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Phase inversion

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Is this right: "Depending on the nature of the interface, i.e., dielectric-conductor or dielectric-dielectric, the phase of the reflected wave may or may not be inverted."? I thought the phase inversion of reflected light depended on the relative refractive index of the interfacing materials. Maybe both are right, but probably the article should be more clear. --Chinasaur 05:42, 4 Aug 2004 (UTC)


The following phase should be the other way round (if not much mistakening): "When light reflects off a material denser (with higher refractive index) than the external medium, it undergoes a phase inversion. In contrast, a less dense, lower refractive index material will reflect light in phase". Kindly verify so. — Preceding unsigned comment added by 2A02:2149:8294:B400:1FC:FCF6:E476:1974 (talk) 07:25, 13 July 2019 (UTC)[reply]

It is correct, but poorly written. We actually have an article on this topic.--Srleffler (talk) 15:32, 13 July 2019 (UTC)[reply]

Question: Reflection vs Refraction

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One thing that this article doesn't explain is why a greater angle of incidents results in more reflection and less refraction. A more fundamental question is, "why does light reflect at all?" If the article answered that question, it might be more obvious how light could be partially reflected and partially refracted in some cases.—Preceding unsigned comment added by 216.206.44.11 (talkcontribs) 18:49, 8 June 2007

Plus the rules of refraction says that transparent refractive substances bend light toward the normal (perpendicular) direction. But I don't see how that can allow an incidence of 180 degree reflection with a single encounter.WFPM (talk) 03:31, 5 September 2010 (UTC)[reply]


I don't understand your question. Note though that light only bends toward the normal if it is entering a medium with higher refractive index. If light is passing into a medium with lower refractive index, it bends away from the normal. In either case, if light is incident along the normal axis, it doesn't bend at all.--Srleffler (talk) 05:33, 5 September 2010 (UTC)[reply]

I'm trying to understand a 180 degree of reflection. Normal glass mirrors would bend the light toward the normal by refraction. but then as you approach a perpendicular direction from the mirror, you get a 180 degree reflection. so how does the reflective material turn the signal around 180 degrees in direction? Can you do that with 1 encounter? or does it take 2?WFPM (talk) 04:03, 6 September 2010 (UTC)[reply]

I think you are confused; I don't understand what you are thinking. You can certainly reflect something straight back with a single encounter. Perhaps it helps to consider a polished piece of flat metal instead of a glass mirror. Light that is incident along the normal reflects straight back. Light that is incident at an angle reflects with an equal and opposite angle from the normal.
A typical household mirror with a reflective coating on the back side of a piece of glass is a little more complicated. When light enters the glass it is bent toward the normal unless it is already along the normal, in which case it is not bent. The light reflects from the reflective backing with an equal and opposite angle. When the reflected light leaves the glass it bends away from the normal since it is moving from a high index medium to a lower index one. The bends when entering and leaving the glass exactly compensate for one another, so that light reflecting from a glass mirror makes the same angle as light reflecting from a bare piece of polished metal.
Does this help??
Or are you just getting confused by the numbers: in terms of the law of reflection, if a beam is normally incident the angle of reflection is actually 0°, not 180°.--Srleffler (talk) 17:43, 6 September 2010 (UTC)[reply]

I'm thinking about a Perpendicular beam coming up to an atom of silver and trying to figure out how the beam is (reflected?) back 180 degrees, which doesn't sound practical, or whether it is reflected 90 degrees by 1 encounter and another 90 degrees by a 2nd encounter. In refraction, the deflection is usually an increased bending of the path around the atom, which increases as the frequency (Stress?)increases. But its hard to understand the mechanics of 180 degree reflection. Or should I say kinematics?WFPM (talk) 01:15, 7 September 2010 (UTC) If we said that the beam went all the way around the atom of silver and then came back out in a perpendicular direction, we might have an explanation for 180 degree reflection. But then we would have a hard time explaining the process of refraction.WFPM (talk) 01:27, 7 September 2010 (UTC)[reply]

