Consider the transition of light from one medium into another, say from air into glass: Conservation of energy and the continuitiy conditons of the electric field require that only a part of the incident radiation is transmitted into the medium while a part is also reflected. The reflection coefficient for the intensity and perpendicular incidence is given by the law of Beer, the refractive index of air is 1, for glass it is 1.5.
We find that about 4% of the incident light intensity are reflected, 96% are transmitted into the glass.
In a material of given thickness the light will eventually travel to the far end and at that boundary again reflection and transmission take place. In fact the light will travel back and forth between the boundaries several times, each time a part is reflected back into the medium and a part is transmitted out of the medium.
In order to find the total intensities which are reflected from and transmitted through the medium we must distinguish between coherent and incoherent light. In the first case we must treat the light as wave and consider interference phenomena. The second case is much easier, we can simply add the intensities. If we deal with sunlight and not with lasers, the coherence length is only a few wavelengths. Thus, in thick media coherence is lost during a round trip of the light, in thin films with thickness in the order of the wavelengths it is maintained.
We could calculate yet another round trip, but that intensity is negligible for the considered case of air and glass. If we add the intensities, the single glass plate yields a total reflected intensity of 7.7% and a total transmitted intensity of 92.3%.
If coherency is maintained, the two red waves will interfere. Depending on the phases, we distinguish constructive and destructive interference, as shown below:
In the constructive case the waves are in phase (have their maxima and minima at the same positions) and the amplitudes of the electric field add up. In the destructive case the waves are of opposite phase, the sum of the amplitudes becomes smaller. We observe that the condition for destructive interference is a phaseshift of half the wavelength. The thickness of the coating must be a quarter of the wavelength because the light travels back and forth (still assuming perpendicular incidence).
The amplitudes of the reflected beams are governed by the indices of refraction of the involved materials (air, coating, glass). For a particular choice they become equal, and we can even extinguish the reflected wave. Going through all the math we would find that the coating must have a refractive index equal to the square root of n0. n2 (n0: air, n2: glass).
In real systems we deal with more than one wavelength because the sunlight is composed of a spectrum of different wavelengths. For a given thickness of the interference coating the condition of destructive interference may apply for one wavelength, but it may become constructive for another. The figure below illustrates the spectral reflection for a coating of given thickness d. The condition for minimum reflection is fulfilled also for smaller wavelenghts, when the phaseshift corresponds to odd multiples of half the wavelength.
We will see this type of diagram again in the next section, but before we proceed we want to see what an antireflection coating actually does for a solar cell.
Silicon has a refractive index of about 4, from Beer's equation we calculate a reflection of almost 40% which is not tolerable for solar cells. We want to reduce the reflection loss with an antireflection coating.
We must take care of two things: The sqare root condition requires a refractive index of about 2, and the thickness of the coating should ensure a minimum of reflection at a wavelength where the solar intensity is highest, i. e. around 600 nm. For destructive interference the optical path length 2. n .d (going back and forth in a medium with index n) should be half that wavelength.
For a coating of TiO2 with its refractive index around 2.3 we get a thickness d somewhere between 60 and 70 nm. Better reduction of the reflection losses is obtained with a double coating of TiO2 and MgF2.
The high reflection for low wavelengths gives rise to the blue
appearance of silicon solar cells. It is due to the fact
that the refractive index of silicon is not constant but
higher than 4 at low wavelengths (see a later section).