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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 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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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 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 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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"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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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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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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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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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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