I see. Your problem is that you are thinking of a single atom of silver. Specular reflection is a collective effect. It requires many atoms to produce it. Snell's law type refraction is too. As the article puts it, "light waves incident on a material induce small oscillations of polarisation in the individual atoms, causing each atom to radiate a weak secondary wave (in all directions like a dipole antenna)." These weak secondary waves are in phase with one another. If the incident light is propagating in a single direction (e.g. a plane wave), the secondary waves interfere with each other and with the incident wave such that the incident wave is extinguished and in fact light is radiated in only two directions: the refracted direction predicted by Snell's Law and the reflected direction predicted by the Law of Reflection. In these two directions there is constructive interference between the radiation from the atoms and the incident beam.--Srleffler (talk) 03:12, 7 September 2010 (UTC)[reply]

Well, I appreciate your attention and consideration of my problem, and with regard to it I have noted that the process of (double) specular reflection is able to explain all of what I see in a corner mirror except the narrow line dividing the left side of the reflection from the right, in the exact location where we need to have this 180 degrees of single reflection. And so I guess I'm spoiled but I hate to have to adopt a plane wave absorption and reradiation theory to explain that small portion of the reflection image. And so I was looking for a way of reflection that would explain that small part by a process of redirection of the light, rather than an absorption and re-emission of it.WFPM (talk) 17:04, 11 September 2010 (UTC) I guess that what I have left out of my concept is your theory that all the redirected incident light is reflected by a process of absorption and re-emission, even the double reflected light. So thank you.WFPM (talk) 17:24, 11 September 2010 (UTC)[reply]

And while we're on the subject matter of reflection and refraction I'd like to ask you if your theory of absorption and re-emission is such as to explain the "first atom slowdown, last atom speed back up" problem that I have with the theory of refraction. Can you use this to explain how a light beam through a diamond, for instance, is able to regain its speed of propagation after it leaves the medium of the diamond and re-enters the faster medium.WFPM (talk) 17:34, 12 September 2010 (UTC)[reply]

Yes, but a detailed explanation of that is too involved for this talk page. Fundamentally, the light never actually slows down. The light that leaves the diamond is not the same light that entered.--Srleffler (talk) 19:32, 12 September 2010 (UTC)[reply]

Yes that was the alternative, that maybe the path got longer (and zig-zagged?) But you're not sounding like absorption and re-emission. So do you have an explanatory link?WFPM (talk) 23:08, 12 September 2010 (UTC). And I appreciate your explanation problem, because I'm trying to understand Feynman's QED, and not having much luck there either. Is that your reference re this matter?WFPM (talk) 21:28, 14 September 2010 (UTC)[reply]

QM reflection

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A quantum mechanical description of reflection would be nice if someone had the ability to explain it, since this article doesn't reveal the reason for which light is actually reflected from a surface.

Seconded! There must be a reasonable/logical explanation for reflection in terms of QM. As it is, the article just refers us to a Feynman book. For an article of high importance on a very common subject, someone with the knowledge needs to get this part up to scratch. I seem to remember Feynman saying something about it in these lectures... [url]http://vega.org.uk/video/subseries/8[/url] but I can't remember if he address QM reflection specifically. David 218.143.30.1 (talk) 05:15, 30 July 2009 (UTC)[reply]

I agree, but a regurgitation of Feynman's QED lectures just wouldn't do, as however great his explainations were, he did not go into the mechanism (microscopically) of reflection, only assumed that it happeded to explain where the light would reflect to. We don't want any stone-counting Mayans on Wikipedia! — Preceding unsigned comment added by 81.51.194.215 (talk) 08:29, 20 August 2012 (UTC)[reply]

Easily Amused

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"all non-shiny objects that are not black."

It's accurate; I just thought it sounded... amusing, probably because of the use of the word "shiny." Not going to edit to "all objects that do not shine and are not black" because that's a tad verbose and there's no real point. Just throwing that out there. --165.134.132.122 22:26, 18 August 2005 (UTC)[reply]

Images

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I have some photographs on my userpage of the famous curved, reflective sculpture Cloud Gate that could be illustrative. Spikebrennan 17:30, 6 July 2006 (UTC)[reply]

A little history?

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I was thinking a little history could be added to this article such as the Euclid's contributions to the field (is his the earliest known studies? Who are the other people noted for their theories on reflection?) and maybe links to and expansion of the term Catoptrics. 69.72.93.142 14:11, 2 December 2006 (UTC)[reply]


Quantum interpretation

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Can someone provide a source for the information provided in this section? Particularly the first paragraph. Thanks!

