Talk:Fabry–Pérot interferometer

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

The existing application note for astronomy mentions etalons with bandwidth between 0.2 nm and 1 nm. I believe that should be 0.2 angstroms to 1 angstrom, or 0.02 nm to 0.1 nm. (It also calls the bandwidth FSR, which isn't correct. The bandwidth is the FWHM of a single peak, not the distance between peaks.)  Lastly, the CaK bandwidth numbers are likely wrong too, but I think they're wrong by more than just a 1/10th conversion (I'm guessing something like 1 to 6 angstroms, not 1 to 2 angstroms or 1 to 2 nm). 192.150.10.200 (talk) 15:48, 31 July 2008 (UTC)
 * Uncited material removed.--Srleffler (talk) 05:16, 1 August 2008 (UTC)
 * "In astronomy an etalon is used to select a single frequency of an atomic transition for imaging. The most common is the H-alpha line of the sun. Etalons are available to visual astronomers with bandwidths (FSR) from 0.2 nm to 1 nm. The CaK line from the sun is also commonly imaged with etalons in the 1 nm to 2 nm range."

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)

Coefficient of Finesse
Shouldn't the coeff. of Finesse be $$ F = \frac{4R}$$ instead of $$ F = \frac{4R}$$? 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)

In response to the above: See page 89 of Introduction to Modern Optics by Grant R. Fowles. The formula on the wiki page for coefficient of finesse stands correct as is.

- Current Optics Student at Stony Brook University — Preceding unsigned comment added by 129.49.227.80 (talk) 17:41, 13 May 2013 (UTC)

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, $$\delta$$, between two successive reflections is $$\delta = 2 \pi \frac{d}{\lambda} $$, where $$d$$ is the distance the beam travels between successive reflections and $$\lambda$$ is the wavelength. The way that $$\theta$$ 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. &mdash; Laura Scudder &#9742; 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 $T_1$ equal in magnitude to the amplitude at point c, but which has been retarded in phase by an amount $k_0l_0$ where $k_0=2\pi n_0/\lambda$ 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)

The equation for phase difference in the "Detailed Analysis" section appears to be wrong, as the refractive indices have diappeared.194.81.223.66 (talk) 14:03, 13 November 2008 (UTC)
 * the refractive index is already in the definition of k, so I think it's correct. Han-Kwang (t) 14:10, 13 November 2008 (UTC)

It's easier to draw a line from point b to light ray right before point c inside Etalon. In that way, it's mush easier to do the phase difference calculation. You will have length difference: d/cos(theta)+(d/cos(theta))* cos(2*Theta))=2dcos(Theta), using cos(2Theta)=2cos(Theta)*Cos(Theta)-1. — Preceding unsigned comment added by 65.167.241.4 (talk) 16:42, 1 September 2011 (UTC)

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)

Finesse versus Q?
This definition of finesse given here is different than what I've seen. According to Haus' Waves and Field in Optoelectronics, the finesse is pi*sqrt(R)/(1-R), and the finesse given in this article is actually more like the Q of the cavity. With the definition I've given, the finesse is equal to the ratio of the resonance width divided by the free spectral range, and I think this is usually what people mean when they talk about the finesse of the resonator. Any thoughts, folks? Birge (talk) 21:50, 6 October 2008 (UTC)
 * I think you misread the article. The formula you give does appear there. Note, though, that this form is an approximation that is only valid for R>0.5. The formula
 * $$ \mathcal{F} = \frac{\Delta\lambda}{\delta\lambda}=\frac{\pi}{2 \arcsin(1/\sqrt F)}$$
 * is valid for all R. Or were you confused by the formula
 * $$ F = \frac{4R}?$$
 * The latter is the coefficient of finesse—not the same thing as the finesse itself. Note that the symbol for the coefficient is an italic $$F$$, rather than a cursive $$ \mathcal{F}$$.--Srleffler (talk) 22:56, 6 October 2008 (UTC)

