electromagnetic-radiationreflection
This video (a bit after 2:00) says that the back of the parabolic dish should be painted black before applying the Mylar over it.
The purpose of the mirror is to concentrate all the light hitting its surface on a single point in order to melt what is there or boil water.
I know that black absorbs all the light so, to me, it looks like a very bad color for a mirror that should reflect light, that is the opposite of absorption.
What am I getting wrong?
Wrong:
"since infrared waves have a shorter wavelength"
Infrared has longer wavelength than visible and visible longer wavelength than ultraviolet .
White is a term for visible light mixed wavelengths. In the plot you can see that almost half of the sun's radiated energy arrives as visible light. The white buildings reflect this visible light which otherwise,impinging on the surfaces would be absorbed and turned into infrared by the interactions, adding to the arriving infrared.
What is absorbed and what is reflected depends on the chemical bonds of the surfaces, whether the incoming radiation can excite molecular states of the materials. Infrared is in frequencies/wavelengths of the black body radiation of bodies in the temperature ranges comfortable for the human body, so they easily raise the vibrational and rotational levels of solids and liquids and the kinetic energy of gasses.
37C curve seen here practically all in infrared, and lower temperatures more so.
Thus white paint will not reflect infrared as efficiently as visible, a large part of infrared will be absorbed as also some part of visible will scatter at the surface and degrade to infrared. Infrared can be reflected by metal mirrors, from the collective fields in metals . If you put aluminum foil in front of a heater you are sheltered from most of the heat which is reflected, but some of it is absorbed as can be seen by touching the foil.
If I expose an object to EM radiation only from the infrared spectrum, will it only reflect back infrared?
Yes, but most of it will be absorbed ( except by mirror metal surfaces) because the materials have the receptors for these wavelengths. This is due to the fact that larger wavelengths have photons with less energy which cannot excite higher energy levels.The energy of the photons goes as h*c/lamda where h is plancks constant, lamda the wavelength and c the velocity of light.
Is this true for other types of EM radiation?
No.Visible and ultraviolet by scatterings degrade their energy down to infrared frequencies, depending on the material.
Is it possible to make an object that looks white and absorbs a lot of infrared radiation?
Usually most of the infrared will be absorbed except by mirror metal surfaces.
If an object reflects most of the EM radiation that it receives of a particular wavelength λ, will it also reflect most of the radiation it receives of wavelengths less than λ (and absorb most of the radiation of wavelengths larger than λ)? Is this why objects that reflect most visible light (and hence look white) also reflect most infrared radiation (since infrared waves have a shorter wavelength)?
There is no such rule. It depends on the material and its chemical bonds.
Ice and snow reflect light quite a bit, so you might be better off focusing the light onto something black and then using the radiant heat from that to melt the snow. But in case you want to do it directly, I would suggest pointing the axis of the parabola at the sun. You don't need a complete parabola (whatever that means!), as any rays that hit it parallel to the axis will end up at the focal point.
A parabola with equation $x = \frac{y^2}{4a}$ (opens rightward) has a focus at $x=a$. So assuming the light comes in along the $x$ axis, the ground would some line passing through the point $(x=a,y=0)$ sloping downward. I.e., $y= \tan\theta\, (x-a) $. You only need the part of the parabola that is to the right of that line (i.e., above ground)
Best Answer
The important thing to notice here is that the layer of black goes behind the Mylar layer. So the light that will be absorbed is whatever penetrated the surface aluminum layer (a small fraction). That is, most of the light will be reflected off the front surface, and only a small amount will be affected by this choice.
Now we ask "Why don't we want that light reflecting?
At this point it helps to have a little understanding of wave optics. The rainbow colors seen on the surface of puddles in the road come from the interference of light reflected from the top of a floating oil layer and that reflected from the transition from oil to water. When you look at a colored band you are seeing the light from some of the visible spectrum that is constructively interfering while light from another part destructively interferes. More interesting still are the black stripes sometimes seen near the edges of the oil slick: in that regime the slick is much thinner than a wavelength of visible and because the front reflection is phase inverting and the back one is not (don't ask, you don't actually care right now) all visible light is destructively interfering in reflection.
The Mylar is designed to be as thin as possible, and the energy from the sun covers a wide spectral band, so in any given direction from the surface some wavelengths will be destructively interfering if reflections from the rear surface are allowed.
You read that right: the "extra" light reflected from the rear surface of the Mylar could result in less energy getting to the collector.
This effect can be even worse if the layer acts as a Fabry–Pérot etalon, because it can magnify the usually small effect of a single reflection interfering (which depends on the amount of light that passes the first surface).