What about absorption?

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If "part of the light is reflected and the remainder is refracted", then I guess a black surface does not absorb light. Does it turn all the light into non-visible electromagnetic radiation? (I doubt it). We all know that a black material heats up more quickly than a white material when exposed to solar lignt. Where does that thermal energy come from? Just from absorbed infrared or ultravioolet radiation? ...

The paragraph about the quantic interpretation says that photons are sometimes absorbed. Do you mean that there are some cases in which the light does not produce heat when it strikes an opaque object, or passes through a transparent medium such as the air or water? Cases in which the light is completely "cold"?

In my opinion, these are questions that one should be able to answer after reading this article, but the article is not clear enough with respect to this topic. I am not an expert in optics, I just wanted to give to the experts an advice for improving the article. Regards, Paolo.dL (talk) 22:26, 6 January 2008 (UTC)[reply]

Diffuse->specular

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I am trying to understand what happens when a surface gets smoother and smoother e.g., when it is polished, but the same idea applies to sound waves, I am sure. What happens to a surface's reflections as it gets smoother? A metal surface with a small-enough pattern of saw-toothed ridges will reflect sound crisply; if you make the pattern small enough, I think you will even get nice sharp specular reflection of light. But how small do they have to be to appear smooth acoustically? Optically? I assume it will be a function of wavelength... —Ben FrantzDale (talk) 00:05, 27 May 2008 (UTC)[reply]

But why?

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Reflection is described in detail in the article, via laws of reflection, Maxwell's equations, etc., but what I don't see is the reason that reflection occurs.

Maybe I just missed it. But otherwise, I think this article would be distinctly enhanced by an explanation of why reflection occurs. In other words, a ray of light strikes a certain medium. There is something special about that medium -- in part its extreme smoothess, of course -- that causes the light to follow the laws of reflection. Can someone enlighten us about why this happens? This should include an explanation of why some materials make better mirrors than others.

Note: One is tempted to say it's just like a ball bouncing off a hard surface. But for one thing, a photon is much smaller than an atom, and from the point of view of the photon I would guess that what appears to us as a very smooth surface would be quite irregular -- after all, the photon could strike an atom at a relatively small or large distance from its nucleus.Daqu (talk) 07:42, 4 September 2008 (UTC)[reply]

Although Compton scattering of photons can occur, classical reflection and refraction of electromagnetic radiation are explained by examining its behaviour as a wave, so there's no analogy with a ball. Electromagnetic radiation has both an alternating electric and an alternating magnetic field. These fields interact with the electrons in the medium they are passing through or into. The effect of this interaction on a macroscopic scale can be deduced from Maxwell's equations and - in the particular case of moving into a different medium - the Fresnel equations as noted in the article. As you can see, the "why" has not got much to do with smoothness, it's a result of the way a wave behaves in a transmittive medium. Obviously a non-smooth boundary between media allows multiple angles of reflection and refraction, allowing dispersion of the wave in all directions, which is why you don't get a clear reflection from a rough surface. --RexxS (talk) 04:32, 8 October 2008 (UTC)[reply]

But photons aren't only waves, so surely Maxwell's equations can't cover it all. — Preceding unsigned comment added by 81.51.194.215 (talk) 08:33, 20 August 2012 (UTC)[reply]

See Correspondence principle. Maxwell's equations are just fine.--Srleffler (talk) 23:21, 20 August 2012 (UTC)[reply]
Complementarity (physics) is also relevant—there is little point in trying to gain an intuitive understanding of a wave-like property of light using a particle-like model.--Srleffler (talk) 23:33, 20 August 2012 (UTC)[reply]

Miraged reflection

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File:Fata morgana of the sun glitter.jpg
Fata Morgana of a sun glitter demonstrates green upper and red lower rims of the reflected and miraged object the Sun

Removed from the article:

On extremely rare occasions a reflection might get miraged, in other words reflection might get refracted. In the process reflection get distorted and magnified. The miraged reflection might show some properties of a reflected object, which cannot be seen well neither in the real object itself nor in the usual reflection of an object. For example, a mirafe of sun glitter clearly shows upper green and lower red rims that the reflected object (the sun) has. A miraged reflection makes the conditions very complex because it creates an interface between not only two different media, as a reflection does, but at least three different media.

