File talk:K-FADOF-spectrum.jpg

Conversion from Wavelength to Frequency and Wave Number

I fixed a problem with the display of the wave-number formula by using a true subscript (reviously my browser rendered it "...cm unintelligiblesquiggle 1"), but formulae have more serious problems. If the frequency were 50 GHz/angstrom, then a wavelength of 1 angstrom (100 pm) would correspond to 50 GHz, and a wavelength of 200 pm would correspond to 100 GHz, etc., which of course makes no sense. I think the author perhaps intended to say that the frequency (in GHz) is 50 divided by the wavelength (in Å), but (and here I am hampered by my lack of knowledge) that seems to conflict with the wavelength formula :$\lambda ={\frac {v}{f}}$, which in the case of electromagnetic waves translates to :$f={\frac {c}{\lambda }}={\frac {0.3Gm/sec}{\lambda }}$, or 0.3 exahertz divided by the wavelength in angstroms, if I have calculated this correctly. A similar calculation would lead to the wave number being 1 cm⁻¹ divided by the wavelength in cm, or 10⁻⁸ kayser divided by the wavelength in angstroms. What am I missing? Peter Chastain (talk) 08:48, 13 May 2009 (UTC)

Also, the discussion mixes nanometers and Å. Can we standardize on nm? Peter Chastain (talk) 08:48, 13 May 2009 (UTC)

Hmm. I do not think your interpretation of the "50 GHz/angstrom" is correct. As I interpret it, this formula does not say that the frequency of the light is 50GHz/angstrom. Rather, it says that the scale on the abscissa may be converted using that relation. So the .1 Å the .2 Å markings could become 5Ghz, 10Ghz ticks, respectively. Because we're talking per angstrom, there is simple constant of proportionality relating the frequency and wavelength of the light. This interpretation does make sense numerically, I think. As the Wikipedia article Atomic line filter says, "The passband of a typical Faraday filter may be a few GHz.".

All of that said, your calculation is still wrong, because the light is not travelling in a vacuum (but a hot vapor), so its speed is reduced (however little) from c. Just FYI ;).

I do not think we can standardise the discussion to nanometers because the graphic itself uses angstroms. We should standardise to that unit, if anything. Or we could leave it: the literature itself mixes the two systems, and they are extremely easy to use side-by-side.

I hope the above is both correct and helpful. If you respond here, shoot me another e-mail, because I don't check my Commons watchlist very frequently. If you have any other questions, or are not satisfied with the above response, please do not hesitate to ask again! -- Rmrfstar (talk) 23:22, 23 May 2009 (UTC)