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If you shine a laser pointer or generally any coherent light at the screen of a fancy, new TV, you get a very sharp X pattern.

enter image description here

I think this is a property of LED screens specifically. This phenomenon has been noted many times (1, 2, 3, 4), but every single answer is unsatisfactory, because they all say "it's a diffraction grating".

If the TV screen were like a two-dimensional diffraction grating, there would be a two-dimensional grid of points in the diffraction pattern; that is the universal pattern you get from scattering off a crystal or a phone screen. But I've never been able to see any points in this grid besides the X itself, even in good photos like the one above; it's impossible for such a sharply non-grid like pattern to be produced by diffraction of a two-dimensional grid.

Searching around, I've seen a lot of discussion on the internet about this, but no resolutions. I suspect the effect is due to something like single slit diffraction off of small features in each pixel. (Since larger things make smaller diffraction patterns, in this case the two-dimensional grid would still be there, but too small to see.) But it may be more complex; these pixels are quite complicated optical components.

Does anyone know precisely what causes this pattern?

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  • $\begingroup$ Do the legs of the X stay put if you view from different angles, or do they move? $\endgroup$ – Ben51 Jan 21 '18 at 1:15
  • $\begingroup$ @Ben51 My TV's back at home, so I can't go check, but I think the X stays put. These TVs are pretty common so hopefully somebody can chime in. $\endgroup$ – knzhou Jan 21 '18 at 14:47
  • $\begingroup$ Relevant?. It looks like the shape of the pixels could be the key. It is not really the "ideal" diffraction grating you may be thinking about...PS take also a look at this: so cool! $\endgroup$ – valerio Jan 24 '18 at 0:12
  • $\begingroup$ LCD, LCD stereo, OLED, plasma, or CRT scrreen? I'm not seeing an X. $\endgroup$ – Whit3rd Jan 24 '18 at 0:20
  • $\begingroup$ @valerio92 I’ve seen that thread, but I don’t think they ever totally resolve the question either! $\endgroup$ – knzhou Jan 24 '18 at 7:53
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As we know, diffraction pattern is a Fourier image of the grating. In your case the pattern looks very much like two crossed patterns from usual 1D gratings symmetric along one of the directions (like in many-slit experiment). If we try to find inverse Fourier transform of an artificially-made pattern similar to the one on the picture in the OP, we'll see:

Diffraction pattern (intentionally padded with empty space to make the domain square):

diffraction pattern

Its inverse Fourier transform:

inverse Fourier transform of the pattern above

So, the math says the same I was guessing above (and what the answer by flippiefanus was trying to tell). Now, what is actually making this pattern — that I can't tell for sure (I don't even know the model of your TV to do any research). My guess is that this might be the arrangement of the conductors powering the pixels. Or it may be the pattern from one of the layers of light spreading system of backlight (if it's not an OLED TV — these don't have backlight).

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  • $\begingroup$ Aha, that's nice! Now if only someone can find out what part of the pixel does this... $\endgroup$ – knzhou Apr 27 at 19:33
  • $\begingroup$ @knzhou do you know the model of your TV? This might give some clue. $\endgroup$ – Ruslan Apr 27 at 19:34
  • $\begingroup$ Unfortunately I don't, but I'll have a look when back home. It does seem to hold for a wide variety of TVs, though. $\endgroup$ – knzhou Apr 27 at 19:35
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Some basic physical considerations may provide a partial (qualitative) explanation. However, perhaps such a partial explanation would suffice here (?). The idea is to use one's basic understanding of diffraction to reverse the process. In other words, the question is what would the input field have to be to produce the observed diffraction pattern?

First, as a kind of reference, let's consider what would produce a grid of points in the diffraction pattern. For this purpose, let's assume the grid is perpendicular. At the end we may consider what would happen with other angles. In this case, one can write the grid of points (crudely) as the product of two functions $$F(x,y)=F_x(x) F_y(y) . $$ Since they are perpendicular, the Fourier transforms that produced the diffraction pattern would also operate independently on the two directions. So the source field would then also be a product of two functions $$F(x,y)={\cal F}\{g_x(u)\}{\cal F}\{g_y(v)\} . $$ The effect of an arbitrary angle can be introduced by an appropriate affine transform on the coordinates. By Fourier theorems, it then follows that one would have the inverse affine transform on the coordinates of the source function.

Now we consider the case where we don't have a grid, but rather an X. Again we start by assuming the legs are perpendicular. In this case, the diffraction pattern is a superposition $$F(x,y)=F_x(x) + F_y(y) . $$ So the source would have to be a superposition as well $$F(x,y)={\cal F}\{g_x(u)\}+{\cal F}\{g_y(v)\} . $$ Going to arbitrary angles, one then finds that the source field is still a superposition, but with the orientations of the independent variables of these functions not being orthogonal to each other. The source field is produced by being reflected off the front of the TV screen, while being multiplied by the reflectance. The reflectance is a periodic function along two directions. So, one can indeed think of it as two diffraction gratings, but instead of a multiplication of their transmission functions (or reflectance functions), we have the addition of their transmission functions.