I have removed this text and the associated image because:

  • the analysis of the image is original research, which is not permitted.
  • the text is not supported by a citation to a reliable source as required by Wikipedia:Verifiability.
  • the content is off-topic for this article. There are reflections, and there are mirages. It's not surprising that one can have a mirage of a reflection, but this isn't really a distinct effect, and doesn't really merit a section in the reflection article. Even if it were on-topic, it certainly wouldn't merit placement above the sections on "other types of reflection".
  • the image, while interesting, is not technically well-suited to Wikipedia. The features of interest are hard to discern, and the green and red rims described in the text are only visible if the full resolution image is loaded and zoomed in (~4× on my monitor).--Srleffler (talk) 04:58, 20 November 2008 (UTC)[reply]
May I please ask what features of interest are hard to discern? The mirage, sun glitter? Even green upper edge is seen in thumbnail.User:Srleffler made it sound like it is very common to see a mirage of reflection The user writes:"It's not surprising that one can have a mirage of a reflection". In reality the mirage like this one is extrimly rare and is very much surprising, and IMO people, who lost the ablility to get surprised would never discover anything new. I am also not sure what original research the user is talking about. I took an image of a phenomenon that I saw with my own eyes. What is really in question in this image, sun glitter, mirage, green rim? I believe both sun glitter, mirage, green rim/flash were recearched long time ago and proven to be exist.IMO the image is good enoug and interesting enough, and on-topic enough to be in the article. We could leave alone (do not talk about green and red edges, if it is confusing), but I believe it is a great image to talk about reflection versus refraction.After all this image depicts one of a very few situations, which show both reflection and refraction together.--Mbz1 (talk) 15:55, 22 November 2008 (UTC)[reply]
The fata morgana is, to the untrained eye, just an indistinct blur across the top of the photo. The colored edges are also not obvious. To understand what one sees in the image, one needs a more detailed understanding of atmospheric refraction than we should expect of readers of this article, nor should this article explain those phenomena in enough detail to make the image interpretable, because that explanation is off-topic. It would be fine for the article on fata morgana, but is not appropriate here. Basically, the image doesn't "say" much about reflection, but it does say something significant about fata morgana mirages.--Srleffler (talk) 17:25, 22 November 2008 (UTC)[reply]

Creating a vapor reflective surface [?]

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The article makes clear that a reflective surface may be solid "mirror" or liquid "water".

I'm curious, though, as to whether it is possible to create a reflective surface with an inert gas, such as argon. And if so, how would one go about doing such? Pine (talk) 19:32, 13 February 2009 (UTC)[reply]

Gasses can reflect, if there is a transition in index of refraction. It's hard to get a large, sharp change in index with gasses, though. This isn't the right place to ask this kind of question. Article talk pages are for discussion related to improving the article. You want the Science reference desk.--Srleffler (talk) 01:09, 14 February 2009 (UTC)[reply]

Third Law of Reflection?

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I am wondering if the third law of reflection, light paths are reversible, should be added to the article? It is sort of important to reflection that people know of that and I do believe that it is a law of reflection. Mcfar54 (talk) 09:31, 25 March 2009 (UTC)[reply]

Can you cite a source that refers to this as a "law of reflection"?--Srleffler (talk) 04:18, 15 May 2009 (UTC)[reply]

Does the speed of the reflecting surface make any difference?

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If the mirror or reflective surface is moving at very high speed, in the direction of its own plane, then do the angles of incidence and reflection still equal each other? Is the frequency of the reflected light the same as that of the incident light? Has this senario been tested either experimentally or via the equations? Norman Sheldon (talk) 14:52, 7 January 2011 (UTC)[reply]

I guess that if we describe this from the point of view of the reflective surface reference frame, we will see a (Doppler-shifted) light coming from a moving source, that will be reflected by the non-moving surface in the usual way, with equal incidence and reflection angles and with no shift in frequency. --GianniG46 (talk) 21:56, 7 January 2011 (UTC)[reply]

And if we describe it from the point of view where the source is at rest and the reflective surface is moving, please? Norman Sheldon (talk) 18:44, 9 January 2011 (UTC)[reply]