Usage of the resulting beams
I'm a bit confused how the interference patterns are generated. I see how the various reflected/transmitted beams will have phase differences, but how are these beams combined to create the resulting pattern? Is this application-dependent? If so, there's a gap on the Wavelength-division_multiplexing page (which is how I got here) -- it kind of hand waves, saying etalons are used, but getting from there to the combined fiber signal is unclear. 71.197.212.174 (talk) 06:43, 3 November 2008 (UTC)
 * The transmittance varies as a function of wavelength. Send broadband (white) light into an etalon, and the spectrum of the light that passes through will look like on these figures. Do you think this needs to be explained more clearly in the Etalon article?
 * I can't comment on how (de)multiplexers work, which is outside the scope of this article. Han-Kwang (t) 14:06, 3 November 2008 (UTC)


 * Assuming the incident light is not a very narrow beam, the various reflected or transmitted beams physically overlap. They have phase differences, but are coherent with one another, so they interfere producing a pattern that varies with wavlength and with angle of incidence on the etalon. Does this answer your question? I'm not sure I fully understand what you are asking.--Srleffler (talk) 04:07, 4 November 2008 (UTC)

Laser Resonators
I think it should be noted that only lasers using flat cavity mirrors are considered Fabry-Perot resonators. These lasers exhibit interference patterns within their beam profile, unlike the TEM mode patterns seen in lasers with one or two concave mirrors. ( Called "Focal" and Confocal" lasers.) Fabry-Perot resonators are often called "Unstable Laser Cavities" because they allow a great deal of the transverse radiation to "walk" off of the beam path, and mirror alignment must be perfect. This is rather misleading, though, because only the radiation which is in perfect alignment with the cavity is amplified, and the beam created is one of the most stable and coherent of any laser. (ie: My fabry-perot YAG laser, unfocused, will burn a 5mm spot into a piece of black rubber from 10 feet away, and burn the same size spot from 100 feet away. Very little divergence.)

Fortunately, for those who have a Fabry-Perot laser, they already have the best tool for checking their mirror alignment: a Fabry-Perot interferometer, as shown by this simple alignment procedure for a military surplus laser. http://en.wikipedia.org/wiki/Image:SSY1_YAG_Laser_Realignment_Procedure.JPG

A good book on the subject is "Principles of Lasers" by Oratio Svelto. Zaereth (talk) 01:18, 14 November 2008 (UTC)


 * In most cases, lasers with flat mirrors are stable, because for most gain media thermal lensing acts as a positive lens. I agree with you that technically lasers with curved mirrors are not "Fabry-Perot". They are commonly called that, however, mostly because their longitudinal cavity mode pattern matches that of a Fabry-Perot cavity. If you want to get really technical, lasers with flat mirrors are only "Fabry-Perot" if both mirrors have the same reflectivity. A laser with one high reflector and one partial reflector would be better described as having a Gires-Tournois cavity, although I have never seen that term used. —Preceding unsigned comment added by Srleffler (talk • contribs)


 * Thank you for your response. I guess if you want to get really, really technical, a true Fabry-Perot cavity would also need mirrors of the same flatness, (ie: two 1/20th wave mirrors), as well as the rod ends, and all other internal optics too. Most sources I've seen specify that a focal or confocal laser is not a Fabry-Perot cavity, as quoted from this encyclopedia: http://www.rp-photonics.com/fabry_perot_interferometers.html, "Strictly, a Fabry–Pérot interferometer by definition consists of two planar mirrors, but the term is frequently also used for resonators with curved mirrors. From a theoretical viewpoint, plane–plane optical resonators are special in the sense that their resonator modes extend up to the edges of the mirrors and experience some diffraction losses."


 * A Fabry-Perot laser is different in many senses. The beam does not have a gaussian profile. The 5mm spot produced by my laser perfectly matches the 5mm laser rod. The circumference has a sharp edge, and power distribution across the beam profile is even. The beam produced is highly collimated. The beam does not produce TEM mode profiles, but instead, misaligned mirrors produce "optical flat" style interference patterns. The output wavelength is slightly tunable by adjusting the distance between the mirrors. (The exact wavelength produced will be an interger, [I think interger is the correct math word], of the cavity length.) The beams produced are "standing wave".


