(Might seem like a duplicate but is not)  The original Young's double slit experiment makes use of a single incoherent source being incident on two different slits to get two coherent sources.  The original source is incoherent.i.e There is no constant phase relation in the light being emitted.So The phase from 2s ago is in no way related to the phase from 0s ago.  But in YDSE.The two waves that interfere were not produced at the same time. So they are not exactly coherent.Also the two slits are not exactly equidistant from the source.Even at 1-2 centimeters pattern is still observed  How can these two beams interfere?  (It isnt  due each wave (emitted by an atom )extending for some time ..So that for small distance difference the phase is still constant,is it)(If it is not..Then could you make sure that  your answer clearly explain the fact  1 source with "Time LAg" would do but 2 lasers wouldn't) do
36516
In , a home experiment is presented which appears to produce a very good double-slit interference pattern with normal sunlight.  The experiment is an empty cardboard box with a visor and a placeholder for a microscope slide with two slits on one side. This is arranged with the slits and visor facing the Sun, so the interference forms on the bottom of the box.    They claim to observe a good interference pattern from the two slits:    Discussions of interference in optics textbooks often stress that coherent light is needed to produce such patterns, and that sunlight and other thermal sources of light do not have such coherence. How, then, is this possible?
102114
I have a little trouble with the . I know of the  which states that it is not possible to duplicate any state. One the other hand, I know about the stimulated emission which out of a photon produce exactly the same (wavelength, polarisation, etc...). Maybe the fact is that the excited atom do a measure. (Let's say it can stimulate emission with only one direction of polarisation.) But now if I have a statistical number of atoms with random &quot;polarisation&quot; direction, I should be able to copy any incident photon. I've made a cloning machine. This cannot be true because of the no-cloning theorem. But I can't figure why.
64486
Consider the scenario where you measure the time it takes for light to travel to the left 10 meters and to the right 10 meters. Both measurements will take the same time, even though we are moving through space at crazy speeds. This must mean that light is not moving relative to "space" as a whole. What does it move relative to? The light emitter? If so, try shooting two beams of light 10 meters from the wall. The first time the emitter is stationary; the second time it is moving at 100 m/s. Am I mistaken in thinking that it would hit the wall slightly faster? Wouldn't this light be moving faster than the light emitted from the stationary source?
8733
In many experiments in quantum mechanics, a single photon is sent to a mirror which it passes through or bounces off with 50% probability, then the same for some more similar mirrors, and at the end we get interference between the various paths. This is fairly easy to observe in the laboratory.  The interference means there is no which-path information stored anywhere in the mirrors. The mirrors are made of 10^20-something atoms, they aren't necessarily ultra-pure crystals, and they're at room temperature. Nonetheless, they act on the photons as very simple unitary operators. Why is it that the mirrors retain no or very little trace of the photon's path, so that very little decoherence occurs?  In general, how do I look at a physical situation and predict when there will be enough noisy interaction with the environment for a quantum state to decohere?
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I am reading Berkeley Physics Course vol. 4 (Quantum Mechanics) , chapter 4 (photons). (1)  Section 46: book says: consider a typical photon emitted by  the source. It can be regarded as a a wave train of finite duration, spreading out in all directions in space and carrying a total energy $\hbar\omega$. If this is true, a photon emitted by a star which is few light years away, would have spread over a sphere of radius of few light years. Then it is not possible to &quot;detect&quot; that photon using a detector.  Then  why does the book say like this?  In section 47, the book gives same example of photon spreading over light years. But in this section, it says that how the people can use this as a paradox; and clarifies that paradox by saying that classical expression for &quot;energy density&quot; refers to a large number of photons, and not for single photon.  So does photon really spread in all directions?   (2)  In Section 38 book says: almost monochromatic photons can not be split into 2 photons of the same frequency which carry only a fraction of the energy of the original photon. While in section 48, it says (with reference to double-slit diffraction experiment) photon came through BOTH the slits.  Does a photon REALLY goes through both the slits? Doesn't it indicate that a single photon can be split into two photons of same frequency ?   Any help will be appreciated. Thanks
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In double slit experiment only small sized particles were used. however if we use large objects such as a tennis ball, the expected pattern is not observed.   Why the wave nature is not visible for large objects in double slit experiment ? How do I then prove that large objects also have wave nature?
