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I'd like to add another take on Cape Code's answer. Holography works because, given reasonable physical assumptions, solutions to the Helmholtz wave equation are uniquely defined by the values of the solutions on one plane. So if we can light a phase / amplitude mask encoding a particular wave equation solution on a plane with a plane wave from a laser, the ...


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If the space is truly "open", then reflection should not play (since there is nothing to reflect off), and refraction can only play if there is a change in refractive index (which "normal air" would not have). That seems to leave two things: the fact that the entire body vibrates when you speak - and thus there is some fraction of the sound being directed ...


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Generally that would be diffraction - the spreading out of a wave. Reflection is bouncing off objects - e.g. an echo of someone's voice. Refraction is about a wave changing direction e.g. water looks shallower than it is when viewed from above in air because the light from the bottom is refracted (bent) at the water/air surface. In this case - diffraction - ...


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Because of diffraction. When the photographic plate is exposed, it blackens and change its refractive index in a spatially varying manner. When illuminated again by the reference beam it can be considered as an amplitude transmittance. In 2D you would define it as the complex function: $$t(x,y)= T(x,y)e^{i\theta(x,y)}$$ Let's say your reference beam can ...


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Picture yourself looking into a large mirror on the wall. Now picture the mirror is made up of smaller, tiled mirrors. You will still see your reflection. If you begin to remove the tiles, so that there are only a few left, you can still use them to reconstruct the image of your face that was given by the original mirror. This is what is happening with ...


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Greatly rewritten based on feedback in comments In order to understand this issue, it is worth considering what a telescope (or any optical / radio imaging system) really does. Taking a simple parabolic mirror, the shape is chosen such that the total path length for all rays "from infinity" to the focal point is the same. By making the path lengths the ...


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x-ray diffraction is not caused by atoms absorbing radiation. x-ray diffraction and diffraction by gratings do have an underlying mechanism in common. In both cases, when two incoming rays of waves (x-rays or light waves, in the case of optical diffraction grating) both rays bounce of the crystal or grating, they have travelled a different distance, say ...


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A crystalline solid is more than a bunch of atoms arranged nicely, because the atoms are "feeling" each other. I don't see scattering at crystals as the interference of absorption/emission of each atom separately (and I don't think that's the usual approach either), but rather as reflection at planes determined by the crystal structure as a whole. So, Yes, ...


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Typically the order of magnitude of angular spread of a fringe in a pattern can be given by lambda/Width. If lambda is much smaller than width the diffraction pattern will be difficult to see. In the other case of lambda much greater than the width the entire screen would be covered by just the first fringe.


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If the slit is much smaller than the wavelength, there will not be significant path-length difference between beams passing through different points inside the slit. If the slit is much larger than the wavelength, most of the beam passing through the slit will not be affected by the edges of the slit. I also guess that Born had in mind Fraunhofer diffraction ...


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The wavelength of sound is of the order of 1 meter. So any objet visible to the eye can deflect it. In the case of light , we need sofisticated equipment to observe the effect of diffraction. Diffraction of light an be best observed when a small slit is used. THe slit should be the order of few microns. Then the light is diffracted which is observed in the ...


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As a general guide, if we consider diffraction of a wave with wavelength $\lambda$ from an object of size $d$ then the characteristic angle of the diffraction is given by: $$ \sin\theta = O\left(\frac{\lambda}{d}\right) $$ where the $O()$ symbol means of order i.e. roughly the same as. So for example in a Young's slits experiment, where $d$ is the slit ...



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