## Light in a fog

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KarenRei
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### Light in a fog

I want to make sure my grasp on this is correct. Envision a fog with infinitely long extinction coefficient (perfectly reflective particles) lit up by the sun with a given solar constant (let's say 1000W/m^2). Inside the fog you place a 20% efficient 1m^2 solar panel that functions equally well from both sides. Does it yield 200W or 400W?

Instinct tells one that it harnesses 400W - that is, 200W from each 1m^2 side. But as per the "fire from moonlight" comic, instinct can be misleading. If you remove the fog, the panel is receiving 200W, from the same light source. All on one side, of course, but that's because the light is isotropic. Does the fog itself really double it? If there was no fog, you could put a mirror behind the solar cell to increase its output (some systems are built around narrow cells with large gaps between them placed in front of a mirror). But of course, the mirror is intercepting a lot more "square meters" of light. Can the fog itself act like the above mirror? If yes, how does it make sense that a fog would really double light? If no, what is the fundamental difference between the fog and the mirror?

Last edited by KarenRei on Sun Jan 08, 2017 4:12 pm UTC, edited 1 time in total.

Tub
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### Re: Light in a fog

Probably neither. I don't think you can assume a constant 1000W/m² radiation from every direction everywhere inside the fog. Even if all the scattering inside the fog evens out, you have a huge portion of the light that'll get reflected back to the sun immediately after entering the fog. Summing up the paths each light ray can take (modelled as a random walk), it would seem that it gets darker when you get deeper into the fog. Even when none of the light gets absorbed.

KarenRei
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### Re: Light in a fog

Tub wrote:Probably neither. I don't think you can assume a constant 1000W/m² radiation from every direction everywhere inside the fog. Even if all the scattering inside the fog evens out, you have a huge portion of the light that'll get reflected back to the sun immediately after entering the fog. Summing up the paths each light ray can take (modelled as a random walk), it would seem that it gets darker when you get deeper into the fog. Even when none of the light gets absorbed.

If we need to alter the scenario, then let us do so. Let's say the sun is brighter than 1400W/m², but a photometer reads 1000W/m² inside the fog after the reflection losses. How does the panel generate in such a situation?

Ed: Assume that the solar panel has no increasing-reflection-with-decreasing-angle losses. Keep it simple, this is an "assume a spherical cow" case.
Last edited by KarenRei on Sun Jan 08, 2017 2:11 pm UTC, edited 1 time in total.

quantropy
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### Re: Light in a fog

66 Watts

Imagine the fog is the form of a cube, with one face lit up by the sun. The light goes into that face, bounces around inside the fog until it's totally confused, and then exits one of the faces of the fog. That implies that the brightness of a face is 1/6 of the brightness of sunlight, and so one assumes that is what a solar panel inside the fog will see. Hence the output is 2*0.20*1000/6 Watts

Of course someone else may have a better answer - it perfectly possible that my argument is out by a factor of 2.

Regarding the mirror, I would think that a 1 m² mirror would be sufficient to illuminate the other side of the panel.

On a related note, we have and east-west facing roof with solar panels on both parts, and the maximum output isn't in a cloudless sky, but when light is coming from the sun and from reflections from clouds elsewhere in the sky.

Copper Bezel
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### Re: Light in a fog

That cube adjoins 6 other cubes, each of which are pumping 1/6 of their wattage into it. So you're right about the factor of two, at least.

There's nothing special about angles of light 45° off one or the other axis to the normal, and real solar panels have different angle ranges at which they'll work with a given efficiency. Not saying that the 1/6 isn't a reasonable approximation, just pointing out for clarity that it is one, rather than a necessity of geometry and physics in idealized conditions.

Edit: And that's still either side by itself, rather than both together, which interfaces weirdly with the original question, but I know the original question was leading for one potential consideration and this is taking a different set into account first.
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quantropy
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### Re: Light in a fog

Copper Bezel wrote:That cube adjoins 6 other cubes, each of which are pumping 1/6 of their wattage into it. So you're right about the factor of two, at least.

