Question

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

Answer 1

For a normal ray, the fraction of reflected incident power (light lost due to reflection) is given by

\(R = \large{\frac{\left(n_1 - n_2\right)}{\left(n_1+n_2\right)}}\)

\(n_1 = 1 \\ n_2 = 3.5\)