Talk:Fabry-Pérot interferometer

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[edit] Applications

I'd like to see discussion of applications. NuclearWinner 23:46, 12 April 2006 (UTC)

Me too - how about something on the FP diode laser? Royzee 21:28, 21 February 2007 (UTC) 20th Feb 07

[edit] Incorrect graph?

Isn't the graph of transmission against wavelength against slightly incorrect. Shouldn't it be F = 1 for the blue line as the full-width-half-maximum equals the distance between peaks. Hence delta-lamda equals 2*pi and hence F=1. Or am I completely incorrect? 129.67.53.158 19:39, 23 April 2007 (UTC)

I don't see your point. The full width at half maximum is clearly about half the distance between the peaks, as it should be for a finesse of 2. (Don't use "F" for finesse. F is the symbol for the coefficient of finesse, which is a different quantity. Finesse is denoted by a curly \mathcal{F}.)--Srleffler 02:57, 1 October 2007 (UTC)

[edit] Coefficient of Finesse

Shouldn't the coeff. of Finesse be F = \frac{4R}{{(1-R^2)^2}} instead of F = \frac{4R}{{(1-R)^2}}? Such as http://scienceworld.wolfram.com/physics/FinesseCoefficient.html --Emplynx 18:43, 28 April 2007 (UTC)

The Wolfram page doesn't give quite the formula you propose. Their formula is different from ours because their r (note the lowercase) is the amplitude coefficient of reflectivity, while our R is the intensity coefficient.--Srleffler 03:03, 1 October 2007 (UTC)

[edit] Phase difference

Maybe I'm being dumb, but in the first equation in this article (which gives the phase difference between successive reflections), shouldn't the cosine factor be in the denominator? The way I see it, the phase difference, δ, between two successive reflections is \delta = 2 \pi \frac{d}{\lambda} , where d is the distance the beam travels between successive reflections and λ is the wavelength. The way that θ is defined in the figure, d = \frac{2l}{\cos{\theta}}, so \delta = \left( \frac{2\pi}{\lambda}\right) \left(\frac{2 l}{\cos{\theta}}\right).

That would make sense and be consistent with the \frac{2 k l}{\cos \theta} in the detailed analysis further down. — Laura Scudder ☎ 18:39, 10 July 2007 (UTC)
If you're dumb then I'm dumb, because this is the fine point that always catches me. From the paragraph above the phase change expression in the "Detailed analysis" section:

The total amplitude of both beams will be the sum of the amplitudes of the two beams measured along a line perpendicular to the direction of the beam(emphasis mine). We therefore add the amplitude at point b to an amplitude T1 equal in magnitude to the amplitude at point c, but which has been retarded in phase by an amount k0l0 where k0 = 2πn0 / λ is the wave number outside of the etalon.

In other words, if you measured the phase difference at the faces then, yes, the phase difference would be
\delta = 2kl/\cos(\theta)\,,
but thats not the phase difference between the two combined beams. The phase difference between two successive beams exiting the interferometer must be measured along a plane perpendicular to the direction of propagation, and it is this phase which is equal to
\delta = 2kl/\cos(\theta)-k_0l_0=2kl\,\cos(\theta).
PAR 11:34, 11 August 2007 (UTC)
Maybe you should add a remark in the article to prevent future confusion. Han-Kwang 12:23, 11 August 2007 (UTC)

[edit] Perot or Pérot?

There seems to be some uncertainty about whether the coinventor's name is Perot or Pérot. I started a discussion on this at Talk:Alfred Pérot. We need a reference that is agreed to be authoritative, that establishes the correct spelling of his name. --Srleffler 02:44, 1 October 2007 (UTC)

"Correctness" is the wrong criterion. And we are concerned here with the correct spelling of "Fabry-Perot interferometer", not so much with his name. In other words it is entirely possible that the correct spelling here could differ from the spelling in the article about him. Gene Nygaard 10:36, 4 October 2007 (UTC)
I agree, but let's keep the discussion together at Talk:Alfred Pérot.--Srleffler 17:32, 4 October 2007 (UTC)