Talk:Brewster's angle

Last sentence

The last sentence: "The refractive index for a given medium changes depending on the wavelength of light, but typically does not vary much. The difference in the refractive index between ultra violet (≈100nm) and infra red (≈1000nm) in glass, for example, is ≈0.01." is only true for dielectric materials, glas, diamond ect. In semiconductors, which have bandgaps (critical points) in the visible or infrared region, the refractive indices can vary quite dramatically and hence the brewster angle.

I've removed most of that statement for a more general one. In fact, it was wrong anyway - most glasses have a bigger variation than 0.01 between the UV and NIR. --Bob Mellish 18:34, 7 October 2005 (UTC)[reply]

ise vs ize

I have converted all occurrences of polarisation with polarization, with apologies to British users. I felt it was confusing that the page used both spellings, somewhat inconsistently. I chose the spelling with z because the page on polarization spells it that way. --Srleffler 15:15, 12 October 2005 (UTC)[reply]

The image spells it 'polarisation' so for consistency I feel one or the other should be changed. It is easier to change the text than the image, and also Brewster himself was Scottish, so I feel it should be consistently 'polarisation' not 'polarization' but either way would be better than it is. Opinions? 62.31.207.26 (talk) 16:22, 4 May 2008 (UTC)[reply]

It's the image that should really change. The normal rule on Wikipedia is that usage should be consistent within an article, and unless the article's topic has strong national ties the article should stick with the national variety of English used by the first significant contributor. The first version of this article was imported from the American Federal Standard 1037C, and used -ize.

Beyond that, the -ize ending is probably a better choice. According to American and British English spelling differences the -ize spelling is not exclusively American, being recommended by many British authorities including the OED. It also prevails worldwide in scientific writing and is used by many international organizations.--Srleffler (talk) 04:08, 5 May 2008 (UTC)[reply]

Vandalism?

Can someone check this edit: [1] and see if it's vandalism. The user is a persistent vandal but I don't know enough about Brewster's angle to know if the change was legit or not. Powers 14:15, 7 March 2006 (UTC)[reply]

I looked at the edit and can confirm that it´s clearly vandalism. The former formulae are correct! waifar 84.178.106.15 11:44, 9 April 2006 (UTC)[reply]

Mechanism of the polarisation of light when incident at the Brewster Angle

It is said in the article that the light reflected from the boundary between the media is polarised because the dipoles that are excited in medium 2 can't emit radiation along the axis of their oscillation.

This explanation at first sight seems logical, however, there is a contradiction. If the experiment is performed with medium 2 as a vacuum, polarised light is still observed. Obviously, if medium 2 is a vacuum there are no dipoles in medium 2 that can oscillate. At the moment I don't have a better explanation for this phenomenon, however I think you should revise the article. Also, the diagram is very misleading, again just image medium 2 as a vacuum.

Hope that helps,

Robert

The refraction and reflection are both caused by the interaction of the light with the material, at its surface. Either way, the dipoles are in the material. The explanation, as written, assumes the light is incident on a medium from outside. The principle is the same the other way, though. If light goes from a medium into a vacuum, the refracted light in the vacuum has to be generated by dipoles at the surface of the medium. Since they can't be in the vacuum, they must be in the medium. At Brewster's angle, the reflected light is parallel to the direction of oscillation that would be required for dipole radiators to generate the p-polarized refracted beam, so p-polarized light cannot be reflected by this surface.

In the end, it's all a crude approximation. Light isn't really produced by "dipoles". The dipole approximation does give a good feel for what is going on, though, and gives the correct results if correctly applied.

Thanks for pointing this out. This paragraph in the article probably needs to be clarified.--Srleffler 22:56, 2 May 2006 (UTC)[reply]

OK, I took a stab at changing the article. It now avoids saying where the dipoles are, other than that they are "at the interface". This is all that is required for a physical description here. It doesn't really matter which medium the dipoles are in, since this is an interface effect. No interface, no Brewster effect.

Typo??

There's a missspelling in the first paragraph: "The polarization that cannot be reflected at this angle is the one for which the electric field of the light waves lies in the same plane as the incident ray and the surface normal." Should be "AS this angle", not "AT this angle, right?—The preceding unsigned comment was added by 134.2.74.148 (talk • contribs) 07:58, October 5, 2006.

