Talk:Fresnel equations

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There seems to be an incoherence on this page: while in the linked articles the transmission and reflection coeficient are defined as ratios of Amplitudes (in this case of the E-Field), the equations in this article seem to describe a ratio of intensities.

I see two possibilities to correct this:

make clear, that the coefficients describe intensities (perhaps remove the links to the other articles or change these):

$R_s = \frac{I_r}{I_i} = \left( \frac{\sin (\theta_i - \theta_t)}{\sin (\theta_i + \theta_t)} \right)^2=\left(\frac{n_1\cos(\theta_i)-n_2\cos(\theta_t)}{n_1\cos(\theta_i)+n_2\cos(\theta_t)}\right)^2$

$R_p = \frac{I_r}{I_i} = \left( \frac{\tan (\theta_i - \theta_t)}{\tan (\theta_i + \theta_t)} \right)^2=\left(\frac{n_2\cos(\theta_i)-n_1\cos(\theta_t)}{n_2\cos(\theta_i)+n_1\cos(\theta_t)}\right)^2$

with the intensities $I i$ of the incident light and $I r$ if the reflected light.

change the equations to fit Amplitudes (I don't know wether this would correspond to the graphs any more):

$R_s = \frac{E_{0r}}{E_{0i}} = \frac{\sin (\theta_i - \theta_t)}{\sin (\theta_i + \theta_t)} =\frac{n_1\cos(\theta_i)-n_2\cos(\theta_t)}{n_1\cos(\theta_i)+n_2\cos(\theta_t)}$

$R_p = \frac{E_{0r}}{E_{0i}} = \frac{\tan (\theta_i - \theta_t)}{\tan (\theta_i + \theta_t)} =\frac{n_2\cos(\theta_i)-n_1\cos(\theta_t)}{n_2\cos(\theta_i)+n_1\cos(\theta_t)}$

with the corresponding amplitudes of the electric field $E 0 r$ $E 0 i$

I would prefer the second possibility since it corresponds to the majority of the literature I have at hand - but I don't have time to check and replace the graphs right now - I'll be back when I have more time and take care of this if nobody else has until then...—The preceding unsigned comment was added by Nanomage (talk • contribs) .

This is probably a conflict between the optical usage, which cares mostly about intensity, and general electromagnetic wave usage, which cares more about amplitude. I've clarified that the equations refer to intensity coefficients. (Note: most literature I have uses small r and t for amplitude coefficients, and large R and T for intensity coefficients, if you decide to change it.) -- DrBob 16:48, 8 September 2005 (UTC)

-Shouldn't also be pointed out that the refractive index used for these equations is a complex number? --an anonymous Raptor 10:00,07/10/05 CET

I think the equation for p-polarization is wrong. From what I've seen in the literature, the following should be the equation:

$R_p = \frac{E_{0r}}{E_{0i}} = \frac{\tan (\theta_i - \theta_t)}{\tan (\theta_i + \theta_t)} =\frac{n_1\cos(\theta_t)-n_2\cos(\theta_i)}{n_1\cos(\theta_t)+n_2\cos(\theta_i)}$

—The preceding unsigned comment was added by 132.206.69.48 (talk • contribs) .

I haven't checked the math here, but be aware that the form of the equations depends on an arbitrary choice of sign convention. Different authors make different choices. --Srleffler 01:01, 18 February 2006 (UTC)

Why not include the equations for both intensity and amplitude and make clear which is which and where each is used?-4.232.0.63 17:06, 7 August 2006 (UTC)

That could work, but the amplitude form is a little more complicated pedagogically. The exact form of the equations depends on a choice of sign convention, and in addition I have seen at least three incompatible mathematical treatments, that essentially define "amplitude" differently. Which treatment is most common depends on what field you're in. The intensity formulism has the advantage of avoiding these difficulties. Differences in sign conventions and definitions often leads to Wiki articles with outright errors, as different editors modify equations without understanding the formalism or realizing that what appears in their textbook might not be correct given the definitions used in the wikipedia article.--Srleffler 00:32, 8 August 2006 (UTC)

1 spelling polarise or polarize
2 Imprecise physics
3 Redirection
4 Metals absorb light?
5 Optical coupling

[edit] spelling polarise or polarize

All references at my disposal use z not s in all forms. --4.232.0.63 16:52, 7 August 2006 (UTC)

"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)

[edit] Imprecise physics

Some physics in the article are imprecise. Approximations or assumptions made for some formulae are not clearly spelled out (e.g. the paragraph on reflectivity of a dual-surface window and the next paragraph on Fabry-Perot interference directly contradict each other), and other assumptions may be misleading (i.e. mentioning dielectric materials, but neglecting parelectric materials). Contrary what is claimed in the article, Fabry-Perot interferometers cannot be used to create perfect mirrors or reflection-free lenses (causality, i.e. the Kramers-Kronig relations, dictates that there is always some absorption.)

I spent a while editing the article, but Wikipedia lost the edited version on trying to preview it. I'm not sure I want to go through that frustration again, so someone else has to correct the article. —Preceding unsigned comment added by 218.186.8.10 (talk • contribs)

Sorry about the lost edit. That rarely happens, although the wiki has been a bit flaky the last four days or so. When you do get an error screen on "preview" or "save" always try your browser's back button. This generally returns you to the edit screen with no loss of text. Thanks for the suggestions.--Srleffler 19:58, 6 November 2006 (UTC)

[edit] Redirection

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).

Sounds good to me. Dicklyon 04:18, 25 January 2007 (UTC)

[edit] Metals absorb light?

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)

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)

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)

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 of crystalline solids produces x-ray diffraction.--Srleffler (talk) 01:58, 1 May 2008 (UTC)

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)

[edit] Optical coupling

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)