 * In focal and confocal lasers, the resonating light within the cavity tends to follow a multitude of paths, but something more resembling the path of light in a ring laser, (or actually, something like the paths a couple of retroreflectors would produce), and the beam is focused into a gaussian profile. These beams are not very collimated, and often require some external optics to even create a straight beam. They are far less touchy to misalignments, which do not produce the severe losses in output power, but instead, misalignments tend to break the beam into two or more gaussian profiles, (called the Transverse ElectroMagentic Modes), the most desireable of which being TEM00.Zaereth (talk) 18:09, 14 November 2008 (UTC)


 * Although, I agree that Gires-Tournois does seem to fit better. But I've never seen it used either. I'd have never looked it up if I hadn't seen it mentioned here.Zaereth (talk) 19:01, 14 November 2008 (UTC)


 * Some of your terminology is incorrect and you're using simplified descriptions of some things, but otherwise I mostly agree with what you have written. --Srleffler (talk) 09:32, 15 November 2008 (UTC)


 * That may be. I'm a self-taught laser enthusiast, and, actually, I kind of like simplified descriptions to go along with the more expansive ones. (Pesrsonally, I think technical articles could use a little more of them.) I think one of the most striking differences about plane-plane resonators is the ability to use the interference patterns generated to achieve perfect alignment, not only between the mirrors, but with the mirrors and the rod as well. Not trying to inject my original research into the article, I simply think the "Laser Resonator" subsection of the "Application" section would be a little more accurate if it were amended to read: "•	Laser resonators are often described as Fabry-Pérot resonators. Although this term is often used for lasers with concave mirrors, a laser with two flat mirrors has characteristics that make the term more suitable. For many types of lasers the reflectivity ...", or something along those lines. Does that sound good? Zaereth (talk) 18:39, 17 November 2008 (UTC)
 * From what you've written, it sounds like your Nd:YAG rod laser has an unstable resonator. Nd:YAG works well in an unstable configuration. As you described, you get a big beam that fills the aperture. A true F-P laser's beam does not fill the aperture,despite what you would think. Diffraction losses create a "mode" that remains smaller than the physical aperture. Just curious: how do you pump the rod? Diode pumping? End or side pumping? --Srleffler (talk) 02:53, 18 November 2008 (UTC)
 * Side pumped, with a flashlamp. 45 joule pulse in 100us pumping a 2 inch long rod.Zaereth (talk) 18:08, 18 November 2008 (UTC)

Low Finesse formula
The formula for the finesse is incorrect. To be convinced, just try to calculate the finesse with a low reflectivity, e.g. R=0.04. The answer is complex. The correct formula is:


 * $$ \mathcal{F} = \frac{\Delta\lambda}{\delta\lambda}=\frac{\pi}{2 \arcsin(1/\sqrt{2+F})}$$.

You can check that this formula equals to 2 when R is 0, and gets closer to the formula written in the wiki when R is close to 1.

Besides, the Finesse is rigorously defined in the frequency domain; not in the wavelength domain.

144.204.16.1 (talk) 14:57, 11 May 2011 (UTC)


 * Got a reference for that?


 * The second equality in your expression is clearly false given the definitions in the article, because the full width at half maximum, $$\delta\lambda$$, ceases to be well-defined for low R. At R=0.04, the transmission never dips below 50% so it is not surprising that $${\Delta\lambda}/{\delta\lambda}$$ has no real value. Your source is clearly defining "Finesse" in a different way, incompatible with the definition used in this article. It might be a better definition, but some care would be required to incorporate it into the article; there is more going on than just an error in the formula.--Srleffler (talk) 16:30, 11 May 2011 (UTC)

Index vs group index
According to Encyclopedia of Laser Physics and Technology, the formula for free-spectral range should use the group index instead of the normal (phase) index. But I couldn't find a discussion of dispersion in reference books. I tried to derive it:

