# Tag Info

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Incomplete answer: assuming the wall is a perfect reflector, we can remove it and ask at what point the sound reaches a second car driving away from the wall to the northwest, in a path reflected from the original car's through the line of the wall. We know the position $O$ of the first car when the driver pressed the horn, and at any given time $t$ we can ...

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Sound leaves the vehicle. When it strikes the wall, it reflects at a 45 degree angle and returns to the car. The triangle you will be using has legs that are length X, with a hypotenuse that is length Xsqrt2. That is for evey 1 foot of length in a leg, I will have 1.414 feet of length in the hypotenuse. What you want to solve for is time based on this ...

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I assume you ask if there is also reflection when refraction occurs. The answer is yes. It happen when light (or any kind of wave) passes from a medium 1 to another medium 2, those medium being characterized with different index (i.e. wave velocity). At the interface between the two medium, in the case of light wave, the electric field has to obey certain ...

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Everything is about electromagnetic waves. At the border of medium one of the components of electric field have to be continuus and other component of electric displacement field. Such calculations leads to Fresnel equations

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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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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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The diagram you want to use looks something like this: Depending on how much attenuation there is in the membrane, you need to consider the potential of multiple reflections (or not). I actually analyzed this problem in some depth - considering not only the intensity of reflections on the different surfaces, but also multiple reflections and even the ...

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Simple answer: no. Reason: Let's say you are a photon. A mirror is placed in from of you. If you do not smash in to it then it is traveling as fast as you. there for, the photon will never reach it to be redirected backwards.

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The reason we do this is that the Poynting vector $\vec{E}\times\vec{H}=\vec{E}\times\vec{B}/\mu$ is in the same direction as the wave vector $\vec{k}$ and shows the wave direction. So, to convert a forwards travelling plane wave to a backwards travelling version of itself, we need to change the sign of only one of the $\vec{E}$ or $\vec{B}$, If we changed ...

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I've made this into an answer because it's too long for a comment, and I really want to show the pictures. It is tempting to think of visible light as "close enough" to (near by wavelengths) and to conclude that "yes, actually, the yellow does affect it. I want a mirror without an obvious tint" However you are wrong, Physics will slap you down. Exhibit A ...

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If you look at the reflectivity of gold (vs silver or aluminum) you can see a plateau at wavelengths below 500 nm source: If blue wavelengths are not reflected as well as other colors, the resulting image will look "more yellow" - which is what you see. At longer wavelengths, gold is a very good reflector (better than the other two above 600 nm). It also ...

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The reflectance of a typical mirror depends on the metallic coating used, but usually it is usually aluminum or silver in more expensive mirrors. Special optical coatings can be used to reflect or scatter EM waves at specific wavelengths. Here is a photo of a mirror reflecting IR (700nm - 1mm wavelength): Mirrors are also used to focus X-rays (.01nm - ...

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You are seeing only one color from each droplet. Even though each droplet reflects the whole spectrum, only one color gets to your eye. The rest of the light from a single droplet is sent somewhere else than your eye (maybe into the eye of someone standing near you). So if you see a thick blue band in a rainbow, the blue band is formed by many reflections ...

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This Wikipedia article gives a LOT of detail regarding rainbow formation https://en.wikipedia.org/wiki/Rainbow To make a long story short, each wavelength (color) reflects off of the back of a raindrop at a slightly different angle, of approximately 42 degrees. To see the rainbow, the sun has to be at your back, to allow the reflected light to enter your ...

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The reflexion co-efficient equation you have given is part of the Fresnel Equations. There are two ways wherein this equation varies with frequency: ultimately both down to the variation of the refractive indices $n_1$ and $n_2$ with frequency. Clearly the equation you have written can vary with frequency if $n_1$ and $n_2$ do so: also $\theta_2$ does ...

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The frequency (wavelength) dependence comes in through the refractive index, and it is specific to the medium. In the lab, the manufacturer of your optics will generally provide this dependence, as in e.g. this resource. Calculating this theoretically is generally a hard problem, though.

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A parabola is a conic section, like circles and ellipses, and all three types of curve can be defined by a focus (or in the case of the ellipse two foci). In the case of a parabola we draw a straight line (the directrix) and choose a point (the focus) and the parabola is the set of points that are an equal distance from the directix and the focus: (image ...

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It ain't necessarily so. Some large telescopes are using silver coatings. For example the Gemini telescopes (Boccas et al. 2006). The reason for going with silver rather than aluminium is a slightly improved reflectivity in the near- and mid-infrared (99.1 per cent at 10 microns), but also that the emissivity of the silver coating is around 38% that of ...

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