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How the Inverse Square gets affected by Diffusion


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I know the inverse square is in reference to point source radiation and you can calculate the dropoff by 1/d^2,  but what about fully diffused 2 dimensional sources like diff frames.  If you have source that is diffused by a 6x6 of magic cloth and the card is filled complete and evenly, should the light emitted from the diffusion fall off slower, faster or at the same rate as the inverse square.  How much does the distance between the unit and the diffusion matter for fall off if the card is evenly lit, would it matter if it was a larger unit further back or several units up close.  I have a white apartment, so its tough to do an objective test.

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The inverse square law holds the same for diffused light as it does for a point source.

The one difference, is that when you place diffusion in front of light, the diffusion surface becomes the light source.  So, you would measure from the diffusion and not the lamp behind the diffusion.

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1 hour ago, Bruce Greene said:

The inverse square law holds the same for diffused light as it does for a point source.

The one difference, is that when you place diffusion in front of light, the diffusion surface becomes the light source.  So, you would measure from the diffusion and not the lamp behind the diffusion.

Man I would kill for an a free hour in a black box studio.  I just can't find anything that verifies that two and three dimensional sources work the exact same way.  Every reference I see of inverse square talks about point sources with equal spherical radiation in all directions. In my mind, if you have a 4x4 even source and a point in space 2 feet from the center of diff.  The corners of the frame will be more than 2 feet away and will light that point in space less than the center.  Only when the point in space gets further back will the distances from each point on the 4x4 be equalish. It can't function like the inverse square up close right? Also I get that the diff is the new source, but so it does not matter if the source behind it was 100 feet away vs led blanket 1 inch away from the the diff?  If the reading in both cases at 5ft is a 2.8, then reading must be the same at 10ft at a 1.4?  If the diff is a reset on the inverse square, does the distance from bounce sources not matter either as long as its evenly illuminated?

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For the purposes of falloff, consider the large area light source as a cluster of infinitely many, infinitely small point light sources side by side. You're right, it's complicated, because the distance to the subject is variable, assuming the diffusion is flat as it usually is, so it ends up being a mathematical integration.

In short, inverse square works perfectly for a single infinitely small light source; it works progressively less well as the dimensions of the light source becomes any appreciable proportion of the distance between the light source and the subject. It's calculable, but it's not as simple as an inverse square.

P

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1 hour ago, Phil Rhodes said:

For the purposes of falloff, consider the large area light source as a cluster of infinitely many, infinitely small point light sources side by side. You're right, it's complicated, because the distance to the subject is variable, assuming the diffusion is flat as it usually is, so it ends up being a mathematical integration.

In short, inverse square works perfectly for a single infinitely small light source; it works progressively less well as the dimensions of the light source becomes any appreciable proportion of the distance between the light source and the subject. It's calculable, but it's not as simple as an inverse square.

P

I respectfully disagree. You are correct that you can imagine a diffuse source as "infinitely small point light sources" and each "source" will spread their light according to the inverse square law.

If we have a point source "x" that has an intensity of 20 at 1 meter distance, it will have the intensity of 5 at two meters distance.  If we add a second point source next to it, there will be an intensity of 40 at 1 meter and an intensity of 10 at two meters. IOW, adding the two intensities together, 20 +20 = 40 (at 1 meter) and 5+5=10 at two meters.  And 10 is also 1/4 of 40.

So if we have "n" light sources (infinitely small point light sources", and 100% intensity at 1 meter = n x point light source, then the total light at 2 meters will be n x (number of light sources) / 4.

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2 hours ago, Ryan Emanuel said:

Man I would kill for an a free hour in a black box studio.  I just can't find anything that verifies that two and three dimensional sources work the exact same way.  Every reference I see of inverse square talks about point sources with equal spherical radiation in all directions. In my mind, if you have a 4x4 even source and a point in space 2 feet from the center of diff.  The corners of the frame will be more than 2 feet away and will light that point in space less than the center.  Only when the point in space gets further back will the distances from each point on the 4x4 be equalish. It can't function like the inverse square up close right? Also I get that the diff is the new source, but so it does not matter if the source behind it was 100 feet away vs led blanket 1 inch away from the the diff?  If the reading in both cases at 5ft is a 2.8, then reading must be the same at 10ft at a 1.4?  If the diff is a reset on the inverse square, does the distance from bounce sources not matter either as long as its evenly illuminated?