Assume normal incidence for simplicity. Then the optical path difference is $$L=2n\ell$$. The peak-transmission criterion is $$kL=$$integer. [I'm using k=1/wavelength, not 2pi/wavelength]. Normally that means $$\Delta k = 1/L$$. But L is wavelength-dependent too. The zeroth-order approximation neglects this: $$(\Delta k)_0 = 1/L_0$$. The first-order approximation of the change in L from one peak to the next is:
 * $$\Delta L = (\partial L/\partial n)(\partial n/\partial k)(\Delta k)_0 =(1/n)(\partial n/\partial k)$$

Then the corrected increment of k is:
 * $$\Delta k = (\Delta k)_0 + \frac{\partial \Delta k}{\partial L}\Delta L = 1/L + (-1/L^2)(1/n)(\partial n/\partial k) = \frac{1}{L}-\frac{\partial n / \partial k}{nL^2}$$

Supposedly this equals $$1/(2n_g \ell)$$. But obviously not -- for one thing, in my formula, $$\Delta k$$ is not inversely proportional to $$\ell$$. Anyone know what's going on here? Is Encyclopedia of Laser Physics and Technology totally wrong? Am I making a mistake or using the wrong approach? Is there a reference that discusses this?

[FYI: For glass, the difference between the Encyclopedia of Laser Physics and Technology formula and the wikipedia formula is about 1%. The difference between my formula and the wikipedia formula is about (1 nm)/L, e.g. 3×10-7 for 1mm thick.]

Thanks in advance!! :-) --Steve (talk) 21:34, 29 August 2011 (UTC)

what about polarization
Hi, just a quick comment: as it is, I see that the discussion is valid for normal incidence, but is it also true for a variable angle of incidence? The way the finesse is defined, it should be dependent on whether the field is p or s polarized. — Preceding unsigned comment added by Anglyan (talk • contribs) 21:02, 3 October 2011 (UTC)
 * The reflectivity R is a function of angle and polarization. See Fresnel equations.--Srleffler (talk) 01:17, 4 October 2011 (UTC)
 * Just so. Shouldn't this be mentioned somewhere in the text? The normal incidence case avoids this problem since both p and s polarizations provide the same result. My concern is that figures represent the oblique incidence case, giving the impression that these equations work for oblique incidence as well, whereas they only apply to purely p or s polarized light. Simply stating: "In the case of oblique incidence, the finesse will depend on the polarization state of the light" would clarify things for the general reader. — Preceding unsigned comment added by 24.11.236.211 (talk) 05:42, 5 October 2011 (UTC)
 * Sounds good. Why don't you go ahead and add that to the article?--Srleffler (talk) 00:42, 6 October 2011 (UTC)

Bad image.
I just noticed that the image at right, which appears in the article, is flawed. It fails to show the reflection from the point where the incident beam strikes the surface. Without including that reflection, one can never analyze the total reflection and get the right result. --Srleffler (talk) 01:24, 4 October 2011 (UTC)


 * I never did like the first "derivation", it seemed to present results via a discussion that was opaque to me. This is why I added the section "detailed analysis". In this section, the incident beam has amplitude 1. When it hits the left face, the reflected beam has amplitude &radic;R, the transmitted beam &radic;T. When the transmitted beam hits the right face, the reflected beam has amplitude &radic;T&radic;R and the transmitted beam has amplitude &radic;T&radic;T=T. etc. etc. The reflected beam is not shown, and probably should be. As for the first "derivation" I don't want to go there. PAR (talk) 04:22, 4 October 2011 (UTC)
 * What confused you is that it is not a derivation, just a presentation of the key results. Wikipedia is not a textbook; derivations are often not appropriate material for WP articles.--Srleffler (talk) 05:14, 4 October 2011 (UTC)

Replacement is available. -Moneco (talk) 10:08, 23 May 2013 (UTC)
 * You're a little late with the correction, but thanks! Stigmatella aurantiaca (talk) 05:01, 24 May 2013 (UTC)

Reflection -> wrong (I think)
Hi,

The second picture in the "Theory"-Section shows two vertical axis: Transmission and reflection, whereas the reflection axis is the inversion of the transmission axis. The calculation for the frequency dependent transmission coefficient is shown in the "Detailed analysis"-section and I can reproduce this calculation and the picture for the transmission. But I can't reproduce the picture for the reflection and also I doubt that this sentence "In the absence of absorption, the reflectance of the etalon Re is the complement of the transmittance, such that Te + Re = 1" is right. I think one gets a similar picture for the reflection not the inversion of the transmission.

