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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?

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    $\begingroup$ Painting the back of the mirror black might have something to do with the purpose of the mirror. The purpose should be stated in your question - it should not be necessary to watch the video to find out what your question means. $\endgroup$ – sammy gerbil Nov 27 '16 at 0:21
  • $\begingroup$ @sammygerbil I added a short explanation. $\endgroup$ – Caridorc Nov 27 '16 at 10:00
  • $\begingroup$ Black also emits more Planck radiation (higher emissivity), which can help prevent the mirror from overheating. $\endgroup$ – k-l Nov 27 '16 at 19:20
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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).

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