Talk:Fresnel equations

message: All references at my disposal use z not s in all forms.
--4.232.0.63 16:52, 7 August 2006 (UTC)
replies: text: "Polarise" is an accepted British Commonwealth spelling. Wikipedia's Manual of Style prescribes that all varieties of English are accepted, although they should not be mixed in a single article. If a particular subject is not directly tied to a particular country, it stays in whichever variety of English was first used in that article.--Srleffler 00:32, 8 August 2006 (UTC)
thread_title:  spelling polarise or polarize 

message: Would it be okay to redirect a Fresnel's law page to this page? —The preceding unsigned comment was added by Wk muriithi (talk • contribs) 13:03, 24 January 2007 (UTC)
replies: text: Sounds good to me. Dicklyon 04:18, 25 January 2007 (UTC)
thread_title:  Redirection 

message: The article says "For materials which absorb light, like metals and semiconductors, n is complex, and Rp does not generally go to zero." But don't clean metallic surfaces essentially not absorb light? Is this right? —Ben FrantzDale (talk) 21:56, 30 April 2008 (UTC)
replies: text: It is correct: metals are absorptive. A lot of light is reflected at the surface, but what gets into the metal is absorbed over a very short propagation distance. I think silver and aluminum reflect 90–95% in the visible. The rest is absorbed (assuming the metal is thick enough.)--Srleffler (talk) 23:22, 30 April 2008 (UTC)
replies: text:  Interesting. Obviously copper absorbs in the blue end of the visible spectrum. This leads me to another question: how does mirror reflection work at an atomic scale? That is, when the continuum assumption breaks down, materials can't have a smooth surface yet light behaves as though surfaces are smooth. If it didn't, one couldn't make a lens and a polished crystal, which has atomic-scale star steps between crystal planes, wouldn't reflect like a mirror but would instead reflect like, well, stair steps. —Ben FrantzDale (talk) 01:16, 1 May 2008 (UTC)
replies: text: Atoms are about 0.1 nm wide. Atomic spacing in solids is maybe a few tenths of a nanometer. Visible light has a wavelength of ~400–700 nm. The light doesn't see roughness below about the 100 nm scale. X-rays have wavelengths comparable to atomic radii and/or lattice spacings, which is why reflecting x-rays off of crystalline solids produces x-ray diffraction.--Srleffler (talk) 01:58, 1 May 2008 (UTC)
replies: text:  Thanks. That's basically what I assumed, but have never had it stated explicitly. Do you know where I would find the quantum-mechanical equations describing light interacting like that with a "bumpy" surface? It seems like one aught to be able to find an "effective surface geometry" for nanoscale-rough surfaces such that physics "acts" like that's where the surface is. —Ben FrantzDale (talk) 11:34, 12 May 2008 (UTC)
thread_title:  Metals absorb light? 

message: I've heard the term "optical coupling" and "wetting out" to describe what happens when optical films come in contact. I assume this is the transition from a film-air-film sandwich to a film-film sandwich as the air thickness starts to get well below the wavelength of the light. Could someone point me to more information on that? 155.212.242.34 (talk) 14:16, 19 May 2008 (UTC)
replies: text: It may be related not to the air film being less than a wavelength, but rather to the gap becoming small enough that intermolecular forces (van der Wahls??) pull the surfaces together. This effect is used in optical contacting to bond optical surfaces together. The surfaces are rubbed together with a fluid in between. Surface tension/capillary forces in the fluid pull the surfaces close enough together that van der Wahls or something similar takes over. It can be used to make both temporary and permanent bonds, but I don't know much more about it than that.--Srleffler (talk) 17:10, 22 July 2008 (UTC)
replies: text:  I think the original poster is onto something with regard to this being a wave-like effect. Fyneman has a great description of how light acts in QED in which you add up all possible routes a photon could take; this explains how a photon can appear to "decide" not to reflect off the first surface of an AR coating. As I understand it, optical coupling should work like this: Suppose you have two pieces of glass with a combined four surfaces. If you bring them close together you'll start getting Newton's rings. At one wavelength apart, you'll get constructive interference from the path that bounces off the 3rd surface. At 1/4 wavelength you get destructive interference because light bouncing off the 3rd surface is 180° out of phase with light hitting the 2nd surface. At less than 1/4 wavelength of air gap, I would guess that the two surfaces would begin to get less and less meaningful. I think that means the evanescent wave from would-be total internal reflection somehow connects ("couples"?) to the second piece of glass (3rd surface). The ~1/4-wavelength size scale seems consistent with the "200nm" quoted in total internal reflection fluorescence microscope. This looks relevant: evanescent wave coupling. —Ben FrantzDale (talk) 00:39, 23 July 2008 (UTC)
replies: text: See also frustrated total internal reflection.--Srleffler (talk) 00:50, 23 July 2008 (UTC)
thread_title:  Optical coupling 