The next thing is to use the qualitative nature of the amplitude variation along the legs of the X to determine the nature of the gratings. It is clear that there is some periodicity in this function, together with an overall envelop. One can model such a case (along $x$ for instance) by $$ F_x(x) = h(x)[p(x)\star C(x)] $$ where $C(x)$ is a pulse train (comb function); $\star$ represents convolution; $p(x)$ gives the shape of each pulse; and $h(x)$ is the overall envelop. The inverse Fourier transform would then give one the grating function $$g_x(u) = H(u)\star [P(u) C(u)] $$ where $P(u)$ and $H(u)$ are the inverse Fourier transforms of $p(x)$ and $h(x)$, respectively, and $C(u)$ is again a comb function. At this point one can start to identify the physical origins of all these functions: $H(u)$ governs the shape of each grating line, which is probably determined by the shape of the pixel of the TV screen; $P(u)$ gives the overall shape of the source field, which is probably related to the shape of the illuminated spot. If one can obtain quantitative information about the diffraction pattern, one can even determine more exactly what the function shapes of these functions are.

Hopefully, this provides a better understanding. The key is to realize that the diffraction pattern can indeed be formed by diffraction gratings, but where their transmission functions are added rather than multiplied.

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  • $\begingroup$ I don't understand the content of this answer. You begin by saying that diffraction patterns are products, which I agree with, leading to the generic prediction of a two-dimensional grid. Then you say that in this case, the pattern is a sum. But that's exactly my question -- why should it be a sum? This answer seems to just re-ask my question in more mathematical terms. $\endgroup$ – knzhou Jan 26 '18 at 13:55
  • $\begingroup$ You were making the assumption that a diffraction grating would always give a grid as diffraction pattern and I showed that one can have diffraction gratings that would produce the observed X pattern simply by having a superposition. $\endgroup$ – flippiefanus Jan 26 '18 at 14:01
  • $\begingroup$ Can you spell that out more? What do you mean by "a superposition"? What is being superposed here? $\endgroup$ – knzhou Jan 26 '18 at 14:02
  • $\begingroup$ Superposition simply means addition as is shown in the expression. So the transmission function of one linear grating is added to that of the other linear grating. $\endgroup$ – flippiefanus Jan 29 '18 at 4:50
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    $\begingroup$ This is analogous to answering “why does gravity pull me down?” with “because the force of gravity points down”. $\endgroup$ – knzhou Jan 29 '18 at 9:10
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I can't give a complete answer and I haven't been able to reproduce the exact effect on any TVs in my home (of varying ages between 1 and about 12 years old) using a plain light (not a laser pen). All my screens are anti-glare rather than gloss.

The patterns I observe are a dominant plus-shaped pattern (which may be due to the LED light source in my case), and a much dimmer and fully symmetrical x-shaped pattern (not the compressed x-shape in your image).

I suggest that what you're seeing is the aggregation of "diffraction spikes" from reflections through the multiple rectangular apertures on the screen. This is a familiar effect in photography, for example.

https://en.wikipedia.org/wiki/Diffraction_spike

The lobes of the diffraction in the links given and the photo posted, are not fully symmetric partly presumably because the apertures themselves are rectangular on those screens - that is, because the pixels are taller than wider, there is a bias towards towards horizontal diffraction rather than vertical.

Incidentally, contrary to your assertion in your question, there are clear "points" in the diffraction which proves it is not from a single aperture. You can see this because there are repeating rainbows of diminishing intensity. The diffraction is obviously frequency dependent which is why the separate wavelengths are split out, but if there was only one point, then there would be at most one rainbow on each lobe.

Nor is it a one-dimensional pattern - it's clearly a pattern in two-dimensions.

EDIT:

Picture added to support comments below.

Diffraction grating

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  • $\begingroup$ Okay, that's interesting! But diffraction spikes as described in the link seems to be a property of a camera, not the object the camera is photographing. But this effect happens on these TV screens and not on phone screens, even if the same camera is used for all photos. I think you're proposing the spikes come from the pixels themselves but I can't see it -- can you explain more? $\endgroup$ – knzhou Jan 29 '18 at 11:25
  • $\begingroup$ The analogy I'm suggesting with cameras is that both have apertures - the camera has an aperture behind its lens, the TV has an aperture in front of each pixel to isolate them from one another and create a crisp image (i.e. as a reflection of its construction). If you shine a light into the TV, then it's going through the pixel aperture, being reflected, and then exiting the aperture again. Also, I'd forgotten that until a few years ago there used to be "plasma screens" that were a different technology from LCD - I don't know whether that is why I can't reproduce the effect myself. $\endgroup$ – Steve Jan 29 '18 at 13:03
  • $\begingroup$ Thinking about it more, I still think this answer is leaving something out. The core problem is the same: a two dimensional grid makes a two dimensional diffraction pattern. If the elements of the grid have these square/rectangular features that make diffraction spikes, then you don't get one diffraction spike, you get a whole two dimensional grid of diffraction spikes. $\endgroup$ – knzhou Jan 29 '18 at 20:39
  • $\begingroup$ That is, if this were the whole answer, I would expect a two-dimensional grid of X's, not a single X. $\endgroup$ – knzhou Jan 29 '18 at 20:39
  • $\begingroup$ @knzhou, but it's not a single X - that's just the aggregate effect. A path from a point source through a single simple prism cannot create the multiple rainbows you see. In fact, I've found an image from Wikipedia (edited into my post) that shows the effect of a diffraction grating in one dimension (and how it creates side lobes with repeating rainbows). The only difference with the Wiki image and the TV is that, in the case of the TV, the light source is not from behind, but is from the front and is being reflected back out of the grating. $\endgroup$ – Steve Jan 30 '18 at 3:49

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