This isn't really the right place for this. Article talk pages are for discussions related to improving the article, not for general questions. The right place to ask a question like this is at the Science Reference Desk. If you post your question there, I'm sure you will get some enthusiastic discussion of it. It's an interesting question.--Srleffler (talk) 20:12, 9 January 2011 (UTC)[reply]
I fully agree with Srleffler, and, moreover, your questions concern relativity, rather than optics. Anyway, just to avoid leaving matters pending, I suppose that the Lorentz transformations would affect in the same way the incident and reflection angles, which make the same angle (in absolute value) with relative velocity, so the two angles will stay equal, while the reflective properties of the surface will be "Doppler-shifted", if they depend on wavelenght. --GianniG46 (talk) 08:45, 10 January 2011 (UTC)[reply]

Suggested merge

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Is there anything at Abnormal reflection that's worth merging to this article? That article has no content and looking at its references suggests that it's not a unique physical phenomenon but rather is a pair of words describing any observations of anomolous reflection effects. --Wtshymanski (talk) 22:02, 30 October 2011 (UTC)[reply]

A short section here discussing abnormal reflection might be a better solution than a separate article covering a variety of disparate phenomena.--Srleffler (talk) 00:04, 31 October 2011 (UTC)[reply]
OK, but I'm at a loss to find something to merge; I suppose one could summarize the "abnormal reflections" in the several items referenced, but there doesn't seem to be any unity of those publications as they are describing different effects. --Wtshymanski (talk) 03:13, 31 October 2011 (UTC)[reply]

Total Internal Reflection

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In the section about total internal reflection, shouldn't the example of X-ray actually be "Total external reflection"? --Cyferz (talk) —Preceding undated comment added 19:39, 16 February 2012 (UTC).[reply]

General Comments

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This physics article presents itself as a general discussion of the physical phenomenon of reflection. The two opening paragraphs give examples of reflection that include sound waves, water waves, electromagnetic waves, and seismic waves. But what follows is a discussion almost exclusively limited to the reflection of visible light, and which includes no examination of the broader principle of reflection as a consequence of the behavior of waves as they encounter constraints on their motion.

The article states, "Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated." This statement can be true, but a change of media is not necessary for reflection unless air at different temperatures or water of varying depth is considered different media. Nowhere in the article is it pointed out that the characteristic that results in reflection and refraction is the change (or rate of change) in propagation speed (or phase velocity) of the wave in the medium.

The "Mechanism" section goes into a level of detail that could mislead the reader into thinking that Maxwell's equations are fundamental to an understanding of wave reflection. It's a bit like using a description of gravity and the properties of water to explain reflection in ocean waves. The truly remarkable thing, and something this article manages to miss, is that the reflection of light as it encounters changes in refractive index, and the reflection of ocean waves that encounter changes in depth are really two manifestations of the same thing. Maxwell's equations do not explain reflection in any meaningful way. Rather, they explain why electromagnetic waves obey the same laws as other kinds of waves. --Alt Livingston (talk) 21:55, 3 June 2012 (UTC)[reply]

These are very good points. If you are so inclined, feel free to try changing the article to present the topic in a more general way.--Srleffler (talk) 02:02, 4 June 2012 (UTC)[reply]


Paddle Image

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Look at the reflection on the paddle -- it is obviously a reflection of the situation in the sky, so this is not a double reflection. 190.141.37.103 (talk) 03:36, 22 April 2013 (UTC)[reply]