If one has no absorption and one interface $$R+T=1$$ (corresponds to intensities) is true, but here we have 2 interfaces. The reflected rays $$R_1$$ and $$R_2$$ have the same phase difference $$\delta$$ as the transmitted rays $$T_1$$ and $$T_2$$. Also $$R_1$$ has the same phase difference to the zeroth reflection (not shown) as $$R_2$$ to $$R_1$$. I did the detailed calculation and found the same picture for the total reflection like for the transmission. But I can have miscalculated something. So can please someone have a look on the reflection of the fabry-perot-interferometer? Why does the reflection show minima where the transmission is maximal. I think its not true. For these frequencies we have for transmission AND reflection destructive interference. The problem is "symmetrical". --Svebert (talk) 20:12, 15 October 2011 (UTC)


 * You have miscalculated somehow. R + T = 1, because energy is conserved. In steady state, the power that is transmitted plus the power that is reflected has to add up to the incident power, and this has to hold for each frequency of light separately.


 * It has been a while since I have looked at the derivation, but my recollection is that when done properly the zeroth order reflection (not shown in the diagram, unfortunately) has the opposite phase to all the other reflections, so reflections R1, R2, ... all add up to destructively interfere with R0. I could be misremembering, however.--Srleffler (talk) 02:51, 16 October 2011 (UTC)


 * What you might have neglected, is that the the incident beam's reflection has no phase shift, while all the subsequent reflections have phase shifts of &pi;. In general, reflected light shifts phase by &pi; when the medium from which it is reflecting has lower refractive index than the incident medium.--Srleffler (talk) 02:58, 16 October 2011 (UTC)
 * Hey! Thanks for the answer! I have checked my calculations and now I get the mentioned result with $$R_e=1-T_e$$. I either lost some minus signs or R-summands in my first attempt. Now everything is fine.
 * But I disagree to the explanation for $$R_e=1-T_e$$ because of energy conservation. Of course we have energy conservation but with destructive interference you can "destroy" energy (you don't really destroy energy but you can make the energy hidden for any measurement). E.g. you have two identical wave senders and you power them both with the same power $$P$$. The first sender sits at $$x=x_1=0$$ and the second at $$x=x_2>x_1$$ and both senders will radiate a plane wave in +x direction with the same frequency (and same $$k$$). If you assign an appropriate phase shift to the second wave, they will interfere destructively (here you have to choose $$k x_1 +\varphi=\pi$$ to satisfy the condition for destructive interference). Both senders you have to power, so you have put the energy $$2P\cdot t$$ into your system after the time $$t$$. If you measure the radiated power at any $$x>x_2$$ you will measure $$0$$. The power is lost because of destructive interference. Of course you can fix the phase of the second wave and with $$k x_0 +\varphi=0$$ you will meet the conservation of energy again.
 * What I wanted to say is: It is not clear ad hoc that no effect like the above mentioned one "violates" the conservation of energy in the fabry-perot-reflection-system.
 * Of course after calculating $$R_e$$ one sees that $$R_e+T_e=1$$ holds but in the article the explanation is the other way round: $$\mbox{energy conservation}\Rightarrow R_e+T_e=1$$. But the truth is: $$\mbox{One calculates }\,R_e+T_e=1\Rightarrow\mbox{energy conservation holds}\rightarrow $$--Svebert (talk) 15:12, 16 October 2011 (UTC)


 * Your model of constructive and destructive interference is incorrect. If two sources are out of phase by 180 degrees, they emit no net power, and so no power input is needed, other than that to make up for inevitable heating losses. Once the two are in phase, at least 2P must be put in to account for the 2P being radiated. PAR (talk) 21:40, 16 October 2011 (UTC)