message: Since the whole article deals with intensities, it is confusing that the picture deals with amplitudes (there is no negative intensity) I added the word amplitude in the caption of the article, I think the article requires more info on amplitudes.  Eranus (talk) 08:26, 22 July 2008 (UTC)
replies: text: Changing the caption was a good call. --Srleffler (talk) 00:50, 23 July 2008 (UTC)
thread_title:  First picture 

message: I think a history section would be a nice addition. I can't find the original form of Fresnel equations when did Fresnel derive them, when where did he publish them? An hour on Google and Google scholar did not help me much in contrast to History of Young's double slit experiment, Snell's law, Fresnel propagation... which are well documented, I would appreciate any refferance anyone may have.Eranus (talk) 17:48, 28 August 2009 (UTC)
replies: text: (Eight-and-a-half years later...)  I serve notice that I intend to add a history section within a few hours. — Gavin R Putland (talk) 07:30, 22 April 2018 (UTC)
thread_title: History

message: I'm wondering if I'm the first one noticing that the formula with the difference of angles fails at normal incidence:
 R_s = \left[ \frac{\sin (\theta_t - \theta_i)}{\sin (\theta_t + \theta_i)} \right]^2
=\ldots
and 
R_p = \left[ \frac{\tan (\theta_t - \theta_i)}{\tan (\theta_t + \theta_i)} \right]^2
=\ldots
both resolve to zero for \theta_t = \theta_i, i.e. normal incidence. This is clearly wrong.
The problem lies in the careless simplification of the correct expression (for R_s, but similar arguments hold for the case of R_p)
R_s =\left(\frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t}\right)^2
 
using Snell's law  n_1 \sin\theta_i = n_2\sin\theta_t. Multiplying the numerator and denominator of R_s with (e.g.) \sin^2\theta_t is only allowed if \sin\theta_t is not zero, which it of course is in the case of normal incidence.
The correct result for normal incidence R_s=R_t=\left(\frac{n_1-n_2}{n_1+n_2}\right)^2 is given further down on the page. My suggestion is to remove the terms with the difference of angles and just keep the following expressions, i.e. change
 R_s = \left[ \frac{\sin (\theta_t - \theta_i)}{\sin (\theta_t + \theta_i)} \right]^2
=\left(\frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t}\right)^2
=\left[\frac{n_1\cos\theta_i-n_2\sqrt{1-\left(\frac{n_1}{n_2} \sin\theta_i\right)^2}}{n_1\cos\theta_i+n_2\sqrt{1-\left(\frac{n_1}{n_2} \sin\theta_i\right)^2}}\right]^2
to
 R_s = \left(\frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t}\right)^2
=\left[\frac{n_1\cos\theta_i-n_2\sqrt{1-\left(\frac{n_1}{n_2} \sin\theta_i\right)^2}}{n_1\cos\theta_i+n_2\sqrt{1-\left(\frac{n_1}{n_2} \sin\theta_i\right)^2}}\right]^2
and correspondingly for R_p.
Oliver Jennrich (talk) 08:05, 23 April 2010 (UTC)
replies: text: Saying the formulas "fail" is maybe a bit strong. When I hear "fail", I think "you plug into the formula and get a prediction which is wrong". That's not what happens. Instead you plug into the formula, and get zero divided by zero which is undefined. So I would say the formulas "don't apply" at exactly zero. But anyway it should be clearer for sure than it is now. My preference would be:
 R_s =\left(\frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t}\right)^2
=\left[\frac{n_1\cos\theta_i-n_2\sqrt{1-\left(\frac{n_1}{n_2} \sin\theta_i\right)^2}}{n_1\cos\theta_i+n_2\sqrt{1-\left(\frac{n_1}{n_2} \sin\theta_i\right)^2}}\right]^2 = \left[ \frac{\sin (\theta_t - \theta_i)}{\sin (\theta_t + \theta_i)} \right]^2
 (The last expression is valid at non-normal incidence, but at normal incidence it equals 0/0.)
or something like that :-) --Steve (talk) 09:00, 23 April 2010 (UTC)

