optical-materialsreflectionvisible-lightwaves
Is it possible to construct a mirror such that it only reflects certain wavelengths of visible light while allowing other wavelengths of visible light to pass through? I was reading about microwave ovens and how the doors are designed to block microwaves from exiting the oven and was wondering if the same principle can be applied to visible light.
Thanks!
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.
Different molecules and different crystalline structures have frequency dependent absorption/reflection/transmission properties. In general, light in the human visible range can travel with little absorption through glass, but not through brick. UV can travel well through plastic, but not through silicate-based glass. Radio waves can travel through brick and glass, but not well through a metal box. Each of these differences has a slightly different answer, but each answer is based on molecular resonance or crystalline structure (or lack thereof) or electrical conductivity.
Bottom line: There isn't one general answer for why $\lambda_A$ goes through material X but $\lambda_B$ doesn't.
Best Answer
You could look at Optical coatings and dichroic filters.
Optical coatings are many thin layers of material of two of more different refractive indices, about one light wavelength in thickness. They structure of the coating can be designed for high reflectivity at some wavelengths and low reflectivity at others, but the switch-over cannot be too abrupt. They are very expensive to make.
Microwave oven doors work on a different principle. The holes are about 1mm diameter. Visible light passes through easily because the wavelength is very much smaller (about 0.5 microns = 0.00005cm) whereas for microwaves the wavelength is very much bigger than the holes (about 12cm). As far as the microwaves are concerned, the grille is not much different from a mirror. See How does the grid on the microwave oven window prevent microwave radiation from coming out?.
If the same principle were used for visible light, it would be difficult to get a relatively "sharp" cut-off between 2 wavelengths. But it could be done, using nano-structured materials. Fine-structuring could produce exotic optical properties. See also meta-materials.