Answer:
t = 121 nm
Step-by-step explanation:
Given:
- Silicon refractive index n_1 = 3.50
- Silicon dioxide refractive index n_2 = 1.45
- The wavelength of light in air λ_air = 700 nm
Find:
What is the minimum coating thickness that will minimize the reflection at the wavelength of 700 nm.
Solution:
- The film’s index of refraction (n_2 = 1.45) is less than that of solar cell (n_1 = 3.50) so there will be a reflective phase change at the first boundary (air–film), and at the second boundary (film–solar cell). The relationship for destructive interference for two reflective phase changes is as follows:
2*t = (m + 0.5)*(λ/n_2) m = 0, 1, 2, ....
- Solve for thickness t where m = 0 (for the thinnest film).
t = 0.25*(λ/n_2)
t = 0.25*(700/1.45)
t = 121 nm ... (rounded to 3 sig. fig)
- This coating technique is important to increase the efficiency of solar cells; If the light can’t reflect, then it must transmit into the solar cell material.