Question

In solar cells, a silicon solar cell $\left( {\mu = 3.5} \right)$ is coated with a thin film of silicon monoxide ${\text{SiO}}\left( {\mu = 1.45} \right)$ to minimize reflective losses from the surface. Determine the minimum thickness of ${\text{SiO}}$ that produces the least reflection at a wavelength of $550\;{\text{nm}}$ , near the center of the visible spectrum.

Answer 1

Hint:- The thin film interference depends on wavelength, refractive index and the thickness of the film. The interference pattern changes as the thickness varies. From the relation between those factors, the thickness can be found.

Complete Step by step solution:
When light is reflected from the surfaces of the thin film it causes thin film interference. The thin film will be having thickness smaller than a few times the wavelength of light. The thin film will be a way to produce both destructive and constructive interference. Hence the path difference will be an integer or half integer of the wavelength.
Given the refractive index of the ${\text{SiO}}$ is $\mu = 1.45$. And the wavelength for least reflection is $\lambda = 550\;{\text{nm}}$.
For least reflection, it would be destructive interference. The condition for least reflection in the destructive interference is given as,
$2\mu t = \dfrac{\lambda }{2}$
Where, $\mu$ is the refractive index, $t$ is the thickness of the film and $\lambda$ is the wavelength.
From the above expression,
$t = \dfrac{\lambda }{{2\left( {2\mu } \right)}}$
Substituting the values in the above expression,
$t = \dfrac{{550 \times {{10}^{ - 9}}\;{\text{m}}}}{{2\left( {2 \times 1.45} \right)}} \\\ = 9.48 \times {10^{ - 8}}\;{\text{m}} \\\ {\text{ = 94}}{\text{.8}}\;{\text{nm}} \\\$
Therefore, the minimum thickness of ${\text{SiO}}$ that produces the least reflection at a wavelength of $550\;{\text{nm}}$ is ${\text{94}}{\text{.8}}\;{\text{nm}}$

Note: We have to note that for constructive interference the path difference will be integral to wavelength. And for destructive interference, the path difference will be half integral of the wavelength.