No, it is correct as written. I'm not sure how you are interpreting it, but you are clearly not interpreting what is written there correctly. If you explain what you think it says, perhaps we can make the wording clearer. --Srleffler 19:38, 13 October 2006 (UTC)[reply]

Unsigned may be understanding the phrase "the one for which" to mean "the angle for which," I suspect it is intended to mean "the polarization for which." Replacing the word "one" with an explicit noun may be stylistically awkward, but would clear up the ambiguity. 138.162.0.44 13:26, 5 March 2007 (UTC)dgmartin[reply]

Revise first paragraph

In the first paragraph, combined with the picture it is very difficult to understand what is reflected and what is refracted, would someone please consider revising this, so that it is easier to comprehend.—The preceding unsigned comment was added by Matthew Rollings (talk • contribs).

Examples for the layman?

Would someone be willing to provide a simple example or two of this phenomenon?

Done. See article. It would be nice to add some example images here. J S Lundeen

External link suggestion

I would like to suggest to add an external link to BrewstersAngle.INFO. This is an easy-to-use Brewster's angle calculator which uses refractive index values of real materials. Wikipedia's policy does not allow me to add the link myself as I am the author of this site. Please, add a link if you find this site appropriate. Thanks! Mpolyanskiy (talk) 00:29, 2 September 2008 (UTC)[reply]

p vs. s

The use of p and s does not seem very clear. Contrast "Light with this polarization is said to be p-polarized, because it is parallel to the plane" with "by definition, the s-polarization is parallel to the interface". After reading three or four times (and having prior knowledge of the physics) it made sense, but it really should be clearer. Saying that s is parallel to the interface when defining it would be better. —Preceding unsigned comment added by 149.169.208.99 (talk) 06:47, 6 January 2010 (UTC)[reply]

Derivation needs a rewrite

The derivation is extremely confusing. The \theta_in + \theta_out=90 is not clear, because brewster's angle only exists for one polarization and not the other. I might get angry and do a rewrite. One really angry guy (talk) 01:30, 16 October 2010 (UTC)[reply]

As the text tries to explain, reflection goes to zero when the electric field direction of light inside the material is parallel to the propagation direction of reflected light outside the material. That occurs when

\theta _{in}+\theta _{out}=90^{\circ }

, and only happens for p-polarization. For s-polarization, the electric field inside the material is never parallel to the propagation direction of the light outside the material. If you think you can describe it better, go ahead and rewrite the derivation. No need to get angry.--Srleffler (talk) 05:26, 16 October 2010 (UTC)[reply]

The derivation can be readily achieved without understanding and evoking the dipole analogy, eg via Fresnel's Equation. Such an article would have an architecture:

reflected percentage is given by this equation (Fresnel).
Solve for reflected percentage equals zero. Get Brewster's angle.
What's the physical interpretation? Bam, oscillation of dipoles.
Alternate derivation via physical interpretation.

I do feel that such a derivation would be a lot more intuitive. Okay, I admit currently the Fresnel's Equation page also looks frightening, but with a bit of revamping, the derivation of Fresnel should be pretty obvious as well.One really angry guy (talk) 07:57, 16 October 2010 (UTC)[reply]

Explaining Brewster's angle in terms of the Fresnel equations would be better, but this should not include a full derivation. By policy, Wikipedia is an encyclopedia, not a textbook. Long derivations are generally out of our scope. In physics articles, even short derivations are really only appropriate when they serve to explain the physics. Where possible, we write for a lay audience, not for physicists. The derivation that is in the article now is not very good, but it does show how the condition described above leads to Brewster's angle.--Srleffler (talk) 21:36, 16 October 2010 (UTC)[reply]

I saw your rewrite. Nice work. I had a few commitments pop up and you beat me to it. I did some minor english edits to make the text more obvious. Other than that, I think it is good - it avoids the long derivation but since it points to Fresnel it is very easy for the reader to see how a mathematical derivation might be made. One really angry guy (talk) 20:56, 17 October 2010 (UTC)[reply]

"That is, if the oscillating dipoles are aligned along the supposed direction of the reflection, no light is reflected at all." I could not follow this sentence. Are we assuming that the dipoles know the angle of incidence and line up with it, or are we assuming that they occur in every direction but deducing that these particular ones are the only ones that can be involved in absorbing and retransmitting light? If so, perhaps the reasoning could be spelled out more completely. — Preceding unsigned comment added by 84.227.237.33 (talk) 16:17, 3 April 2014 (UTC)[reply]

The derivation is poorly written and needs a rewrite, but I can't attempt that tonight. The dipoles are excited by the incident light, and have their moments parallel to the electric field of the light in the medium (i.e. the refracted light). It happens that the dipoles can't emit light along their moment axis, so if the dipoles happen to be lined up with the direction in which the reflected light has to go, that light can't be produced.--Srleffler (talk) 06:38, 5 April 2014 (UTC)[reply]

I see that I wrote a clearer explanation in 2010 but somebody overwrote it.--Srleffler (talk) 07:01, 5 April 2014 (UTC)[reply]

I replaced the bad explanation with the old one. As it was, the explanation was flat out wrong on several points.--Srleffler (talk) 07:12, 5 April 2014 (UTC)[reply]

The definition is ambiguous as well

n1 and n2 are not defined, and since the argument of arctan can be either greater or less than 1 we cannot deduce which medium has refractive inded n1 and which n2. This renders the definition meaningless.