I was assuming the cube represented the whole of the fog. In essence light enters going in one direction, but in the fog there are 6 directions it can go in, so it will be 1/6 of the intensity in each direction.

Copper Bezel
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### Re: Light in a fog

If you just had a layer of perfectly scattering fog in between the sun and the solar panel, then yes, 1/6 of the light will leave the layer in 1/6 of the solid angle available, because it's evenly distributed across all angles and we've decided that 1/6 of them are useful, while the other 5/6 of the light leaves the fog in useless directions and is never reflected back.
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KarenRei
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### Re: Light in a fog

Copper Bezel wrote:If you just had a layer of perfectly scattering fog in between the sun and the solar panel, then yes, 1/6 of the light will leave the layer in 1/6 of the solid angle available, because it's evenly distributed across all angles and we've decided that 1/6 of them are useful, while the other 5/6 of the light leaves the fog in useless directions and is never reflected back.

And if it was a tetrahedron would there be 1/4th of the light inside the fog? If it was a sphere would it be perfectly dark? That's not logical.

But back to the question.... inside your fog, a photometer reads 1000W/m^2. What does your double-sided 20%-efficient 1m^2 solar panel generate? A 1x1x1m cube? A 1m radius sphere? I'm increasingly feeling that it generates 400W. That you're a 2m^2 object, akin to a cube with the Z axis infinitely small, or a 2m^2 sphere.

Copper Bezel
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### Re: Light in a fog

KarenRei wrote:And if it was a tetrahedron would there be 1/4th of the light inside the fog? If it was a sphere would it be perfectly dark? That's not logical.

I've said twice now that quantropy's choice of 1/6 is arbitrary. They're attempting to model how much of the light that enters the fog is actually scattered in a useful direction, which was relevant to the original question.

If we're starting from a photometer reading, we don't need to worry about that; your fog is effectively glowing of its own volition and the original light source is irrelevant, but that's where we are if we've already decided that the original question is too hard.

If your photometer reads 1000W/m^2, and the light is uniform in all directions, then a 20% efficient solar panel will generate 200W/m^2. Your double-sided square solar panel has a surface area of 2 m^2, your cube has a surface area of 6 m^2, and your sphere has a surface area of 4π. It is literally not any more complicated than this.

Edit: Bears noting that an ordinary photometer reading of an ordinary directional source is itself directional, and would be a given number of W/m^2 at a given ideal angle and decrease from there as it was rotated away from being normal to the light source.
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KarenRei
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### Re: Light in a fog

Indeed, that's what intuition says. The thing that makes me hesitate is that if you put the same double-sided solar cell in 1000W/m² direct sunlight, you only get 200W, not 400W. There's a naggling feeling that the two situations should be equivalent and I'm trying to figure out how to convince myself that they shouldn't.

Copper Bezel
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### Re: Light in a fog

Well, again, direct sunlight is directional, and you're not exposing both sides of the solar panel to it. The light in your fog is omnidirectional, and we've magically set it to be 1000W/m2 on any plane you choose.

I still don't think I'm picturing the geometry correctly, and I'm not sure whether quantropy's factor of six (or three) is right, but the irradiance of the sunlight that is pumping light into the fog is definitely much higher than the irradiance on any plane you place within the fog and is what accounts for this difference.
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morriswalters
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### Re: Light in a fog

Well here goes. Assuming a collimated source. The solar panels will act as a shadow source. I think this will create a cone of shadow over the solar panel opposite the source. Assuming the panel itself is non reflective, no light can shine through it. So it can only see light from the edges initially. I think the controlling factor is density. My thinking is that in the overall scheme of this scenario there are more ways to reflect away from the panels then towards them. So less then the rated power of one panel depending on how close to the boundary layers the panels are.

ucim
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### Re: Light in a fog

If there's no fog, then the light can only reach the front of the solar panel, but all the light (aimed at it) reaches it. Light that is not aimed at it is lost to infinity. So, you would get the full amount of energy that aimed at the solar cells, and none of the energy that missed it.

