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52

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 ...


46

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 ...


3

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 ...


2

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 - ...


2

Short answer: Yes Slightly longer answer: If you scatter the wavefunction of a propagating electron from a potential (surface of a material for example), it generally splits into two parts - a transmitted part and a reflected part. As the names indicate, the reflected part represents a 'reflected' electron, the transmitted part a transmitted one. However, ...


2

The Poynting vector is useful not because we say so, but because of Poynting's theorem, which in essence states that the Poynting vector can usefully model how electromagnetic energy is moved around a system of changing electromagnetic fields. More precisely, you can define a quantity $$ u=\frac12\left(\varepsilon_0\mathbf E^2+\frac1{\mu_0}\mathbf ...


1

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 ...


1

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 ...


1

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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