text: I removed the offending forms. They weren't really wrong. As Steve notes, they are technically undefined for the case of normal incidence. I haven't checked, but I expect that they are in fact correct in the limit as the angle of incidence goes to zero. Still, they didn't really add much value to the article; better to remove them to avoid confusion. Normal incidence is kind of an important case.--Srleffler (talk) 02:11, 24 April 2010 (UTC)
replies: text: Fine with me :-) --Steve (talk) 05:55, 24 April 2010 (UTC)
replies: text: I think they should be put back since they are a lot simpler than the other formulas, and therefore the form one preferably uses. They are not incorrect, just not well defined for normal incidence, and they do converge to the correct values in the limit as the angle of incidence goes to zero. Ulflund (talk) 08:24, 15 September 2011 (UTC)
thread_title:  Normal incidence problems 

message: I suggest, if possible, that the autor of the graphics modifies them by plotting the coefficients for a refraction index of 1.5, more typical af a glass, which is the main material dealt with in the article.  --GianniG46 (talk) 07:52, 15 June 2010 (UTC)
thread_title:  Reflection coefficients graphic 

message: I just wanted to direct your attention to a recent disambiguation implemented in transmittance involving Fresnel equations: see here. Thanks. Fgnievinski (talk) 22:30, 9 March 2011 (UTC)
thread_title:  Transmittance to/through a medium 

message: Why is there an empty section called "Special cases"?  — Preceding unsigned comment added by 129.242.204.238 (talk) 07:47, 25 October 2016 (UTC)
replies: text: It's not empty. It contains several sub-sections.--Srleffler (talk) 16:53, 25 October 2016 (UTC)
thread_title:  Special cases 

message: On 2017-11-05 I have replaced the plots for coefficients versus angle of incidence for air to glass and vice versa, for both amplitude and power coefficients, with substantially improved ones. The old plots by Ulflund (power coefficients) were very good, but showed only the values for reflection. The old plots by Kohlik17 showing the modulus of the amplitude coefficients were of low resolution and, in the case of glass-to-air, suffered from a limited vertical range hiding the important t>1 regime, and showed nonzero transmission in the TIR regime, which is just wrong physics. The new plots add power transmission, Brewster's angle for power plots, as well as expressions for the normal incidence limit as a function of n. I tried to keep as many of the good things from Ulflund's plots as possible. Numerically, the new plots agree with the old ones by Ulflund, and for the amplitude coefficients I have checked them against 4th Ed. of Hecht. They are submitted in SVG format. The generating Python code will be shared when I get around to it.
I hope this edit is in line with how things are done here! Otherwise, please do let me know. Suggestions for further improvement welcome. Sbergjohansen (talk) 11:50, 5 November 2017 (UTC)
replies: text: Thanks. These new graphs look much better. You're right there were some serious problems with the old amplitude graphs. One important question: Have you used the same formulas for amplitude coefficients as given in this article? There are several different conventions used in the literature, so if you used formulas from some other source the results won't match the formulas given here.--Srleffler (talk) 16:02, 5 November 2017 (UTC)
replies: text: Yes - the new graphs use the sign convention of the article! And thank you too, Srleffler, for keeping a close eye on this and many other important details. I will post the sources as soon as I have figured out the best way to do it. Sbergjohansen (talk) 18:02, 5 November 2017 (UTC)
thread_title:  Plots of amplitude/power coefficients versus angle of incidence 

