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Light is at first a wave, and like any wave the simplest description of his propagation is the Huygens-Fresnel principle : at any point a new wave is re-emitted in every direction. When you sum the contributions, in free space the only location that does not vanish trough interferences is the next location in straight line direction. But if there is a hole ...


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The light waves from the many slits overlaps and you see the result of that overlapping. Along certain directions (at certain angles) the waves of a particular wavelength from all of the slits overlap in such a way so as to reinforce one another to produce light of high intensity. That direction depends on the wavelength of the light and so for different ...


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Yes it would depending on how thick the smoke was. Photons intercepted by the smoke would be scattered or absorbed but the photons that still have a clear shot to the screen would contribute to the original fringe pattern. The pattern would still be the same but not as clear.


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Essentially, the smaller the aperture gets the more it starts acting like a point source. As mentioned by dominecf, a point source creates circular wavefronts therefore resulting in a "bending" of light. A quantum mechanical explanation of the phenomenon has to do with the Heisenberg uncertainty principle which relates the uncertainties in position and ...


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In the case of refraction we have to know that electromagnetic waves travel at a different velocity in different mediums, for example they propagates faster in air that they do in water. So, if we have a ray of light which is emitted from below the surface of water it will accelerate when it reaches the surface of the water. Now, If the ray hits the surface ...


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Neither image displays diffraction. They both illustrate shadowing. The second photo was apparently shot on a clear sunny day. All of the rays come from the sun. Some are blocked, those that aren't go straight down to the floor. The second was probably taken on a hazy day. Rays originate from the haze; they come from all directions, and make their ...


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The trick is that because the "slit" is infinitely wide, you shouldn't work in the far-field approximation (Fraunhofer difraction integral which leads to fourier optics), but with distances computed to the square order (Fresnel diffraction integral). The results are appropriately named Fresnel integrals (this time this is a name of a special analytical ...


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any angle over theta 90 will mean that the diffraction will be going behind the diffraction gratings which is impossible. so 90 is the maximum that you can get this is why you have to round down the decimal answer you will get.


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It is obvious that moving wave fronts will made moving fringes on an observer screen, will they not? If the detector screen is in the yz plane and the slits are at particular y values and the light is originally going in the x direction, and you had light polarized in the z direction then indeed the electric field hitting a single point on the screen ...


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For a classical double slit, there is a classical electromagnetic field and so there is an electric and a magnetic field at every point in space. But the phase that is drawn is more like a tracer. If you had a wave like $A\sin((x-vt)/\lambda)$ then it has an amplitude, a speed, and a wavelength. You could draw the place where it is $+A$ and see how that ...


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Of course they move. How could they not? I suspect that what is confusing you is that you are mis-interperting what a bright fringe on the screen is. If you are imagining that a bright fringe represents a point of instantaneous high amplitude then you have the wrong idea: a bright fringe is a point of maximum optical energy delivery (power). The actual ...


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A slightly more general construction than Wendy or Bill's is "A wavefront is a contiguous region of constant phase". Wendy's choice of "crest" is just the selection of a particular phase ($0 \pm 2\pi n$ for wave represented with $\cos$ for instance), and that is the usual choice when visualizing wavefronts, but there is nothing that prevents you from ...


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What represents the lines are coherent photons. The only thing oscillating are photons, not waves. The photons are coherent because their all emitted in phase from the same point source. Because they are in phase gives them the appearance of a wave like this sketch:


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Here is an updated video concerning single-slit diffraction (functions). Waves: Light, Sound, and the nature of Reality https://www.youtube.com/watch?v=Io-HXZTepH4


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A wavefront is the crest of the wave. When you go down to the beach, and see those things called waves, the front is the whole line that is at the same height. In electromagnetics, it's the same thing. It's the points that are at the same height. In your diagram, the black curves represent a unit of cycle, and the two waves through b and c can either ...


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Strictly, you can't just say the light behind a slit is coherent. You can say it has a certain coherence time $\tau$, meaning it can interfere with a copy of itself which was delayed by time $\tau$ it has spatial coherence with respect to the light behind another slit, meaning they have a (somewhat) fixed phase difference and can interfere with each other ...


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A point source has a spherical wave front with the intensity falling as 1/r^2. The fronts are in constant phases because there is no dependence on theta and phi in the intensity. For an aperture with a width see the question and answer here and links therein. It depends on the width of the slit, the frequency and the coherence length.


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Actually what's necessary for interference is a constant phase difference and nothing else. When we have a constant phase difference the super position of the waves gives us the interference pattern but when the phase difference is not constant the superposition just averages out to give a uniform intensity. The same frequency constraint is a consequence ...


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From the link you shared Far away from the aperture, the angle at which the first minimum occurs, measured from the direction of incoming light, is given by the approximate formula $$\sin\theta\approx 1.22\frac{\lambda}{d}$$ for small $\theta$ $$\theta\approx 1.22\frac{\lambda}{d} $$ where θ is in radians, λ is the wavelength of the light and d is the ...


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Ground glass is a very good diffuser, which means you not only have light being re-emitted in a near-Lambertian pattern but with a pretty decent decoherence as well. So not only do you need an angular dependence (solid angle) for the amplitude, but also a random phase angle. I don't know how much local phase angle correlation (i.e. the likelihood of the ...



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