Look again. The image in the water, which is a reflection of the sky, is inverted. If the image in the paddle were a direct image it would be inverted too. It isn't, because it is a reflected image of the reflection off of the water. --Srleffler (talk) 03:47, 22 April 2013 (UTC)[reply]
Perhaps the paddle is concave, then the reflection doesn't have to be inverted. —Kri (talk) 17:47, 30 April 2013 (UTC)[reply]
The paddle would have to be convex. The trees and sun are too far from the paddle to form an upright image, if the paddle is concave. See Curved mirror. The image does appear to be reduced in size, but it could be either an inverted double reflection or an upright direct reflection.--Srleffler (talk) 00:50, 1 May 2013 (UTC)[reply]
What are you talking about? You can't invert reflected in a mirror by making the mirror convex, but you can invert the image by making the mirror concave. The article you linked to contains several images, how could I possibly know which one of them you are referring to? On the other hand, all photographs in the article are of convex mirrors, and in none of those has the image that is reflected been inverted. Are you sure you didn't mix up convex and concave?
Furthermore, it can't be too far to a reflected object for it to be inverted in a concave mirror; on the other hand the reflected object can be too close to the mirror (closer than the focal point) to be inverted, but that would not be the case here since, as you say, the trees and sun are far away from the paddle. —Kri (talk) 11:24, 3 May 2013 (UTC)[reply]
I wasn't referring to any of the images in that article, but rather the text, which explains under what conditions curved mirrors form upright or inverted images. Anyway, we seem to agree about when one gets an upright or an inverted image from a given type of mirror, but one of us is misunderstanding the other about the application to the image in this article. The anonymous poster above suggested that the reflection in the paddle is a direct reflection of the sky, not a reflection of the image reflected in the water. The image reflected off of the water is upside-down. The image reflected off of the paddle is rightside-up. If the latter is a direct reflection of the sky, then the paddle is forming an upright image of the sky, which is possible only if the paddle is convex. I'm not sure what you have in mind or where you disagree with this (if you do).--Srleffler (talk) 11:44, 3 May 2013 (UTC)[reply]
On second thought, I think you are right. I was confused by the fact that the reflection in the water is upside-down, but it is not an "inverted" image in the optics sense (a plane mirror forms upright images). --Srleffler (talk) 03:40, 4 May 2013 (UTC)[reply]

I agree; it doesn't look like a double-reflection to me. It looks like the paddle is reflecting the sun directly. Even if somehow this is not the case, I suggest a less ambiguous image. 174.103.118.211 (talk) 20:56, 21 March 2016 (UTC)[reply]

Bragg Reflection

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Bragg Reflection is either a different type/mechanism of reflection or an interesting special case of one - or just a different level of abstraction.

I've been studying it to understand play-of-color in precious opal - including contra-luz opal -- but the Braggs's discovery was for/with X-Rays.

It seems like this article could benefit a lot by including info on that somewhere. PMH232 (talk) 18:12, 29 May 2015 (UTC)[reply]

Status March 2017

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This is among the worst technical articles I've had the misfortune to find on Wikipedia. Apparently, at some point it included some actual MODERN physics, but if so, it has been removed. (I came to this article expecting to find, at least, a reference to the quantum mechanics behind the 17th Century geometric optical "science". Nothing.) The first sentence insists that reflection must be defined in the optical media in which the "wavefront" travels. Really? OPTICAL media?? Wrong. In the 2nd paragraph, we learn "many" types of EMR reflect. Really? Not "all" but "many"? Rubbish. In the next paragraph we read:"In specular reflection the phase of the reflected waves depends on the choice of the origin of coordinates, but the relative phase between s and p (TE and TM) polarizations is fixed by the properties of the media and of the interface between them." Of course, no explanation, not even a definition, of s, p, TE, and TM is provided. GARBAGE. (Of course, if a reader has to spend 20 minutes somewhere else learning what a "phase" is, nevermind a "wavefront". The next paragraph informs us that a mirror is a model for specular light reflection. A model, not an example, but a model. Brilliant. I could go on, and on, and on, but if anybody thinks this article is useful, perhaps they could spend some time cleaning it up, personally, I think it is so bad that it is beyond redemption.40.142.182.216 (talk) 16:20, 20 March 2017 (UTC)[reply]

Information icon Thank you for your suggestion. When you believe an article needs improvement, please feel free to make those changes. Wikipedia is a wiki, so anyone can edit almost any article by simply following the edit this page link at the top.
The Wikipedia community encourages you to be bold in updating pages. Don't worry too much about making honest mistakes—they're likely to be found and corrected quickly. If you're not sure how editing works, check out how to edit a page, or use the sandbox to try out your editing skills. New contributors are always welcome. You don't even need to log in (although there are many reasons you might want to). --Srleffler (talk) 01:28, 21 March 2017 (UTC)[reply]

The formation of image

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Bc — Preceding unsigned comment added by 197.231.239.109 (talk) 08:37, 14 August 2022 (UTC)[reply]

The redirect Reflection (physics has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at Wikipedia:Redirects for discussion/Log/2024 March 3 § Reflection (physics until a consensus is reached. Utopes (talk / cont) 18:28, 3 March 2024 (UTC)[reply]