How do anti-reflection coatings in solar cells make light stay inside a solar cell?

Question

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.

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.

So,

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

Answer 1

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) .

Answer 2

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.