Let's assume that solar light reaches a silicon solar cell with an angle of incidence of θi=0o. For simplicity, let's consider the refractive index of silicon to be nSi=3.5. The refractive index of air is nair=1. What percentage of light would be lost due to reflection at the air-silicon interface? Assume that the solar light is randomly polarized.
To determine the percentage of light lost due to reflection at the air-silicon interface, we need to use the laws of reflection and Snell's law.
1. Law of Reflection: The angle of incidence (θi) is equal to the angle of reflection (θr) at the interface between two media.
2. Snell's Law: The ratio of the sine of the angle of incidence (θi) to the sine of the angle of refraction (θr) is equal to the ratio of the refractive indices (n1/n2) of the two media.
Mathematically, this can be written as sin(θi)/sin(θr) = n2/n1.
In this case, the light is incident on the silicon solar cell with an angle of incidence of θi=0°. When the angle of incidence is 0°, the angle of reflection is also 0°. Therefore, all the light is reflected back into the air.
In terms of the percentage of light lost due to reflection, it would be 100%. This means that none of the light is transmitted through the air-silicon interface and is lost as reflected light.