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Dispersion question |
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| Feb2-13, 10:34 AM | #1 |
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Dispersion question
From the elementary texts, dispersion is the phenomenon where the refractive index of a medium depends on the wavelength of electromagnetic radiation through it. From what I've read, it is the wavelength of the radiation in vacuum. Also, it is said that the refractive index increases with decreasing wavelength, and increasing frequency.
My question is simple, by wavelength, are they referring to the 'new' wavelength as the light passes through the material or are they referring to the wavelength in vacuum? About frequency, I don't really have much question about it, at least if I believe that the frequency stays constant even after going through different mediums. If we go by the equation [tex]\lambda=\frac{\lambda _o}{n}[/tex] I can see that they must be referring to the 'new' wavelength, but this is still a bit baffling for me. I'd like to ask more but maybe after someone replies on this. I probably just missed something. Edit: I'm starting to get the hunch that I'm might be using the wrong model/mathematical expression for this. |
| Feb2-13, 11:15 AM | #2 |
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If we measure the wavelength of a wave before and after entering a medium we will find that it has changed. The amount of change lets us calculate n, the refractive index. Let's say it gives us 1.1 for n. Now, we shoot a shorter wavelength through. We measure again and find that n now equal 1.10002. The refractive index is different for different wavelengths. Not by much, but it is. The greater the difference in your refractive index between different wavelengths the greater the dispersion.
Your equation is simply the relationship between the wavelength before entering the medium, after entering the medium, and the refractive index of the medium. |
| Feb2-13, 11:18 AM | #3 |
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Looking back, I think the texts are actually referring to the wavelength in vacuum. The expression above seems to be valid at least if we are concerned with the wavelength of the light as it passes through the medium. I'm now half convinced that we cannot conclude anything about the dependence of the refractive index in the frequency of the specific radiation from the expression above.
I actually tried to express the the vacuum wavelength in terms of speed of light and its frequency, but I'm getting contradicting results. |
| Feb2-13, 11:26 AM | #4 |
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Dispersion question
You can use the expression above to find the refractive index of a material for different wavelengths. That will then allow you to find the dispersion.
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| Feb2-13, 11:29 AM | #5 |
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Thanks anyway, I think I got it, the expression is valid only for a specific frequency of a wave. Now, I guess my real question is, why does the refractive index increase with increasing frequency of the wave (and consequently with decreasing wavelength of it in vacuum)? |
| Feb2-13, 11:31 AM | #6 |
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| Feb2-13, 11:40 AM | #7 |
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