# What's Wrong with my Experiment?

by FeDeX_LaTeX
Tags: dispersion, refractive index
 P: 422 Hello; For my coursework I am studying how changing the wavelength of a beam of light affects the index of refraction. According to the results I have collected, wavelength and refractive index are inversely proportional. However, my physics teacher says that this should not be the case. Assuming my physics teacher is correct, why do my experiments show this? I can't find any sources which tell me the answer, but if I am correct in that they are supposed to be inversely proportional, is there a general formula linking wavelength with the index of refraction? Thanks.
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 For visible light, most transparent materials (e.g., glasses) have: $$1 < n(\lambda_{\rm red}) < n(\lambda_{\rm yellow}) < n(\lambda_{\rm blue})\ ,$$ or alternatively: $$\frac{{\rm d}n}{{\rm d}\lambda} < 0,$$ that is, refractive index n decreases with increasing wavelength λ. In this case, the medium is said to have normal dispersion. Whereas, if the index increases with increasing wavelength the medium has anomalous dispersion.
 In general, the refractive index is some function of the frequency f of the light, thus n = n(f), or alternatively, with respect to the wave's wavelength n = n(λ). The wavelength dependence of a material's refractive index is usually quantified by an empirical formula, the Cauchy or Sellmeier equations.
http://en.wikipedia.org/w/index.php?...y%27s_equation
http://en.wikipedia.org/w/index.php?...meier_equation
 P: 422 Thanks. Is it also correct to say that, because; $$v = f\lambda$$ and $$n = \frac {c}{v}$$ Therefore; $$n = \frac {c}{f\lambda}$$ So n is inversely proportional to the wavelength. This is correct too, yes?
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## What's Wrong with my Experiment?

 Quote by FeDeX_LaTeX Hello; For my coursework I am studying how changing the wavelength of a beam of light affects the index of refraction.
That's kind of backwards: the index of refraction depends on the material and is defined by the response of the material to an electromagnetic field with frequency $\nu$, or $n = n( \nu )$

 Quote by FeDeX_LaTeX Is it also correct to say that, because; $$v = f\lambda$$ and $$n = \frac {c}{v}$$ Therefore; $$n = \frac {c}{f\lambda}$$ So n is inversely proportional to the wavelength. This is correct too, yes?
Not really: it's true that c = n v, but because of the frequency dependence of n, $v( \nu ) = \nu \lambda (\nu)$

I wrote this out in terms of the frequency dependence because the frequency of a photon is the same regardless of the medium it travels through (frequency = energy, wavelength = momentum), which is not the case for wavelength.

The index of refraction has a complicated structure given by the Kramers=Kronig relations: the real part of the susceptibility (absorption) has local maxima near resonances and the imaginary component (refractive index) exhibits anomalous dispersion (n increases with increasing wavelength).
 Sci Advisor P: 2,424 I'm assuming you are working with visible spectrum. If so, you are only probing a relatively narrow window of material's transparency range. And within any small enough range of wavelengths, there is a linear dependence between index of refraction and frequency. This is sufficient to find a fit for your data to an inverse relation. So yes, your experiment will agree with hypothesis that index of refraction is inversely proportional to the wavelength, but this is because you simply cannot probe sufficiently wide range of frequencies to see any "interesting" regions. I'm not really sure exactly what your teacher's complaint is. What you report is consistent with the physics of the problem within the studied region. Maybe you can ask him what he expected to see that's different.
 P: 4,667 Andy Resnick is correct. The index of refraction for most glasses is a complicated function of wavelength, but it is certainly not (and should not be) inversely proportional to wavelength. There is no theory that predicts this inverse relationship. See thumbnail. The variation of index of refraction in the visible range is controlled by the Kramers Kronig relations, and by strong absorption in the UV (short wavelength) region. If there were no absorption in the UV, then the index of refraction would be constant, independent of wavelength, in the visible range. The Kramers Kronig relations link the real part and the imaginary part of the refractive index through causality. Bob S Attached Thumbnails
 P: 2,258 can this relationship between freq and refractive index be related to the response of an electronic circuit consisting of an inductor and a capacitor to varying frequencies?
 PF Patron P: 1,890 Hello Fedex.You may find it useful to google "Cauchys dispersion formula" and the "Sellmeier equation".
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 Quote by granpa can this relationship between freq and refractive index be related to the response of an electronic circuit consisting of an inductor and a capacitor to varying frequencies?
Correct. See Bode's book "Network Analysis and Feedback Amplifier Design". The "Real Part Sufficiency" theorem is a direct consequence of the Kramers Kronig relations. There is a full page of dispersion relations on about page 330 linking the real and reactive impedances of passive circuits.