 * Interference never violates conservation of energy. The input energy always comes out somewhere, or else input of energy to the system is suppressed as in this case. In most physical cases, constructive interference in one region of space compensates for destructive interference elsewhere, such that energy is conserved. --Srleffler (talk) 23:08, 16 October 2011 (UTC)
 * Ok, then tell me where is the error in my simple example? I mean, I have to put 2P into the system to measure $$0\cdot P$$ at $$x>x_2$$. There is no energy flow in any other direction unless +x.
 * Conservation of energy holds in a closed system. Now My closed system is a small box around the sender 2, excluding sender 1. The energy balance is the following: The energy flow into the system is: Energyflow of the plane wave from sender 1 plus (e.g.) electrical power supply for sender 2. The energy flow out of the system is nothing. But conservation of energy states that the same amount of energy flows out of the system as comes into the system. Obviously the conservation of energy is somehow violated. Of course you can fix the phase to achieve energy conservation.
 * @PAR why is my model incorrect? It is not true that my system doesn't need input power. If I cut the electricity supply of sender 2 my system doesn't work anymore, to achieve destructive interference in my case I have to put energy into my sender 2. — Preceding unsigned comment added by Svebert (talk • contribs) 10:03, 17 October 2011 (UTC)
 * The conditions you have in mind are unphysical. If sender 1 and sender 2 are resonant with each other in the way you describe, electrical power flows out of sender 2, not into it. An ideal lossless source is also an ideal, lossless detector. If you drive power into sender 2 instead of allowing power to flow out of it, its output amplitude will be higher than that of sender 1 and there will be net emission of radiation from your system.--Srleffler (talk) 17:43, 17 October 2011 (UTC)
 * I don't understand this argumentation... Since when depends the power onto the phase of the emitted wave? Sender 1 doesn't know anything of sender 2. So if I turn off sender 2, then sender 1 gets input power p and I can measure this energy flow as plane wave anywhere on the right hand side of the sender 1. Also the other way around, if i switch off sender 1 and turn on sender 2. Why should there something be different in the powering of the senders if I turn them both on? They are just senders, no detectors. I assume that both senders accelerate the electrons in a thin rod up and down. And I assume that the wave from sender 2 doesn't affect the electrons in sender 1. But I assume that the waves produced by the senders interfere. Therefore I put 2P into the system and 0P gets out.
 * So, I don't assume any resonance of the detectors. I don't care whether this pathological example could be working in reality or not. I just want to say that it is not obvious that interference is energy conservative.
 * Assuming the electrons in sender 2 are affected by the wave from sender 1, than in the "destructive interference case" the electrons are pushed for instance up by the wave from sender 1 but dragged down by the power of sender 2. Therefore they come to rest. But I have to effort energy to de-accelerate the electrons.
 * In your "resonance-case" the sender 2 absorbs all energy from wave 1, therefore the electrons have to oscillate in phase with the wave. If I switch off the sender 1, the electrons doesn't oscillate anymore in sender 2 (after some time: x2/c). Now I don't have any wave anymore on the right hand side of sender 2. Therefore your "resonance-scenario" is not the same as the scenario I mentioned. When I switched sender 1 off I measure the energy coming from sender 2 at x>x2.--Svebert (talk) 11:27, 18 October 2011 (UTC)


 * I agree that it isn't obvious that interference is energy conservative, but it is true. I tried to show you why your example does not contradict this, but you do not seem to have understood my explanation.


 * Let me try to run through this again, using similar terminology to your last reply: Sender 1 does not know anything of sender 2, but sender 2 does know of sender 1; it is affected by the waves emitted by the latter. If sender 1 is off, the driving circuitry causes the electrons in sender 2 to oscillate so as to produce a wave with power P, 180° out of phase with the wave sender 1 would emit if it were on. Now, to make your example work you have to assume that the driving circuitry will maintain this same electron oscillation in sender 2 regardless whether sender 1 is on or off. I trust you won't have a problem with that assumption. Now, consider the case when sender 1 is on: the wave it emits has the same power as the wave that would be emitted by sender 2 on its own, but it is 180° out of phase. This means that the wave from sender 1 does work on the electrons in sender 2. If the driving circuitry is set up to ensure that the oscillations in sender 2 do not change, it can only achieve that by allowing energy to flow out of sender 2 into the driving circuitry. If everything is lossless, the electric energy that flows out of sender 2 equals that which flowed into sender 1, and energy is conserved. If, instead, the driving circuitry continues to put electrical power P into sender 2, the electrons in sender 2 will have the wrong amplitude of oscillation to produce the wave that cancels that from sender 1. --Srleffler (talk) 04:32, 19 October 2011 (UTC)


