note that we are talking about really high frequencies here. The region, where the refractive index is below one is usually in the x-ray region at frequencies higher than all bound resonances in the system. Here, the physics is somewhat simple. The incoming light wave will cause the electrons the perform oscillatory motion, which will in turn result in the electrons emitting light again. The light wave behind the material will be the superposition of the initial light field E_0 and the re-emitted light field by the electrons E_s. As you can see from the phase lag of 180°, this means that in the high frequency region, you will get something like E_s= - E_0*k, where k is some complex value that includes plenty of effects: wavelength, density of electrons and so on. If you do the math, you will find that light field behind the material is roughly speaking something like E= E_0 (1-k), which directly translates to a refractive index slightly below one.

If this is still puzzling, have a look at the Feynman lectures, chapter 31 (http://www.feynmanlectures.caltech.edu/I_31.html).
 

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