Is optical reflection loss through multiple objects added or compounded?

Question

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.

Answer 1

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

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

Answer 2

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 $57.0\%$)