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In addition to the quantum mechanical model of electrons being in bound energy states , in orbitals around an atom, the band theory for solids, a quantum mechanical model, is necessary to explain the interaction of metals with light. The electrons around an atom occupy more and more bound energy states. A photon that does not have the appropriate energy to ...


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The following diagram may be helpful: If you have an incident ray that is polarized with the E field up and down (in the plane containing the incident ray and the normal to the surface), then when that ray is refracted, it contains a component of electric field that is perpendicular to the refracted ray (and still in the same plane). The reflection is ...


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A more accepted term for sound treatment is acoustic treatment. Acoustic treatment deals with the quality of sound, whereas sound proofing deals with the attenuation of sound. A recording studio may need to be acoustically treated in order to increase the fidelity of the recording equipment. It also may need to be sound-proofed in order to avoid ...


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The answer is "Yes" but not the way you might expect. It is possible to construct a telescope mirror from rotating liquid metal.Mercury used to be used but something like Gallium is safer and better. So print a cradle for it, put in the Gallium, raise the equipment past the melting point (about 30 degC), spin gently to get a parabolic surface, and then ...


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If the wall is made of a perfectly reflective material (such as a mirror) no, you won't see the dot. However, most walls are covered with paint or made of a diffusive material : when the laser beam hits the wall, its light gets diffused in all directions. Thus you are able to see the laser spot on the wall. To summarize : reflection depends on the ...


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The answers here is beautiful. But, i’ll give another simple example. Just take a glass square. The critical angle of glass is 42 degree. So pass a light ray through a very small slit , such that it strikes the side of the square glass slab, at 45 degrees. It will reflect and will ALWAYS strike the faces of the slab at angles greater than 45 degrees. And ...


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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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The conditions for total internal reflection from an air-common glass interface are: The light has to be travelling in the glass. The angle of incidence has to be larger than the critical angle.


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For a pane of glass, the two surfaces are parallel to each other. That means that the exit angle from the 2nd surface will be the same as the entrance angle of the first. Whatever light enters (and isn't absorbed) will exit: $$n_{air}\sin\theta_{1}=n_{glass}\sin\theta_{g1}$$ and $$n_g\sin\theta_{g2}= n_{air}\sin\theta_2.$$ A quick sketch of two parallel ...


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As far as I understand wave physics, the only way to reflect a wave is to have a change in the medium they are travelling through. Two waves can have constructive or destructive interference however but to reflect each other isn't possible.


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What you are observing is the apparent total brightness reported to you via the optical sensing portion of your brain. A lot of things go into that subjective feeling: how wide your iris opens, how many rods/cones in the retina are stimulated, and what the relative peak and minimum intensities are in your field of view. If you in fact used a calibrated ...


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Think of the rays at the extremities of the cone of light. The angle of incidence on the surface of the ceiling from these rays and so the scattering angles from the surface of the ceiling will be greater that those rays which hit the ceiling at smaller angles (you concentrated cone). So there will be more light heading towards the walls with your wide ...


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Yes, it's necessary for example on some telescopes to keep the image the same way up on a camera as the telescope tracks across the sky. VLT naysmyth focus There are a couple of optical designs, using either rotating prisms or a rotating set of mirrors. Look up field rotator. eg http://www.ing.iac.es/~eng/optics/documents/OPT-WHT-001.pdf


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Yes you can. Moreover, you don't need any lenses and such things, you only need a medium with Refractive index $n(\vec r)$. Using Fermat's principle (https://en.wikipedia.org/wiki/Fermat%27s_principle) you can calculate the path of the light travelling through a medium with $n(\vec r)$. You just have to choose a proper $n(\vec r)$ function.


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"When light is normally incident on a glass surface, about 4% gets reflected and the rest is transmitted. The reflected wave is 180° out of phase with the incident wave." In my opinion, this statement is only true if the slab is finite, That statement comes from Fresnel equations, which say that when a wave pass from $n_1=1$ to $n_2=1.5$, the ...


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Recall that from energy and momentum conservation when the Snell's law is applicable, then R=0. But when R=1, then i=r.


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Yes. When you look at a (Schwarzschild) black hole with a sufficiently good telescope (according to GR), you see arbitrarily (infinitely) many "Einstein rings". Between each concentric pair of rings, there is an image of the entire surrounding sky. These images correspond to the light paths (null geodesics) that are unbound but very close to the photon ...


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Color is just a "projection" of spectrum to what we can perceive. So... almost every color can be reproduced in ifinitely many combinations of different wavelengths! Essentially, your real physical state is a function $j(\lambda)$ that for every wavelength $\lambda$ tells you the intensity of the light. You have an infinite amount of information here, as the ...


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It is simple : When you put all the color from visible spectrum on a surface (so it reflects all wavelengths hitting it), You will get White Color When you extract all the color from visible spectrum on a surface (so it absorbs all wavelengths hitting it), You will get Black Color Newton demonstrated that white light could be broken up into its ...


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It depends on your definition of a "color". There is a crayon in my box of coloring pencils that is "white" - you could call that a single color. But if I look at the spectral components of a light source that my eye perceives as white, there could be many different compositions, all of which look "white" to me. It might be "all the colors of the rainbow" ...



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