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cjbroadbent
30-Oct-2010, 10:52
The ISL is my enemy and I fight back with:
Fresnels with scrim flags,
Neutral density wedges worn sideways,
Very large window-lights with monstrous jagged-edge foam panel gobbos.
Window-size beehive grids,
Backgrounds (and foregrounds) painted dark on the near-side and light on the far-side.
Black flags, Barn-doors, cardboard and anything available that might gradually feather off the light without screwing the shadows (shadows get screwed whenever the flag or barn-door happens parallel to the far side of the window.

That's why round sources or sources with central hotspot are so much kinder to noses, boxes or anything else, for that matter).
In fact, an umbrella with the lamp stuck well inside to avoid spill, and pointed across the subject does a better job with the inverse square law. It is round, has a hotspot (better penumbra and brilliance) and the edges of the brolly feather nicely all by themselves.

But here is the question.
Using a real window and a clear sky. There should be no Inverse Square Law dropping a stop or two at between nine and twelve feet from the window. The sky is at infinity. But it does drop. Like when the window is covered with a sheet of frost. Why?

Jimi
30-Oct-2010, 10:58
The light coming in through the window from the outside is just a source of light, and the further you move from that source, the less light you get. No escape. :)

No, I can't explain it better than that, sorry.

jp
30-Oct-2010, 11:04
You are much better at lighting things than most of us; hopefully I'll be understood as just sharing my understanding rather than teaching.

Looking out my window I see more trees and branches than sky. So the trees are reflecting and filtering the light coming in. Someone else might be half lawn, half sky. If you have a sunbeam coming in, that is of course coming from practically infinity and will not exhibit ISL. I would pretend the normal diffuse illumination from the window is a big rectangular softbox with a light in it; a natural rembrandt lighting source.

Policar
30-Oct-2010, 11:18
The ISL is my enemy and I fight back with:
Fresnels with scrim flags,
Neutral density wedges worn sideways,
Very large window-lights with monstrous jagged-edge foam panel gobbos.
Window-size beehive grids,
Backgrounds (and foregrounds) painted dark on the near-side and light on the far-side.
Black flags, Barn-doors, cardboard and anything available that might gradually feather off the light without screwing the shadows (shadows get screwed whenever the flag or barn-door happens parallel to the far side of the window.

That's why round sources or sources with central hotspot are so much kinder to noses, boxes or anything else, for that matter).
In fact, an umbrella with the lamp stuck well inside to avoid spill, and pointed across the subject does a better job with the inverse square law. It is round, has a hotspot (better penumbra and brilliance) and the edges of the brolly feather nicely all by themselves.

But here is the question.
Using a real window and a clear sky. There should be no Inverse Square Law dropping a stop or two at between nine and twelve feet from the window. The sky is at infinity. But it does drop. Like when the window is covered with a sheet of frost. Why?

Hard light coming through a window does not suffer from ISL (well, except to negligible extent that across the room is farther from the sun than less far across the room is). Soft light through windows is 100% bounced light and suffers from ISL the same way bouncing any other source with a soft bounce card (white side, not mirror) would. I'm not sure what the physics are here and it's probably not strict ISL, but this is what's happening.

ic-racer
30-Oct-2010, 11:21
The closer the subject is to the window, the larger the angle of view out the window, which allows more light rays to hit the subject. As you move back, the angle of view diminishes. Eventually you will be so far back that you have only collimated rays from the window hitting the subject. These few rays are going to be only a very small subset of all the angled rays hitting the subject when the subject was close to the window, and therefore dimmer.

Jim Michael
30-Oct-2010, 12:36
Under some circumstances (e.g. w/fresnels) you may calculate the position of a virtual point source placed at a different distance than the actual source in order to see a result that maps distance in f-stop feet to falloff in f-stops.

Policar
30-Oct-2010, 12:44
The closer the subject is to the window, the larger the angle of view out the window, which allows more light rays to hit the subject. As you move back, the angle of view diminishes. Eventually you will be so far back that you have only collimated rays from the window hitting the subject. These few rays are going to be only a very small subset of all the angled rays hitting the subject when the subject was close to the window, and therefore dimmer.

I believe this is it; it's an issue with fast fall-off from all soft sources due to the varying angles of incidence, not ISL.

cjbroadbent
30-Oct-2010, 13:22
I believe this is it; it's an issue with fast fall-off from all soft sources due to the varying angles of incidence, not ISL. That is, Ic-Racer has it right, the nearest side gets more sky. So the further the subjects are from the window the less difference there will be. Just like Inverse Square Law unfortunately.

