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In a more realistic way you would draw wave beams instead of rays, like in the image below. And here is the same situation as an animation. (animated image from Wikimedia - File:Internal-reflection.gif) So you are right. The reflected wave and the incident wave indeed interfere with each other. Where the incident and reflected beams overlap, we get kind ...


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If the incident beam had nonzero diameter, or were a plane wave, then they would interfere to form a standing wave. But rays are an idealization, which assumes a zero-diameter beam. In this approximation, the incident and reflected waves don't overlap (except at the exact point where they meet the surface) so there is no standing wave.


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The link below shows water diffraction better than the images you found. https://www.quora.com/What-is-single-slit-interference For water we have diffraction but there is no interference for the single slit, as you mentioned. In your images there is a slight pattern but that is likely due to the thickness causing extra reflections. Water waves and light ...


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You won't get an interference pattern. This was very nicely explained by @Vivekanand Mohapatra, but there is a more simple reason why. The interference never happens. In the original double slit experiment, you get the interference when the electron's wavefunction crosses both slits and interferes with itself, which causes a change in the probability ...


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Uncertainty principle is not about whether we can calculate the physical observables simultaneously or not. It works for the non-commutating operators only, which means we cannot have simultaneous eigenstates for those operators. Okay, there are many "quantum mechanical terms" which you should know before. you understand why your solution won't work. The ...


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That is a lot of questions, but they are all strung together sensibly, so I will try to answer all of them. Yes, two point sources if they are mutually coherent and are stationary will produce an interference pattern where their beams overlap. If you shift the phase of one of the sources (e.g., by interposing a thin sheet of glass to retard one of the ...


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The way you measure which path the photon takes involves interactions with the photon. You have to operate on it. Your observer on another planet could only observe it if something happened to scatter energy and/or information their way. This would read as noise in the experiment. You would not be able to determine the source of the effect, you would ...


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Writing your second equation using the same symbols of the first equation $(s\rightarrow d)$: $$ w = \frac{m \lambda D}{d}.$$ Then, from geometry, $w = D \tan\Theta$, so you end up with: $$ d\tan\Theta = m\lambda $$ for the double slit diffraction pattern. Which, as you are saying, is different from the general diffraction grating formula: $$ d\sin\Theta = ...


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The width of the slits has nothing to do with the separation of the fringes. Doubling the width of each of the two slits leaves the separation of the fringes the same but each fringe will now have twice a much light forming it. So the answer is $2I$.


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Actually, the most important thing about the light source in thin-film interference is its spatial coherence: the range of angles at which it illuminates a point on the thin film. If that range is small, (and if the spectral bandwith of the source covers the visible spectrum), then the colors due to thin-film interference are easy to see. If that angular ...


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The phase difference depends on the wavelength of light. So the different coloured interference patterns don’t overlap perfectly. This makes it harder to resolve between intensity peaks reliably.


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Monochromatic light is light of a single wavelength, or at least very narrow bandwidth. This makes for very sharply defined interference bands. This is pretty much essential for a good two slit result. As it is usual to use lasers for such experiments, the light will also be coherent. That is, all the photons in the light are in phase across the beam and ...


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I woke up today with the realization that the populations of photons hitting each slit are identical. This in fact renders unnecessary any concern about what frequencies or phases the photons are. Of course this presumes a wave interpretation. No matter what crazy population is passing through a slit at a given instant, the SAME crazy population is hitting ...


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Historically photons are said to interfere ever since the days of Huygens but we physicists know that photons do not truly interfere as that is a violation of conservation of energy. But interference is still taught classically and does a good job of approximately modelling the concept of the slit and the thin film mathematically. So what is really ...


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There is a very simple answer. Interference (in this case) occurs between each individual photon and itself-- not between different photons. To calculate the interference pattern, you simply add the patterns due to the individual photons. When this is done, you obtain the pattern that is actually seen.


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Let the x-axis be perpendicular to the screen passing through the center of both lenses and the y-axis be on the screen on some arbitrary direction. Let $\theta$ be the angle formed by the ray and the x-axis after the first lens. Let's think it in 2D (xy plane). The interference of the ray with angle $\theta$ will be imaged on the point who forms an angle $\...


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The most significant recent development in the double slit experiment was running the experiment with single photons (1960s - you can google it). What's remarkable is one still gets the "interference" pattern ... but it seems impossible, how did a single photon manage to "interfere". You will read statements like the photon interferes with itself, or the ...


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Is the peephole even necessary? Because the slit by itself would be narrow enough to create coherent light? To see interference effects reliably, the phase of the light should be in step everywhere across the hole. In other words, the spatial coherence length should be larger than than the width of the hole. In general, passing light through a small hole ...


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There exists two types of coherences: temporal coherence, which describes how the field amplitude $u$ at time $t$ is correlated with the field amplitude at time $t+\tau$, $$\langle u(x,t)\cdot u(x,t+\tau) \rangle = \lim_{T\to \infty} \frac{1}{T} \int_{-T/2}^{T/2} u(x,t)\cdot u(x,t+\tau) \,dt $$ When considering temporal coherence, we are interested in the ...


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You are partly right but it also comes down to the different energy levels of the electrons. It is a random process as to when electrons will fall to lower energy levels releasing their energy. The direction that energy is released is also random.


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Because filament lamp isn't just a single source, but consists of enormous numbers of different sources. Different atoms emits light in random manner. There is no particular pattern in emission of light from these sources. And that's why at some point, the phase difference of different light waves are not constant. That's why light emitting from filament ...


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The slit does not make the light beam coherent. The beam's coherence is primarily a function of the source. You are right that light from a filament is incoherent. It is temporally incoherent (multi-wavelength) and spatially incoherent (spread out in space). A slit does increase the spatial coherence of the light: if you consider the slit to be a source, ...


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Just as it a choppy (incoherent) sea there are clear wavecrests the light emmited from the incandescent lamp has locally clear wavecrests (although on a much shorter scale). We say that the incoherent light has short range coherence. The pinhole samples only a small part of the wavefront from the light emitted from the lamp, and as long as the ...


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Coherency is a very important property of waves, whether they are light, sound or water waves. It is a complex subject especially as physicists have tended to apply the wave properties of water to light ever since the time of Huygen. But there are important differences, for example water waves through a pinhole do diffract but there is no interference ...


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Imagine that you had two sources emitting waves of exactly the same amplitude $A$, same frequency with a constant phase difference. Where there is superposition of these wave there will be an interference pattern. At some positions the waves from the two sources will arrive in phase with one another and the resulting amplitude will be $A+A = 2A$ with ...


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