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I was wondering why a laser beam diverges. If all the photons are in the same direction, I would imagine that it would stay that way over a long distance. I am aware that a perfectly collimated beam with no divergence cannot be created due to diffraction, but I am looking for an explanation based on  photons rather than wave physics.
91372
I just want to first say that I'm aware I am asking a question due to my own confusion and ignorance and not because of anything to do with special relativity. I hope that's alright.   What I'm confused about is whether two beams of light moving in opposite directions away from eachother, have a relative speed of $c$, or $2c$.   The thing is, we can always say it's $c$, but then when we look at the relative distance travelled for the time elapsed, the result is $2c$.   Can anyone explain what is going on?   Response to Alfred Centauri 22/08/2016   I question your reasoning for two reasons. Firstly to photons moving in opposite directions was one of Einstein's examples of the counter-intuitive implications. So what you're saying wasn't around then, and Einstein himself was misconceived. In fact you can find instances opposite direction light beam as recently as 10 years ago. Phillip Green in his popular science book uses that very example in his introduction. So the question is whether this is a formal adjustment to SR, backed by a publication and a reasonable consensus. Is it?   The second reason is to do with your reasoning. You say it is not a legitimate measurement because there is no inertial frame in which one or the other photon is at rest.  That at-rest criteria as you define the constraints, is equally applicable to two beams coming toward eachother. In fact almost every problem setup has no way to assume that one of the two frames is stationary. The way it's handled, I believe, is mathematically by the simple addition of the equal and opposite motion. This is also implicit in the Laplace transforms, I think.  The reason this can be done, I think, is because all inertial frames are assumed to be equilvant....they are part of a global reference frame.   A further matter derives from the fact the relative motion of two frames can be represented as tangential and radial components. If what you are saying is correct, then one of the two components cannot be measured in almost all cases.   I definitely want to make clear I am a novice in the theory, and you probably are not. It's possible this is all misconception, except for the historic part of my response which can easily be confirmed. Please do let me know if you can as I am trying to learn and need my mistakes corrected, especially the long running ones.
57794
How can a light passed though a single slit produce a similar interference pattern to the double-slit experiment? How does the diffracted wave produce the points of cancellation and reinforcement, if there is only one wave?
56669
So I've been reading a bit and watching some videos about the double slit experiment, and therefore the wave particle duality; I've also read this "implies" that a particle can be in two places at the same time. Can you clarify this, and how could that be possible?
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I'm not a physicist; I'm just curious about what would happen.  According to this video "" (at 2:50):     "Somebody decided to leave the detectors on but just not take any data. [...] So if it was also going to a magnetic tape - there was no magnetic tape loaded. [...] And what do you think happed? They got this. They got the diffraction pattern."   So the act of measuring (beeing concious of the data) destroys wave-form pattern?  If so lets consider these cases:   What if I'd throw a dice after the experiment to decide if I should look at the data or to destroy it? (e.g. look at the data if the number on dice is even) What if the information about measurement would need to travel longer period of time than the particle would need to hit the wall?    To above cases: Would it mean that information travelled instantly - faster than light or that "the particle can look into the future"?
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If we take the double slit experiment as a way of demonstrating the wave-particle duality, which types of particles would show an interference pattern?  For example, I know that electrons show such a pattern. But do protons, too? What about atoms? Where is the boundary between "wavey particles" and "classical particles"?