Tobias Key
30-Oct-2010, 15:10
Don't be fooled with all this the sun is 93 million miles from the earth crap. The window is an aperture and as such behaves in the same way as your camera aperture or any point source of light does - the position of the aperture is your light source and all calculations start from that point not not the position of the sun. Diffraction works from that aperture or window in the same way it works in a camera lens, otherwise you'd never get any light fall off from any daylight scene shot at close focus, after all what's an extra couple of inches when the sun is 93 million miles away?

As an experiment take light meter readings from a window at doubling distances, the inverse square law will apply exactly the same as if the window was a flashgun set at that spot.

Jim Michael
30-Oct-2010, 16:14
Not so simple as that. There are multiple possible scenarios and one of them is the sun shining directly through the window onto the subject, which is equivalent (approximately, not counting ambient from other sources) to being outside. The window acts more like a scrim; since the rays are parallel there is no effect of limiting the number of converging rays. Not sure what you're referring to WRT diffraction as it relates to the exposure reading of the subject.

Otherwise, it's as ic-racer describes when the window is transparent, and more like a broad source when the window is translucent or has a curtain etc. For a broad source, since ISL is based on a point source, you have to integrate over the the x and y dimensions to find the total intensity at the point where you're interested, i.e. sum the point sources that make up the broad source.

I posted my calculations on the virtual point source for sources with a specified beam spread on the cinematography list some time ago, may have that archived somewhere if there's any interest.

Richard Mc
30-Oct-2010, 22:08
A point source of light (eg. the sun) puts out a constant stream of photons in all directions. The intensity is light at any point in space looking back at the source is related to how closely the light rays are packed together. Picture a sphere (the sun) with 1000 evenly spaced arrows emanating out radially from the center. Close to the sun, the light rays are very closely spaced (ie. the arrows seem very closely packed here) and further out, less so. It is this exact phenomenon that gives us the maths for the inverse square rule as the surface area of a sphere is given by 4∏r*r....ie. proportional to the square of the radius. As photons get further out from the source they must spread out over an area that is proportional to the square of the distance from the source, hence the intensity (photons passing through unit area) is inversely proportional to the square of the distance from the source.

There is a very good explanation of this with diagrams at :

http://en.wikipedia.org/wiki/Inverse-square_law

Now, strictly speaking, this equation only applies for a point source radiating equally in all directions in a vacuum. Using this equation for non-point sources is only ever a rough approximation but it is helpful at times in pointing us in the right direction. Light modifiers (eg. reflectors etc.) and atmospheric conditions among other things cause the strictness of the equation to fall down to some degree.

Ignoring any atmospheric effects , the intensity of light from the sun at any two points on earth should be pretty similar, as the distances of these two points to the sun will be pretty similar, as the distance to the sun is large compared to the distances between any two points on earth.

Now once the photons hit something on earth (the side of the building that you can see out of the window or maybe the window shade) then their evenly spaced radially radiating relationship is no longer valid. They scatter in all directions (but not equally), so they become something like a new point source (or thousands of new little point sources) but, as they are not emanating from a single point and equally in every direction, the inverse square law (using the distance from the "new" light source to the subject) will only ever be a poor approximation.... but it will be more accurate than trying to apply the inverse square law using the distance to the sun as this no longer applies.

The light coming in through the window will probably obey the inverse square law to about the same degree as out photo lights and reflectors etc..... none of these strictly obey the inverse square law but it is a useful approximation to nudge us in the right direction. You'll just have to make some determination as to where the light in question is coming from and figure that into your approximation. If its direct sunlight, then the intensity won't change much if the subject is 1 foot from the window or 10 feet from the window (as long as they are still in direct sunlight). If it is reflected/scattered light from the sky then there will be slightly greater effect on moving the subject, and if the light is reflected/scattered from a nearby object (building, window shade), then the light will drop off much more with distance from the window.

The main use to me for the inverse square law is just to say that light drops off A LOT with distance. In the end, a bit of trial and error and a few meter readings usually gives us our correct settings.

jim kitchen
31-Oct-2010, 01:24
Ladies and Gentlemen,

I mention ladies if they decide to contribute or have contributed, but from what I remember about my practical physics in university and any subsequent light wave propagation, diffraction propagation or reflected light studies that I participated in while studying architecture, happens to be that I was taught that the Inverse Square Law did not have a corollary, unless a new science surfaced in the last few years that now indicates how the projected surface area measured from any light source type does not change with distance from the light source. I then believe that your hypotheses, your personal modifications to the ISL, or your interim interpretations of the ISL, whether it is a window, a door and, or a reflected source that allows a light source to strike the subject differently than what ISL emphatically states are absolutely incorrect.

