Really stupid physics question

[unpopular]

I feel like I should be ashamed for not realizing the solution to this, but why is it that inverse square doesn't apply to diffuse light? Why don't things get darker as you view them from a distance the same way illumination does?

Does inverse square apply to specular?

 

[KmH]

Who says the Inverse Square law doesn't apply to difffuse light.
The Inverse Square Law applies to any energy, force, or conserved quantity that is radiated in 3 dimensions radially outward from a point source. Diffused light is just a collection of many point sources aimed in many different directions (scattered). If you are looking at a diffused light source, a lot of the scattered light isn't entering your eye.

Things do get darker as you view them from a distance, which is why astonomers have to use long exposure times (sometimes days) to image intrinsically very bright objects that are very far away.

While the sun is very bright here on earth, at the orbit of Pluto, the sun is no longer the brightest star in the sky, as seen from Pluto.

 

[Helen B]

The reason that things don't look darker further away is because the image is smaller in the same proportion as the light fall-off, therefore the image stays at the same brightness. (Double the distance, quarter the light, quarter the image size) The moon is quite a long way away (or so I am told) but it looks as bright as any sunlit landscape. Point sources, and sources that are effectively point sources (like stars) behave differently, because the image area loses its relationship to distance.

 

[Ysarex]

I believe the basic formula as we know it can be applied to point source lights but requires modification for light sources like soft boxes and umbrellas. I know I once saw a modified formula that could be used which plugged in the surface area of a softbox. I'll ask my son (physicist) and get back to you.

Joe

 

[unpopular]

No need, Helen adequately answered the question. Thanks Helen, as usual you're explanations are concise and clear.

Then if you took a telephoto lens, would exposure time be shorter with a wide angle lens under similar magnification (i.e. closer to the subject), provided that the f-ratio stayed constant?

 

[Helen B]

That's why the f-number is tied to the focal length. The image is larger with a longer lens, but the hole the light passes through is also larger in the same proportion (at the same f-number).

 

[unpopular]

Oh ok. That does make sense, actually. Thanks Helen.

It's very interesting that everything works so proportionally.

 

[Ysarex]

No need, Helen adequately answered the question. Thanks Helen, as usual you're explanations are concise and clear.

Then if you took a telephoto lens, would exposure time be shorter with a wide angle lens under similar magnification (i.e. closer to the subject), provided that the f-ratio stayed constant?

OK, but I had already sent the email -- so what the heck, here's his answer.
My son teaches physics at the Univ. MN and is a Phd candidate working in bio-physics (proud dad here).

Joe

--------------------------------------

The inverse square law is true for a light source that is a point - infinitely small. More precisely, the inverse square law is an approximation that is valid if you're far enough away from the light source that the physical extent of the source is negligibly small. So if you have a small flash, say 1 square centimeter, and you're 10ft from it, the inverse square law will give you reasonably accurate results. If you have a larger source, and you're not super far away from it, then you're dealing with something more complicated.

You can think of the large source as being composed of many dim point sources. (You can build something like a light box by gluing 500 small LEDs onto a piece of plywood.) Mathematically, to derive the "modified" inverse square equation, you would add up (integrate) the intensity contribution from each little light bulb. We know that the contribution from each little bulb is = (bulb intensity)/r^2, where r is the distance from a light bulb to the point you place your exposure meter. r is different for EACH little bulb. After doing the geometry and the integral, you will end up with an equation that depends on how far you are from your light box and the shape of the light box (ie, circle, triangle, rectangle, etc). Most light boxes have lengths that are approximately the same as their height (3' x 4'), so we might approximate all light boxes as something simple like a circle or a square. We can then express our approximate formula in terms of the surface area of the light box.

One interesting note: If you have a spherical light source (like one of those oriental paper lanterns), it still follows the inverse square law no matter how close you are to it. You can think of this as a quirk peculiar to spheres. Of course, this isn't going to be too interesting to photographers since most light boxes you'd encounter in a studio have flat surfaces.

