Question

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? answer is 121 nm.

Answer 1

Concepts and reason

The problem deals with the concept of the destructive interference in the double slit.

The reason for destructive interference is that the double slit between each peak of constructive interference is a single location where destructive interference takes place.

Fundamentals

The condition or the formula to determine the destructive interference is as:

$2nt = \left( {m + \frac{1}{2}} \right){\lambda _{vaccum}}$

Here $n$ is the refractive index, $m$ is the order of the fringe, $\lambda$ is the wavelength, $t$ is the thickness.

The condition for the destructive interference in double slit is as:

$2nt = \left( {m + \frac{1}{2}} \right){\lambda _{vaccum}}$

Rearrange the formula to determine the thickness as:

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

Substitute the value for the refractive index of the silicon dioxide $n = 1.45$ , the order of the fringe $m = 0$ since the reflection should be minimum and the wavelength in the vacuum ${\lambda _{vaccum}} = 700{\rm{nm}}$ in the formula of the thickness as,

Put the values in the formula as:

$\begin{array}{c}\\t = \frac{{\left( {m + \frac{1}{2}} \right){\lambda _{vaccum}}}}{{2n}}\\\\ = \frac{{\left( {0 + \frac{1}{2}} \right)700{\rm{nm}}}}{{2\left( {1.45} \right)}}\\\\ = \frac{{350{\rm{nm}}}}{{2.9}}\\\\ = 120.68 \approx 121{\rm{nm}}\\\end{array}$

Ans:

The thickness of coating of silicon dioxide required to minimize the reflection is $121{\rm{nm}}$ .