0
$\begingroup$

I was watching some media on how solar cells work, and the description of the antireflective coating confused me.

According to this video and this Physics Asylum video, the antireflection coating is optimized to cause destructive interference between reflections from the top and bottom interfaces of this layer. This light is now not being reflected, we get more transmission of light into the PV that can be converted into energy.

Is this really how it works? If two reflected photons' phases cancel out due to destructive interference, why would that lost intensity then be added to the portion of light that transmitted through?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Do you understand how a quarter-wave plate works? $\endgroup$
    – Jon Custer
    Commented Jun 17, 2022 at 0:01
  • $\begingroup$ Don't get too stuck on "light is photons" . Try to view this as a couple standing waves and you'll see what's going on $\endgroup$
    – Carl Witthoft
    Commented Jun 17, 2022 at 13:20

2 Answers 2

Reset to default
2
$\begingroup$

In wave interference, the energy has to go somewhere. So, if interference cancels the radiation in a particular direction it must enhance it in another.

Improve this answer
$\endgroup$
1
$\begingroup$

Its best to think of this in terms of transmission and intensity. Suppose that light traveling through air encounters a material with index of refraction $n_m$ The transmission of of light through the material can be described as

$$I_0 = I_{R} + I_{T}$$ $$I_{T} = I_0 - I_{R}$$ where $I_{R}$ and $I_{T}$ are the reflected and transmitted intensities.

Suppose now, a different material of given thickness and different index of refraction is deposited on top of the original material (anti-reflection coating AR), then the transmission of light through this material can be described as:

$$ I_0 = I_{AR(reflected)} + I_{AR(transmitted)}$$ and then that light is reflected from and transmitted through the material/AR boundary $$I_{AR(transmitted)} = I^{'}_{AR(reflected)} + I^{'}_T $$ Finally

$$I_0 = I_{AR(reflected)} + I^{'}_{AR(reflected)} + I^{'}_T$$

Clearly if the two reflected components could cancel each other out$$I_{AR(reflected)} + I^{'}_{AR(reflected)} = 0$$ then 100% transmission would be possible $$I^{'}_T = I_0$$

This is what AR coating design takes advantage of and how the destructive interference that cancels these reflected waves increases transmission.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.