Isaac

 

[Helen B]

Isaac,

That applies to illumination at a location, not to image brightness. When we view the brightness of a surface with our eyes, we should be considering image brightness, not the illumination falling on our face.

 

[Ysarex]

Isaac,

That applies to illumination at a location, not to image brightness. When we view the brightness of a surface with our eyes, we should be considering image brightness, not the illumination falling on our face.

Hi Helen,

My fault there. When I first saw Unpopular's question I zeroed in on the first sentence and thought of it as a question about light sources and that's what I passed on to Isaac. Later I did a better job reading Unpopular's second sentence.

Joe

 

[JAC526]

The reason that things don't look darker further away is because the image is smaller in the same proportion as the light fall-off, therefore the image stays at the same brightness. (Double the distance, quarter the light, quarter the image size) The moon is quite a long way away (or so I am told) but it looks as bright as any sunlit landscape. Point sources, and sources that are effectively point sources (like stars) behave differently, because the image area loses its relationship to distance.

I would like to challenge the TPF forums to try and stump Helen B.

Just saying. She has an intelligent answer for questions on so many threads.

 

[unpopular]

Isaac,

That applies to illumination at a location, not to image brightness. When we view the brightness of a surface with our eyes, we should be considering image brightness, not the illumination falling on our face.

Hi Helen,

My fault there. When I first saw Unpopular's question I zeroed in on the first sentence and thought of it as a question about light sources and that's what I passed on to Isaac. Later I did a better job reading Unpopular's second sentence.

Joe

Yeah, I should have been more clear, I was asking about diffuse reflected light, not diffused light sources. Still, your son's reply is an interesting read.

 

[table1349]

The reason that things don't look darker further away is because the image is smaller in the same proportion as the light fall-off, therefore the image stays at the same brightness. (Double the distance, quarter the light, quarter the image size) The moon is quite a long way away (or so I am told) but it looks as bright as any sunlit landscape. Point sources, and sources that are effectively point sources (like stars) behave differently, because the image area loses its relationship to distance.

I would like to challenge the TPF forums to try and stump Helen B. Just saying. She has an intelligent answer for questions on so many threads.

If a chicken and a half lays an egg an a half in a day and a half, how long would it take a monkey with a wooden leg to kick all of the seeds out of a dill pickle?

Question complements of Harry Anderson of night court fame.

 

[Ysarex]

The reason that things don't look darker further away is because the image is smaller in the same proportion as the light fall-off, therefore the image stays at the same brightness. (Double the distance, quarter the light, quarter the image size) The moon is quite a long way away (or so I am told) but it looks as bright as any sunlit landscape. Point sources, and sources that are effectively point sources (like stars) behave differently, because the image area loses its relationship to distance.

I would like to challenge the TPF forums to try and stump Helen B. Just saying. She has an intelligent answer for questions on so many threads.

If a chicken and a half lays an egg an a half in a day and a half, how long would it take a monkey with a wooden leg to kick all of the seeds out of a dill pickle?

Question complements of Harry Anderson of night court fame.

Depends. Is the monkey being paid time and a half or just half time?

 

[table1349]

I would like to challenge the TPF forums to try and stump Helen B. Just saying. She has an intelligent answer for questions on so many threads.

If a chicken and a half lays an egg an a half in a day and a half, how long would it take a monkey with a wooden leg to kick all of the seeds out of a dill pickle?

Question complements of Harry Anderson of night court fame.

Depends. Is the monkey being paid time and a half or just half time?

My first inclination was to ask this quantum physics question:

The electrons going around the nucleus can only exist in certain orbits - and they don't actually travel from one orbit to the next. Instead, they disappear in one orbit, only to reappear in the one above or below. Where are they when they're in between? Since they're like waves, are they actually *cancelling one another out*, like the crest and trough of a wave do?

I figured that this one would be too easy for Helen B. She is a world of knowledge, that is undeniable.