Chapter 19.1, Problem 9PP

To determine

The thickness of the nonreflective layer of silicon monoxide.

Answer to Problem 9PP

  $t=95.7\text{ nm}$

Explanation of Solution

Given:

Refractive index of air is ${\text{n}}_{air}=1.00$

Refractive index of silicon monoxide film is $n_{film}=1.45$

Refractive index of silicon is $n_{silicon}=3.5$

Wavelength of light is $\lambda \text{ = 555 nm}$

Formula used:

There will be a phase inversion in the light rays reflected from the interface between air and silicon monoxide layer because of $n_{film}{\text{> n}}_{air}$ and also there will be a phase inversion in the light rays reflected from the interface between silicon monoxide layer and the silicon because of ${\text{n}}_{silicon}>n_{film}$ .

The equation for destructive interference to keep the yellow-green light from being reflected is given by,

  $\text{2t = }\left(m+\frac{1}{2}\right)\frac{\lambda }{n}$ $\left(1\right)$

Where, $\text{t }$ is the thickness of the film $\lambda$ is the wavelength of the light in the film $n$ is the refractive index of the film, In this problem $n=n_{film}=1.45$ $m$ is the order of the fringe,

Calculation:

For the thinnest film, $m=0$

Substituting for $m$ and $n$ in equation $\left(1\right)$ ,

  $\text{2t = }\left(\frac{1}{2}\right)\frac{\lambda }{n_{film}}$

  $\Rightarrow \text{t = }\left(\frac{1}{4}\right)\frac{\lambda }{n_{film}}$ $\left(2\right)$

Substituting the numerical values in equation $\left(2\right)$ ,

  $\text{t = }\left(\frac{1}{4}\right)\frac{\left(555\text{ nm}\right)}{1.45}=95.7\text{ nm}$

Conclusion:

The thickness of the nonreflective layer of silicon monoxide layer on the silicon solar cell to keep yellow green light with a wavelength of $\text{555 nm}$ from being reflected is $95.7\text{ nm}$ .