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Last night my daughter was asking why a mirror "always does that" (referring to reflecting a spot of light). To help her figure it out, I grabbed my green laser pointer so she could see the light traveling from the source and reflecting off the mirror.  But as we were playing, I .  Rather than one spot, there were several. When I adjusted the angle to something fairly obtuse    The effect became quite pronounced    And when you looked closely, you could actually see several beams   (Of course, the beams actually looked like beams in real life. The picture gives the beams an elongated hourglass shape because those parts are out of focus.)  I made these observations:   The shallower the angle, the greater the spread of the split beams and resulting dots. The directionality of the reflection is due to the orientation of the mirror, not the laser pointer itself. Indeed, by rotating the mirror 360° the string of dots would make a full rotation as well. I can count at least 8 individual dots on the wall, but I could only see 6 beams with the naked eye. If you look at the split beam picture you can see a vertical line above the most intense dots. I didn't observe any intense spots of light there.   And when I looked closely at the spot where the beam hit the mirror    you can see a double image. This was not due to camera shake, just the light reflecting off the dust on the surface of the glass, and a reflection of that light from the rear surface of the mirror.  It's been a few years since college physics, I remembered doing things like the double split experiment. I also remembered that light seems like it does some strange things when it enters liquid/prisms. I also know that the green laser has a certain wavelength, and you can .  Why does the mirror split the laser beam? How does that explain the effects that I saw? Is there any relation to the double split experiment, or the wavelength/speed of light?
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In the double slit experiment we have something like this     Light waves from two sources interfere at screen to give us this pattern. To reach the middle (that is a point on the perpendicular bisector) of measurement screen waves will travel equal distances (at sources they are in phase because of the same main source) and hence will be in phase at the middle of the screen and hence a constructive interference will occur.    We can write things a little more mathematically, if at time $t$ the crest of the light wave is at the sources (secondary) then at time $t+nT$ ($T$ is the time period) crests will be at the middle of the measurement screen and hence a maximum will occur. Here, I have assumed, for simplicity, that distance between the source $S_1$ and the middle point M is $n\lambda$.    Now, my question is when there was a crest at $S_1$ and $S_2$ then we had a maximum at $M$, but when there is no disturbance at sources (I mean $E(t)=0$ and $B(t)=0$) then there should be no disturbance at M too and therefore we should obtain a dark fringe (no light) for this particular interference. Why does the interference pattern remains static? The bright fringes should be turning on and off periodically.    My question comes down to normal light also, why do we see the light even when the electric field and magnetic field components reach their zero position periodically? We should see light as being turning on and off, off when $E(t)$ and $B(t)$ becomes zero and on for every other time.    P.S. :- My $M$ in the figure doesn't coincide with the maximum but by $M$ I meant the middle of the screen where the first maximum gonna occur.
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I had always kind of wondered why we didn't see interference in things like windows -- we were taught that the condition is that the thickness of the film/slab/medium just has to be an integral number of wavelengths, $d=m\lambda$. My logic was, if visible light has wavelengths ~400-780nm, any window pane has got to be an integral number of wavelengths (or close to integral) for some of the wavelengths in the visible.  For example, if a window is, $6mm = 6\times10^6nm$ thick, then you should see interference for wavelengths (and some range around them) 200nm, 250nm, 300nm, 400nm, 500nm, 600nm... Because 6mm is an integral number of each of those.  Obviously that would be really annoying and isn't the case. I found  (warning: PDF) source that seems to explain it and I was wondering if what they're saying is sound. On the 10th page (p171 for them), it says that the reason we don't see this interference is that no window is that smooth, it has a variance in the thickness of at least several wavelengths, so the interference effect is effectively average out among all those thicknesses, which usually gets rid of it. They do the math and show that.  Is that the reason?  If so, that leads me to two questions: Could we, if we were really careful and used something like MBE, make a glass window with that sort of accuracy, and then see interference effects in the optical range? I know that quartz crystal monitors can easily give you nanometer accuracy for evaporators and sputterers, but they also have pretty limited growth rates.  My second question is, then, do we see interference effects in glass windows with electromagnetic radiation of a comparable wavelength, like microwaves?  Thanks!
2134
Yesterday I read that we can affect the path and the 'form' (particle or wave) of a photon after the fact ().  Part of what is puzzling me is the beam-splitter. Are the individual photons actually being split into two new photons of lesser energy?  This  implies that you cannot split a photon but it seems that beam splitters do exactly that.
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When I (beginner) learnt Michelson Morley experiment, it was expected that there should be a shift in the fringe pattern when the setup was rotated by 90 deg if the presence of ether is true. But why should there be a shift in fringe after the setup was rotated, assuming the presence of ether?