Bob S
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 Quote by K^2 I'm assuming you are working with visible spectrum. If so, you are only probing a relatively narrow window of material's transparency range. And within any small enough range of wavelengths, there is a linear dependence between index of refraction and frequency. This is sufficient to find a fit for your data to an inverse relation.
There's a difference between "linear" and "the graph lies on a straight line"

linear means there exists a relation n = c*f. This means that the index of refaction is twice as large for blue as for red light. Such large differences in refractive index do not occur for any material, and I can't see how these could ever have been measured without some
gross error.
The graph of refractive index vs. frequency could certainly have looked like a straight line
however, and any high school/undergraduate experiments are likely not accurate enough to see the difference
 P: 422 Hello; Thank you for all of your replies. I am unable to reply on a frequent basis as I still do not have my own access to the internet. In the written work for my investigation I have stated that only testing the range of visible wavelengths in the EM spectrum will of course not be a wide enough range to deduce a conclusion that encompasses all wavelengths. The material I am using is borosilicate glass block (BK7). This material exhibits normal dispersion according to my results... What is meant by the 'imaginary part' of the refractive index? Are you saying that the index of refraction can be complex as well as real? As for my physics teacher, he says that he expected me to not see any correlation (though I am getting a very clear relationship of inverse proportion). I have researched Cauchy's dispersion formula and Sellmeier's equation; they have been very helpful, thanks. Bob S, when you say that the index of refraction for most glasses is a 'complicated function of wavelength'; could you elaborate, please? Regarding Sellmeier's equation - once I have simplified the right-hand side, I solve for n by square rooting, yes? So I can square root $$n^{2}(\lambda)$$ to get $$n(\lambda)$$ ignoring the +/- value? I have looked at the thumbnail, and doesn't the graph show an inversely proportional relationship between the index of refraction and wavelength in the region of visible wavelengths? Thanks.
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 Quote by FeDeX_LaTeX Bob S, when you say that the index of refraction for most glasses is a 'complicated function of wavelength'; could you elaborate, please?
Look at the two plots in the graph at

http://www.phy.duke.edu/~rgb/Class/p...19/node50.html

The Re(ε/ε0) plot represents refraction index plotted vs. frequency (not wavelength). The region to the left of big resonance is normal dispersion in visible light for glasses.

The Im(ε/ε0) plot represents absorption. It increases dramatically at resonance.

Re(ε/ε0) and Im(ε/ε0) are coupled by the Kramers Kronig relations.

Use the Previous button in html twice to go to derivation of complex index of refraction.

 I have looked at the thumbnail, and doesn't the graph show an inversely proportional relationship between the index of refraction and wavelength in the region of visible wavelengths?
Look at the two plots in above html. The index of refraction increases at higher frequencies (increases at lower wavelengths). But is not a hyperbola (inverse proportional relationship between refraction index and wavelength).

Bob S
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 Quote by FeDeX_LaTeX I have looked at the thumbnail, and doesn't the graph show an inversely proportional relationship between the index of refraction and wavelength in the region of visible wavelengths?
No. Inversely proportional is when the graph of the index of refraction vs. (1/wavelength) is a straight line throught the origin.
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 Quote by willem2 No. Inversely proportional is when the graph of the index of refraction vs. (1/wavelength) is a straight line throught the origin.
If for example, the index of refraction n increases at shorter wavelengths λ, and

n = const/λ, this is inversely proportional.

This can be rewritten as

nλ = const

which is a hyperbola.

But this has no relation to the physics of dispersion.

Bob S