 * Here is a kind of artificial example which is mechanical. Suppose you have a swing set, the kind you find in a child's playground. Everybody knows its harder to swing when the frame is not anchored but if you get a friend to swing with you out of phase, its much easier. Suppose the frame of the swing set is not anchored to the ground, but is instead attached to some ratcheting mechanism which lifts a weight. Rocking the frame and lifting the weight is the analog of emitting radiation. If a child sits on a swing and starts swinging, they will have to exert enough power to not only overcome friction, but also the energy needed to lift the weight, since their swinging will tend to rock the frame. Call that extra energy P. If two children start swinging in phase, the weight will be lifted twice as fast, and the total power output of the two kids will be 2P plus twice the friction. If they swing 180 degrees out of phase, the frame will not rock back and forth, and the weight will not be lifted (no power radiated). The power that each kid will need to output in order to obtain the same amplitude of swing will drop to zero, plus the small amount of power needed to overcome friction. If they input the same power as when they were in phase, the amplitude of their swings will greatly increase, but still the weight will not be lifted (i.e. no power radiated) and all of that power will go to overcoming the increased friction at that higher amplitude. PAR (talk) 09:30, 19 October 2011 (UTC)

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Incorrect expression in part 1. of Theory section
The factor 4 in the denominator of the last equation of part 1. of the Theory section is incorrect, as can be easily seen by inserting the expression from the line above:

$$ \gamma_{q,L} (\nu) = \frac{\pi}{2} \Delta \nu_c \tilde \gamma_q (\nu) = \frac{\pi}{2} \Delta \nu_c \frac{1}{\pi} \frac{\Delta \nu_c / 2}{(\Delta \nu_c / 2)^2 + (\nu - \nu_q)^2} = \frac{(\Delta \nu_c / 2)^2}{(\Delta \nu_c / 2)^2 + (\nu - \nu_q)^2}. $$

The same error is present in equation (17) of the citation given for this section (Opt. Express 24, 16366-16389 (2016)). 130.183.91.36 (talk) 12:42, 15 October 2018 (UTC)

Answer by Pollnau: I saw this entry only today. In fact, there is no error in equation (17) of Opt. Express 24, 16366-16389 (2016). Both equations are equivalent, as one can easily see by multiplying the half-width-at-half-maximum equation by 4/4 to obtain the full-width-at-half-maximum equation:

$$ \frac{(\Delta \nu_c / 2)^2}{(\Delta \nu_c / 2)^2 + (\nu - \nu_q)^2} = \frac{4(\Delta \nu_c / 2)^2}{4(\Delta \nu_c / 2)^2 + 4 (\nu - \nu_q)^2} = \frac{(\Delta \nu_c)^2}{(\Delta \nu_c)^2 + 4 (\nu - \nu_q)^2} $$

I will add the original equation 17 to part 1.

Description of the Fabry-Perot resonator in wavelength space: free spectral range (FSR) of the etalon, Δλ
Free spectral range is slowly varying. For a limited band, FSR can be considered fixed i.e. FSR is approximately invariant over many FSRs of a practical resonator.

I do not understand therefore the origin of the cosine in the denominator. I suggest that the approximation to the right would be correct if the cosine were deleted. I cannot give a correct expression for the exact equation to the left, but do not believe it should contain a cosine. Could somebody please either explain or revise? — Preceding unsigned comment added by Oh, I'm just here for (talk • contribs) 11:35, 26 February 2021 (UTC)


 * The free spectral range varies with the angle of incidence of light on the etalon, hence the $$\cos \theta$$.--Srleffler (talk) 02:53, 1 March 2021 (UTC)


 * Thanks Srleffler, that is obvious now! Apologies Oh, I&#39;m just here for (talk) 14:56, 6 April 2021 (UTC)