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Solar cells are given antireflection coatings to maximize their efficiency. Consider a silicon solar cell (n = 3.50) coated with a layer of silicon dioxide (n = 1.45). What is the minimum coating thickness that will minimize the reflection at the wavelength of 700 nm, where solar cells are most efficient?

Respuesta :

Answer:

t = 121 nm

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.

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