message: and anyone else listening: I'm a bit busy and will respond quickly to some of your points:
* I pointed out a couple places that I would have used a different convention, but really none of these are too important. A person capable of understanding the math can easily adapt to the reciprocal of impedance or rotated axes. The one that I would have found the most difficult if I were reading it from scratch is having to remember i=1 and t=2. A derivation on Wikipedia doesn't need to follow the conventions of any author, but I try to avoid clashes with common conventions such as the orientation of the coordinate system. But again, anyone who could understand it still can, and there's probably no point in changing a convention once this has all been written.
 For linear polarization, a complex vector is unnecessarily general; it suffices to use a constant unit vector multiplied by a complex scalar.
* No, actually you are wrong. A complex epsilon or mu (attenuating medium) leads to a k which is a general complex vector (a real vector plus j* a different vector in a different direction). The derivation you wrote already does that, you just didn't realize it! (It's buried in the "complex angles," which I abhor!). For instance, with TIR consider the field in medium 2 (kx real, kz imaginary). When epsilon or mu are complex the medium supports inhomogenous waves in which the direction of wavefronts and the direction of power flow are different. And all that works fine now. My only point was that it is more confusing to have a quantity X which multiplies a unit vector defined as -yhat rather than just saying -X times yhat or -X being the y component of something. There wasn't any real need for the e_k unit vector so I took it out.
 OTOH, some people expect that x is to the right and y is up. But I can change two diagrams if I have to.
 I did it that way in order to re-use the red arrows. 
*Those comments seem to indicate you didn't understand my remark about the coordinate system. But as I said, you could probably leave it the way it was written.
* A lot of your writing is really long-winded (I understand as I have the same problem!). So in a number of places I've made it more concise or removed content that was duplicated elsewhere or offloaded content where wikilinks suffice. In a few cases I identified content that I'm hoping YOU will fix, such as section 6.10 which should just be a short paragraph and the ending moved to the history section. I hope you have been looking at the edit summaries, not just the result.
*Yes, you got me! Capacitance is a common admittance-type term.  
* I was wrong about what I imagined to be the complexity of including mu≠1. Indeed your results appear fine, I apologize. I also had never realized (of course it's a situation not to be seen in practice) of the unusual results when mu~1/epsilon, n=constant across an interface! 
* Forget what I said about Brewster angle when mu changes. It's really messy (including s-polarized Brewster transmission) and of course of no practical importance.
 .... especially when that breaks the last subsection.
* I don't know what you're referring to here. If I "broke" something, then fix it! Nothing I do is irrevocable. If you don't like a change, you can revert it (up to 3 times a day, without getting into trouble). Or better, discuss it, or just change it "back" in a way that addresses the reason that someone else did it differently. But I don't know what I broke so I can't fix it myself.
 OK, but one doesn't solve that problem by deleting the one subsection that refers to impedance
* As I said in the edit summary, having a section just explaining how to use the reciprocal of a quantity isn't needed for an article/section that could only be understood in the first place by someone conversant in such math. There was no actual "problem", just my suggestion that "impedance" is a much more accessible term.
* Gavin: I'd appreciate it more if you could put all your replies in one block rather than interspersed with what you're replying to. Otherwise I don't find it easily. If needed, requote the content you're replying to (as I have here). Thanks again for your hard work! Interferometrist (talk) 16:31, 22 May 2018 (UTC)
thread_title: Latest replies

message: The formulae after ‘Making this substitution, we obtain equations using the refractive indices:’ seem to me to only depend on n = n₂ / n₁ and not on n₁ and n₂ separately.
:Not quite. Take a closer look.--Srleffler (talk) 22:12, 25 September 2021 (UTC)
replies: text: Yes, you're right...I didn't look at it closely enough.--Srleffler (talk) 21:24, 26 September 2021 (UTC)
thread_title:  n₁ and n₂ 

message: Please add a photo on 's and p polarization' to describe it. Himadrichakrabortydip (talk) 15:05, 15 May 2022 (UTC)
thread_title:  It's a request.... 

message: In the "Complex amplitude reflection and transmission coefficients" section, the given equation for power transmission coefficients for complex media is
T = \frac{n_2 \cos \theta_\text{t}}{n_1 \cos \theta_\text{i}} |t|^2
But I don't think this equation can be correct. If
T_\mathrm{s} = 1 - R_\mathrm{s}
and
T_\mathrm{p} = 1 - R_\mathrm{p}
is true, and it must be to fulfill the conservation of energy, and if
R_\mathrm{s} = |r_\mathrm{s}|^2
and
R_\mathrm{p} = |r_\mathrm{p}|^2
is true, then
\frac{n_2 \cos \theta_\text{t}}{n_1 \cos \theta_\text{i}}
would necessarily have to be real for the initial equation to be true. But since , ,  and  can all be complex, this fraction will generally not be real-valued.
Based on my own derivation, I believe the correct expression (applicable to either polarization) is
T = \operatorname{Re}\left(\frac{n_2 \cos \theta_\text{t}}{n_1 \cos \theta_\text{i}}\right) |t|^2
I'm hesitant to change the page based on my derivation alone, but in the books I have and the bit of googling I did, I didn't find anything on this topic. I don't have access to the source quoted here (Hecht, 2002, p.120, eq.(4.57).).
Would anyone else have any sources that confirm (or deny) my assertion? Perhaps someone has a copy of that Hecht book and could double-check what it says. I'd be happy to post my derivation here if anyone would like to see it. Homm2trschell (talk) 17:11, 14 July 2023 (UTC)
replies: text: I don't have time to look into this right now, but you are most likely wrong. The definitions and sign conventions mess everyone up the first time through this material.--Srleffler (talk) 20:59, 15 July 2023 (UTC)

text: I have an earlier edition of Hecht. The formula at the top is correctly transcribed.--Srleffler (talk) 21:04, 15 July 2023 (UTC)
thread_title:  Correct power transmission coefficients for complex media