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I am taking a solar cell class and came across a question that deals with reflection losses as light passes through multiple materials before it reaches the solar cell. I tried to look up optical properties and laws, but can't seem to find the one that deals with this.

This isn't the problem I'm working on, but an example.

A solar panel is in a glass case. The glass case reflects 8% of the light. The silicon is reflecting 38% of the light.

Is it 100% - 8% - 38% for a total of 54% of the light being transmitted into the cell, or

100% - 8% = 92% - 38% = 57.04% of the light being transmitted into the cell?

This is really confusing me.

user1362058
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  • I don't really think this belongs to math stack exchange. If you had the mathematical model for reflection losses you would see that the mathematical part of the problem is quite trivial. – skyking Mar 22 '17 at 06:54
  • ...that is as written. I'd suggest that when asking physics questions here you include the mathematical model for the physics. – skyking Mar 22 '17 at 07:08
  • @skyking Thanks. I didn't have any models for the reflection. That is all the info I had. – user1362058 Mar 29 '17 at 02:54

2 Answers2

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The process can't be subtractive. Imagine you had three layes with the same reflectance as silicon. A subtractive formula would give

$$100\%-38\%-38\%-38\%=-14\%,$$ which is meaningless.

The process is multiplicative. After the first reflection, what remains passes the next diopters and undergoes other reductions in the same proportion,

$$100\%\to62\%\to62\%\cdot62\%\to62\%\cdot62\%\cdot62\%=23.8\%.$$

Hence

$$(100\%-8\%)\cdot(100\%-38\%)=57.0\%.$$

  • Thanks. I thought the same thing logically as well. Turns out after I got the assignment back that they had normalized the light to give reflectivity based on a 100% scale or something like that. So no matter how much material you lined up in front of each other it would never add to 100% percent unless you specifically wanted to block all light. Very confusing since it wasn't defined that way and it goes against, what I would consider, conventional thought. – user1362058 Mar 29 '17 at 03:00
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I think you should have included the mathematical model for losses that you use. The mathematical problem arises when you take into account that reflections are not actual losses - it just makes the light go away in another direction. I will use a model where there is no actual losses (except that the silicon absorbs all non-reflecting light). Also for making the explanation simplier I will ignore internal reflections in the glass.

The reason it's somewhat mathematically interresting is that light reflected of the silicon will hit the glass again and we can assume that it reflects $\alpha=0.08$ of this light too (back at the silicon). The silicon then reflects $\beta=0.38$ of this light(*).

Note that the above assumptions may or may not be physically correct. For one it basically assumes that there's air between the glass and the silicon, otherwise the reflection here will be different, but in principle the calculations follow the same reasoning.

Let $I_0$ be the incoming light, $I_1$ is what hits the silicon and $I_r$ is what's reflected back from the silicon. Now we get the equations:

$$I_1 = (1-\alpha)I_0 + \alpha I_r$$ $$I_r = \beta I_1$$

which gives:

$$I_1 = (1-\alpha)I_0 + \alpha\beta I_1$$

Then solving for $I_1$ we get

$$I_1 = {1-\alpha\over 1-\alpha\beta}I_0$$

Finally we note that $(1-\beta)I_1$ actually enters the silicon. So you get a factor of

$${(1-\alpha)(1-\beta)\over 1-\alpha\beta}$$

So it's actually a bit more than both your alternatives, if you insert the values you get $58.8\%$.

(*) If we don't include back reflections in the glass screen the problem is reduced to the problem where $I_1 = (1-\alpha)I_0$ and then only $(1-\beta)I_1$ enters the silicon giving the factor $(1-\alpha)(1-\beta)$ which would be your second alternative (ie $57.0\%$)

skyking
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  • If I am right, you are assuming that $\alpha$ holds both for the reflection on the air to glass diopter and backward, glass to air. I am not sure this is correct. In any case, your denominator should be $1-\alpha\beta$ (which is coherent with the fact that internal reflections increase the total transmission), though you made the computation with a minus. –  Mar 22 '17 at 07:31
  • @YvesDaoust That's the model I've used here yes. That may not be physically correct, but I mean on math stack exchange we should therefore be provided with a physically correct model (that is that we work with the model provided - right or wrong). I'll update the answer however... – skyking Mar 22 '17 at 07:53
  • I don't think that the OP ever thought of internal reflections and his model is the concatenation of two transmissions $(1-\alpha)(1-\beta)$. –  Mar 22 '17 at 08:11
  • Yes, however with that model it becomes much of a non-question here IMHO - that's why I included back reflections from the glass screen. Btw is it you who downvoted my answer? – skyking Mar 22 '17 at 08:36
  • Yep, for the first reason I explained. –  Mar 22 '17 at 08:39
  • @YvesDaoust Don't you think I've addressed that objection in my updated answer? Or have I still missed something? Note that I exclude internal reflections, the back reflection is again from air-to-glass. (For the record on normal incidence the reflection is the same for internal and external reflection according to the Fresnel equations) – skyking Mar 29 '17 at 06:10