If you put a mirror off to the side, and behind the solar cells, and aimed that mirror properly, you would (almost) double the output, because you would be capturing the energy that was not aimed at the solar cells but would have been lost to infinity. Instead, it gets reflected (for free) back to the backside of the solar cells, where there are more solar cells to receive it. (Putting a mirror in the shadow of the array would be silly. This will become important.)

As I see it, what the fog does is provide a (fuzzy) way for the light that was not aimed at the solar cells to reach it anyway. If you in fact have a uniform illumination of the fog, then the backside gets the backscatter, which (except for the shadow) would be equal to the frontscatter. So, the question would be just how much the shadow of the solar cells diminishes this. It's not by zero, since if the solar cells were big enough, the entire backside would be in shadow; that is, the illumination on the front side comes in part unimpeded from the light source, but this is not true of the rear. All of the light from the rear is backscatter, and all of the backscatter is from outside the shadow.

So... somewhere in between. Just how much in between is left as an exercise for the reader.

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KarenRei
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### Re: Light in a fog

Bringing up a mirror just complicates it. Because sure, in the direct-collimated-sunlight case I could add a mirror beside the original solar panel that reflects light onto the back of the panel and causes its generation to double. But I could also add a mirror that would make its generation increase 5x. Or 0.1x. Or 100000x. Or whatever number I wanted, depending on what sort of mirror I placed. The fog is doing just "2x", and there's just something that kind of feels weird about that (since you're starting from the same light level).

It's probably not a big deal, but I was just hoping to be able to get rid of that feeling there.

I still don't think I'm picturing the geometry correctly, and I'm not sure whether quantropy's factor of six (or three) is right, but the irradiance of the sunlight that is pumping light into the fog is definitely much higher than the irradiance on any plane you place within the fog and is what accounts for this difference.

No, I can think of real-world examples where that sort of ratio given (1400W/m² sun, 1000W/m² fog) can be found. For example, Venus's cloudtops, in the ~48-60km height range. They're not perfectly scattering (there's about 80% as much nadir light as zenith in the middle cloud layer), but it doesn't take 3x, 6x, or whatever times more light from the light source to pump a fog to near isotropy.

Also, fig. 14.1 from this book, p. 284, for zenith vs. nadir light:

But again, I'd rather stay away from the real world and all of its complications and stick with simplified scenarios, such as a perfect fog with direct in-situ measured illumination.

But then again... hmm. Maybe this right there is the answer. Maybe it's impossible for backscatter to increase at a rate faster than zenith light drops off. Perhaps the problem was that I was simplifying the scenario too much, and that the real-world scattering factors are what explain it? Does Venus match such a hypothesis, perhaps?

gmalivuk
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### Re: Light in a fog

At the very least, you're simplifying directionality in a way that feels like too much.

If you posit at the beginning that the radiant flux through, say, a window at any orientation into a fogless room is exactly 1000W/m^2, and then you posit a solar panel that has no dependency on the direction of the incoming light, then of course 2m^2 will get twice what 1m^2 gets, however each square meter is oriented, and then you multiply that by the overall efficiency of the panel. (If the fogless room had windows on two sides instead of just 1, it would get twice as much light.)

But in the real world I suspect it makes a big difference that the incoming light in the fog is diffused across all angles (from the perspective of a point on the solar panel), instead of from a tiny circle half a degree across nearly perpendicular to the panel. (I know most of the discussion of directionality in solar panels has to do simply with maximizing the area in a plane perpendicular to the sun's rays, but I suspect that even when you correct by a factor of the cosine of the angle of incidence, panels are still less efficient for low-angle incoming light.)
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