To my knowledge, ISL does not have a corollary, no matter what the light source happens to be, whether it is diffused, diffracted, reflected, a single point source and, or whether the light source happens to be fifty feet away, or equal to the distance from our friendly neighbourhood sun.

I may be wrong after all these years complete with an aging memory, and although you may be able to apply ISL to celestial bodies where a mass in motion might have a Newtonian corollary, I would happily read your ISL dissertation or another physicist's paper, regarding the modification of ISL physics with interest... :)


jim k

Jack Dahlgren
31-Oct-2010, 01:46
Ladies and Gentlemen,

I mention ladies if they decide to contribute or have contributed, but from what I remember about my practical physics in university and any subsequent light wave propagation, diffraction propagation or reflected light studies that I participated in while studying architecture, happens to be that I was taught that the Inverse Square Law did not have a corollary, unless a new science surfaced in the last few years that now indicates how the projected surface area measured from any light source type does not change with distance from the light source. I then believe that your hypotheses, your personal modifications to the ISL, or your interim interpretations of the ISL, whether it is a window, a door and, or a reflected source that allows a light source to strike the subject differently than what ISL emphatically states are absolutely incorrect.

To my knowledge, ISL does not have a corollary, no matter what the light source happens to be, whether it is diffused, diffracted, reflected, a single point source and, or whether the light source happens to be fifty feet away, or equal to the distance from our friendly neighbourhood sun.

I may be wrong after all these years complete with an aging memory, and although you may be able to apply ISL to celestial bodies where a mass in motion might have a Newtonian corollary, I would happily read your ISL dissertation or another physicist's paper, regarding the modification of ISL physics with interest... :)


jim k

Not sure what you are going on about. As an example. the intensity of light at a point 4 inches behind a 100mm lens in bright sunlight is enough to kill an ant, a creature impervious to ordinary sunlight.

The window, if acting as a diffuser through presence of dirt, curtain, etc. will do the opposite, diffusing the light to a varying extent.

jim kitchen
31-Oct-2010, 09:12
Not sure what you are going on about. As an example. the intensity of light at a point 4 inches behind a 100mm lens in bright sunlight is enough to kill an ant, a creature impervious to ordinary sunlight.

The window, if acting as a diffuser through presence of dirt, curtain, etc. will do the opposite, diffusing the light to a varying extent.

Dear Jack,

I can only agree with your examples, but what I seem to be going on about, as you politely put it, happens to be about the known physical characteristic of the ISL, where the light source's intensity is inversely proportional to the distance from the source. I believe that no corollaries exist that modify the original law.

Your examples talk to a light source's modified intensity over a modified surface area. Your first example points to and introduces a physical change and reconcentration to the original light source intensity through an optical device, where the light source's intensity is obviously dispersed within a smaller concentrated surface area as it strikes the poor ant. Your first example will surely follow the ISL as the refocused and reconcentrated light source intensity leaves the lens's rear element.

The intensity of the light source within the ISL, as you state in your second example, will surely change too as the light source becomes diffused, dispersed and, or diffracted over a larger surface area. Every light source and modified light source complete with its inherent intensity will always follow the ISL, whether the light source passes through a window, a door opening, or happens to be reflected off a surface material and, or diffused by any translucent material.

I thought my ISL comments were clear about its known characteristic, but then again, I apologize if I was not clear enough... :)


jim k

cjbroadbent
31-Oct-2010, 10:03
We are talking about a window, a large rectangular light source facing the north sky. The conclusion, with regard to the Inverse Square Law, seems to be that the origin of the source is at the window frame, regardless of whether the window is covered with a diffuser or left wide open. Is that it?

cowanw
31-Oct-2010, 12:39
Yes. (in the northern hemisphere)
Regards
Bill

Ben Syverson
31-Oct-2010, 13:50
The soft light from a window is the collected reflections of the outside world, sky, etc. The sun and sky obviously don't suffer from ISL (at least on Earth). So what gives?

Well, imagine the window from the perspective of someone standing directly in front of it. From their perspective, the window is huge. Their face gets light from many different directions (top of the window, bottom, left, right), so the quality of the light is soft and bright. If they look out the window, they can see the landscape and clouds.