A "3 dimensional" light source?  If you mean a big white ball, than that complicates the math as each point on the surface of the ball has a different distance to the subject.  But practically speaking, at a far enough distance you can "view" your ball as a point source.  After all, the sun is a sphere, but it is so far away that we "see" it as a point source.

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So then a booklight really has no utility, unless you need more output than what a litemat/litetile can provide, but at the same time do not have the distance to go straight through the diff evenly.

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3 hours ago, Bruce Greene said:

If we have a point source "x" that has an intensity of 20 at 1 meter distance, it will have the intensity of 5 at two meters distance.  If we add a second point source next to it, there will be an intensity of 40 at 1 meter and an intensity of 10 at two meters. IOW, adding the two intensities together, 20 +20 = 40 (at 1 meter) and 5+5=10 at two meters.  And 10 is also 1/4 of 40.

Yes, that's absolutely right. I think what we're discussing here is the situation that arises when the size of the light source becomes a significant proportion of the distance between it and the subject. In that situation, the outer edges of the diffuse light source may be considerably further away from the subject than the centre.

This might well happen if I was trying for a particularly beautiful cosmetics-commercial sort of look on someone and put a 4x4 diffusion a foot away from her. The centre of that diffusion is a foot away but the outer edges are a lot more than a foot away, so the resulting intensity at any point on her face is slightly complicated.

P

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13 hours ago, Phil Rhodes said:

Yes, that's absolutely right. I think what we're discussing here is the situation that arises when the size of the light source becomes a significant proportion of the distance between it and the subject. In that situation, the outer edges of the diffuse light source may be considerably further away from the subject than the centre.

This might well happen if I was trying for a particularly beautiful cosmetics-commercial sort of look on someone and put a 4x4 diffusion a foot away from her. The centre of that diffusion is a foot away but the outer edges are a lot more than a foot away, so the resulting intensity at any point on her face is slightly complicated.

P

If one is shooting with the light so close to the subject, why worry about the inverse square law at all?  At least as it applies to the subject.  The background will be far enough away that one might take it into consideration however, and at that distance, the size of the diffuser won't matter.

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18 hours ago, Ryan Emanuel said:

So then a booklight really has no utility, unless you need more output than what a litemat/litetile can provide, but at the same time do not have the distance to go straight through the diff evenly.

Yes... a booklight is a method of filling a large diffusion frame more evenly by bouncing the light first.  If you had a Litemat of equal size as your booklight then you wouldn't need to use a booklight probably since by the nature of an LED panel, the light is evenly distributed edge to edge. Of course, some would argue that the color and warmth of a powerful tungsten light and a muslin booklight set-up is more attractive on a face -- but at the cost of a lot of power used and heat generated.

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On 7/28/2019 at 3:14 AM, Phil Rhodes said:

 I think what we're discussing here is the situation that arises when the size of the light source becomes a significant proportion of the distance between it and the subject.

If I remember right, as long as the size of the light source is less than 20% of the distance to the subject, the inverse square law is still useful.

So, light coming from a 4x4 diffusion frame at 20ft from the subject would behave according to the law, but as you moved it closer it the law would become less and less accurate.

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10 hours ago, Marcos Cooper said:

Or watch this, starring the most patient woman in the universe.

 

I'm not sure it helps, I don't know how they filled the card, unless you have magic cloth I don't know any diffusions that would fill the card evenly without a double break, unless the source was very far back.  Full grid still might need a double break with something light to make the point source evenly light the frame. A 8x8 litetile would be a cool comparison with a booklight, but if done correctly they'll look the same as far as softness is concerned

 

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  • 4 weeks later...

Was finally able to test this yesterday on set.  I had a leeko 15 feet away from a 4' 216 frame, and compared inverse square with a 4x4 fabric led 2 inches from the frame and the inverse square incident readings were exactly the same.  Bye bye book light!

 

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Ryan, when you say you compared the inverse square, are you saying you measured the same falloff with both sources? As in, you took a measurement, x=distance from the 216 frame, then doubled that distance and it was 1/4 of the brightness?

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20 hours ago, Bryan Fowler said:

Ryan, when you say you compared the inverse square, are you saying you measured the same falloff with both sources? As in, you took a measurement, x=distance from the 216 frame, then doubled that distance and it was 1/4 of the brightness?

yes at 6 and 12 feet

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