I wonder the reason for TiO2 thin films to be transparent for visible light but not for UV. I made a quick search and I found that it is due to the band gap of TiO2. It absorbs UV light but not visible light. I imagine this occurs because of the different wavelengths of these two types of radiation. But what is the relation between the wavelength of a certain type of radiation and the width of the band gap a semiconducting material? And how does this effect its optical properties?
Electromagnetic Radiation – Why Titanium Dioxide is Transparent for Visible Light but Not for UV
electromagnetic-radiationsemiconductor-physics
Related Solutions
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.
I think the point you are making is why isn't the n-CdS/p-CdTe junction inverted to be p-CdTe/n-CdS. As you say this would allow the "high mobility" CdTe layer (it's not really high mobility, it more that any carriers generated in CdS recombine instantly) to be placed first and absorb a little extra light. Something on the order of $7mA/cm^2$ of photocurrent is lost due to parasitic absorption in the CdS layer, so this would seem like a good idea.
I am not an expert on thin-film cells but I believe the problem boils down to suitable workfunction materials for the electrodes. If you invert the cell then you also need to have p-layer contact which is transparent. If the workfunction is a poor match then its possible to setup a space charge which prevent carrier collection.
Another option would be to do a n-CdTe/p-CdS design however I think there are problems with doping CdTe as a donor, or at least it can only be done to a low level ~$~10^{-14}cm^{-3}$, which will give you a small built-in field.
So in summary the design of the thin-film solar cell is probably evolved due to material constraints.
This is in an interesting question, I will look more into it and update. Hopefully this gives you some pointers until then.
Best Answer
The energy per photon of light with wavelegth $\lambda$ is given by:
$$ E = \frac{hc}{\lambda} $$
If the energy per photon is smaller than the band gap the light cannot excite electrons from the valence to conduction band so it will pass through the material without being absorbed. If the energy is larger than the band gap the light will excite electrons and will be (partially) absorbed. The cutoff wavelength is given by simply rearranging the above formula to get:
$$ \lambda \approx \frac{hc}{\Delta E} $$
where $\Delta E$ is the band gap. I've used the approximately equal sign because band gaps are rarely sharp and the light absorbtion will increase over a wavelength range of around the cutoff wavelength. If you want to establish the band gap accurately you'd use a Tauc plot.