Now imagine the window from the perspective of someone standing 100 feet away (we'll assume it's the only window in a great hall). From their perspective, the window is tiny. Their face is only getting light from the very specific direction of the window, so the quality of light is hard, like a bare bulb. The light is also dimmer, which is NOT because the window light itself is suffering from ISL.

The light is dimmer because the percentage of reflected light in the world that can reach the subject has plummeted. The person standing at the window can see buildings, landscape and clouds, which all contribute reflected light to their face. The person at the end of the great hall perhaps only sees the mountaintop or church steeple through the window. The only light that's reaching them is the meager amount of light reflected off the steeple.

So it has nothing to do with ISL!

jim kitchen
31-Oct-2010, 14:44
We are talking about a window, a large rectangular light source facing the north sky. The conclusion, with regard to the Inverse Square Law, seems to be that the origin of the source is at the window frame, regardless of whether the window is covered with a diffuser or left wide open. Is that it?

Dear Christopher,

That is correct, and to add to your comment, a new and totally unique ISL is created from every reflective surface, if one could completely isolate, identify, and measure the reflected light source... :)

jim k

Jim Michael
31-Oct-2010, 14:49
The window scenario is open to a lot of hand waving, so I'll just assume a broad source where the window is providing the north light illumination. If you have a different interpretation then the following won't work.

Broad sources do follow ISL but it's complicated by the fact that you have to account for more than one point source, hence my reference to integrating over the source. For a source orthogonal to a line extending from the source to the subject, let d0 be the distance from the center of the source to the subject and d1 be the distance from the source at a point x units from the center, then d1 = sqrt(d0*d0 + a*a) and the change in intensity of the light originating from d0 to d1 follows the inverse square law. The intensity at the subject is the sum (integral) of those intensities.

And, ISL is nothing more than saying that as the light leaves a point source the area of the sphere onto which the light falls at a distance gets bigger by a predictable amount (look up area of sphere and calculate change in area between two radii of f-stop feet like 8 and 11). You just have to decide from where the light originates.

I looked up what I wrote before re the sources with a stated beam spread, e.g. fresnel:

Assume a source with a beam spread angle of phi and intensity I. At distance d, the intensity is I/area or I/pi(d tan phi)^2. The ratio of intensity at distances d' and d is i'/i = ((d' tan(phi))^2)/((d1 tan(phi))^2) so i' = i((d' tan(phi))^2)/((d1 tan(phi))^2) or i'=i*d'^2/d^2. Picking a couple of f-stop feet, 8 and 16, i' = 4.

And re the virtual point source:

Assume a beam from a source of radius r and beam divergence phi. We can think of this as a point source originating at a distance d0 behind the source where d0 = r/tan phi. Then when comparing illumination at a distances d' and d'', use the distance to the virtual point source in place of the physical source. In the case of a perfectly collimated beam the virtual point source is at infinity, so intensity i' = i''.

Check my math.

cjbroadbent
1-Nov-2010, 02:48
Jim K, thanks, you always have a balanced and well-informed reply to our primitive questions.
Jim M, Wow! I'ts going to take me awhile to work that out. I'll print it and hang it on the wall. The great thing about LF users is their wide knowlege base from other fields.

ki6mf
1-Nov-2010, 03:41
All the more reason to have a hand held meter to measure the quality of light from all directions so as to create dramatic lightning by moving camera subject lights etc! There are times when I use my Sekonic for incidence over my Pentax V spot meter

Robert Hughes
2-Nov-2010, 12:09
I believe the inverse square law refers to how light intensity drops off with the square of distance from the light source to the illuminated object. As the sun is 93 million miles away, a couple extra feet into the room is not going to make any difference whatsoever.

ki6mf
2-Nov-2010, 13:23
Inverse Square Law would apply from the point where light enters. In this case a window is the source where light enters and the square of the distance determines how the light falls off in intensity. With the subject 10 feet from the window the light is 100 times weaker even though the sun is 93 million miles away! This is why you can shoot a digital picture of a subject with on camera flash in your point and shoot camera and if the wall behind your subject is 10 feet away the background goes dark!

Bruce Watson
2-Nov-2010, 14:33
The window is an aperture and as such behaves in the same way as your camera aperture or any point source of light does - the position of the aperture is your light source and all calculations start from that point not not the position of the sun.

+1. Exactly right.

Bruce Watson
2-Nov-2010, 14:36
We are talking about a window, a large rectangular light source facing the north sky. The conclusion, with regard to the Inverse Square Law, seems to be that the origin of the source is at the window frame, regardless of whether the window is covered with a diffuser or left wide open. Is that it?

That's it.