Andrew Smith — Preceding unsigned comment added by 82.32.48.177 (talk) 17:20, 20 January 2012 (UTC)[reply]

Fixed. The answer was common sense, though: n₁ is the index of the first medium through which the light propagates.--Srleffler (talk) 22:59, 22 January 2012 (UTC)[reply]

Needs general EM formulation for parallel and perpendicular polarization

This page is appropriate for optics, but brewsters angle applies to all EM waves, including radio, microwave, etc. It needs its EM formulation, and a description of what happens to parallel and perpendicular e-field vectors. 166.19.102.20 (talk) 21:08, 27 January 2011 (UTC)[reply]

Sounds like a good idea. Feel free to write something about it and add it to the article.--Srleffler (talk) 04:56, 28 January 2011 (UTC)[reply]

First sentence

Where it reads "Brewster's angle ... at which [light] with a particular [Polarization] is perfectly transmitted through a transparent [dielectric] surface, with no [Reflection]"

shouldn't it read;

Brewster's angle ... at which [light] with a particular [Polarization] is perfectly reflected from a transparent [dielectric] surface, with no [Transmission] throgh it"?

... isn't the whole point that it is the reflected light that ends up perfectly polarised? — Preceding unsigned comment added by 78.86.7.250 (talk) 00:08, 23 November 2012 (UTC)[reply]

You achieve perfect polarization of the reflected light by perfectly transmitting the other polarization. The polarization that is reflected is not perfectly reflected: some of that polarization is transmitted as well.--Srleffler (talk) 05:37, 23 November 2012 (UTC)[reply]

I have read this sentence several times, and I cannot see the logic that justifies the use of 'therefore':- "Brewster's angle (also known as the polarization angle) is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. When unpolarized light is incident at this angle, the light that is reflected from the surface is therefore perfectly polarized. " I propose the following definition: "Brewster's Angle is that angle of incidence of unpolarised light on a dielectric surface at which the reflected light is perfectly polarised. The polarisation of the reflected light is such that its E-field is parallel to the reflecting surface". g4oep — Preceding unsigned comment added by G4oep (talk • contribs) 18:10, 9 May 2018 (UTC)[reply]

The "therefore" is justified. Breaking it down:

One polarization is completely transmitted through the surface,
therefore there is no reflection of that polarization,
therefore all of the light that is reflected has the opposite polarization,
therefore the reflected light is perfectly polarized.

The definition in the article is better than yours because it describes both of the important functions of a Brewster surface, and how they are related. The fact that one polarization has precisely zero reflectance is at least as important technologically as the fact that the surface can produce a perfectly polarized reflection. While understanding that these two properties are in fact the same thing may take a bit of thought, it is necessary to appreciate both to fully understand Brewster's angle.--Srleffler (talk) 02:23, 11 May 2018 (UTC)[reply]

year of discovery

Is there any publication before 1813? The one I found predates the one used in the article by two years .. doi:10.1098/rstl.1813.0016. {{cite journal}}: Cite journal requires |journal= (help); Missing or empty |title= (help) --Stone (talk) 09:42, 9 January 2013 (UTC)[reply]

Not on subject

I meant to write in the edit summary (but must have hit the wrong key and it got published before I could finish!) that the section on polarizing sunglasses and camera filters doesn't really have to do with the Brewster angle, but is about reflections at large incidence angles, period. So I didn't think it belongs in this page, but didn't want to argue about it and just improved what was there. If someone wants to remove it for that reason (and perhaps incorporate any of it in a more relevant page), please go ahead.Interferometrist (talk) 22:33, 5 November 2021 (UTC)[reply]

Dead link in page; found one site with respective document

https://sites.esm.psu.edu/~axl4/lakhtakia/Documents/No170(Optik).pdf 2607:FEA8:5420:680:F3C0:9271:A531:9CA2 (talk) 03:37, 23 February 2024 (UTC)[reply]

Fixed. Thanks,--Srleffler (talk) 19:14, 24 February 2024 (UTC)[reply]