We know that silicon is too shiny to absorb incoming light that's why anti-reflection coating is needed to make the incoming light stay inside the cell.

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However, the problem is, even though the cell is covered with anti-reflection coating, the silicon material inside is still shiny and it still reflects the incoming light that passes through the anti-reflection coating so,

How come an anti-reflection coating be useful on a solar cell?

They explain that the coating has such a thickness that consecutive lights cancel each other that's why light waves do not go out, but canceling light waves mean that they were going out of the cell and they were canceled out. So lesser amount of light wave remains inside.


As they claim, how come light stays inside the cell?

  • $\begingroup$ From reading on the comments it seems you have the interference concept a bit wrong. Interference is about the phase of the light. It's a complex construct. We show the "reflected rays" just as hypothetical. The whole AR coating is a fixed interferometer. Light is a complex field, which is described by amplitude and phase. In this case, more important is phase. Also, you need to look at the system as a whole (you do not have just the reflected ray from the top layer, and the reflected from the lower, you have both at all times. $\endgroup$ Sep 3, 2021 at 18:04

2 Answers 2


In your picture the light going through is not in the picture. Since in ideal case al the reflected light is cancelled out, the through going light is the only one which reaches the silicon amplified. Whenever you have cancelling interference in one part, you have amplifying on the other side. If nothing comes out, everything stays in (law of conservation of energy) .

  • $\begingroup$ however we know that they are deleted, they are cancelled out on the outside, they deleted each other. We know what happened to them. So where is the law of conversation? We know that the energy has been disappeared. $\endgroup$ Sep 2, 2021 at 15:03
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    $\begingroup$ the law of conservation says, if the energy is not outside it must be inside. $\endgroup$
    – trula
    Sep 2, 2021 at 15:13
  • $\begingroup$ But we know the amount of light waves going inside the silicon after passing through the coating. It is less than the amount of light coming onto the solar cell because of reflection and deletion. So, how are those deleted energy carried inside(according to law of conservation of energy) despite of the delete? How does less light produce the same thing after delete? $\endgroup$ Sep 2, 2021 at 15:22
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    $\begingroup$ deletion means in reality nothing is reflected, all goes thru. this is easier to understand, if yo look at the 2 slit experiment. some places don't get any light. it is "annihilated" not "destroyed" in other places the intensity is higher so energy is conserved. $\endgroup$
    – trula
    Sep 2, 2021 at 15:25

You have a misunderstanding of destructive interference. It was framed as two waves that go from an initial inbound state to a reflected outbound state such that:

$$ \psi_1^{\rm in} + \psi_2^{\rm in} \rightarrow \psi_1^{\rm reflect} - \psi_2^{\rm reflect} = 0 $$

and two "rays" of light just disappear.

That's not how interference works.

Instead, there is a transition from the initial state to two possible final states (reflected or absorbed). There is one path (to 1st order) for absorption. For reflection, there are two paths: reflection from the top layer or the bottom layer. To find the transition amplitude to each final state, you need to add all paths coherently.

Reflection from the bottom layer picks up a phase $e^{i\pi}=-1$ so:

$$ \psi^{\rm reflect}_{\rm top} + \psi^{\rm reflect}_{\rm bot} = \psi^{\rm reflect}_{\rm top} + e^{i\pi}\psi^{\rm reflect}_{\rm top}=0$$

while the transmission does not:

$$ \psi^{\rm absorb} = \psi^{\rm in} $$

Now it seems like you could write

$$ \psi^{\rm absorb} = (1- \psi^{\rm reflect}_{\rm top})(1- \psi^{\rm reflect}_{\rm bottom})\ \ \ \ {\rm ?}$$

because that is how probabilities work (and it seems the only way to preserve causality), but we're dealing with probability amplitudes (in the photon picture), or electric fields in a classical view, and it's just different.

  • $\begingroup$ Can you elaborate? Your explanation is not clear enough to me I guess. $\endgroup$ Sep 2, 2021 at 16:59
  • $\begingroup$ Describe how two photons enter the material, how that are reflected, by the top and bottom, and why they are not reflected out to infinity. $\endgroup$
    – JEB
    Sep 2, 2021 at 17:59

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