# Really stupid physics question



## unpopular (Jun 26, 2012)

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 (Jun 26, 2012)

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 (Jun 26, 2012)

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 (Jun 26, 2012)

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 (Jun 26, 2012)

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 (Jun 26, 2012)

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 (Jun 26, 2012)

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

It's very interesting that everything works so proportionally.


----------



## Ysarex (Jun 26, 2012)

unpopular said:


> 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 (Jun 26, 2012)

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 (Jun 26, 2012)

Helen B said:


> 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 (Jun 26, 2012)

Helen B said:


> 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 (Jun 26, 2012)

Ysarex said:


> Helen B said:
> 
> 
> > Isaac,
> ...



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 (Jun 26, 2012)

JAC526 said:


> Helen B said:
> 
> 
> > 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.
> ...


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 (Jun 26, 2012)

gryphonslair99 said:


> JAC526 said:
> 
> 
> > Helen B said:
> ...




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


----------



## table1349 (Jun 26, 2012)

Ysarex said:


> gryphonslair99 said:
> 
> 
> > JAC526 said:
> ...




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.


----------



## pgriz (Jun 26, 2012)

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



And who also manages to suffer the fools among us and not get unhinged.  Another plus.


----------



## unpopular (Jun 26, 2012)

Oh but when she does it is very, very impressive ....


----------



## pgriz (Jun 26, 2012)

Ah yes, I remember that incident.  Made a mental note NOT to pick a fight with HelenB.  There's just no win in that.


----------



## tris_d (Nov 13, 2012)

Ysarex said:


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




- "_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._"


Did this guy just say that spherical light source if photographed from, say two meters away, will look not only smaller, but would also be four times less bright than when photographed from one meter away?


----------



## christop (Nov 13, 2012)

tris_d said:


> Did this guy just say that spherical light source if photographed from, say two meters away, will look not only smaller, but would also be four times less bright than when photographed from one meter away?



No.


----------



## tris_d (Nov 13, 2012)

christop said:


> tris_d said:
> 
> 
> > Did this guy just say that spherical light source if photographed from, say two meters away, will look not only smaller, but would also be four times less bright than when photographed from one meter away?
> ...



What then? What else could he mean by "follows the inverse     square law"?


----------



## unpopular (Nov 13, 2012)

light is focused to to a smaller region proportional to it's distance. This is why it's relative brightness appears similar and unaffected by distance - because our eyes and cameras alike have lenses that focus the light. What that light illuminates does not focus it, this is why inverse square appears to apply to ambient illumination, and directly imaged - such as the candle illustration you had in your other threads. The more distant, the smaller the projection and more densely distributed. Unfocused, the light falls off by inverse square as expected.

It's hard to imagine, because we always assume that light is packaged into nice little bundles that are easily determined from one another. This isn't how light behaves - we've developed lenses so that we can determine objects. We think of a cameras lens as a sort of artificial aid to make light behave as it "truly is". But light in it's natural form is just ambient radiation, scattering out in all directions. It's not focused into neat little packets and projected into distributions relative to distance, simply because we need it to.

All light follows inverse square. Lenses just take that dispersed light and focus it back into a representation of the source. Light radiates off the source by emission or reflection, spreads out according to inverse square, some portion of it enters the lens and is focused back down to a region we call "bright": a relatively dense region of photons.

In fact, if the back focus distance is too great, you again loose brightness. In photography (since ypu're clearly not a photographer), this is compensated by a bellows extension factor.


----------



## tris_d (Nov 13, 2012)

unpopular said:


> light is focused to to a smaller region proportional to it's distance. This is why it's relative brightness appears similar and unaffected by distance - because our eyes and cameras alike have lenses that focus the light. What that light illuminates does not focus it, this is why inverse square appears to apply to ambient illumination, and directly imaged - such as the candle illustration you had in your other threads. The more distant, the smaller the projection and more densely distributed. Unfocused, the light falls off by inverse square as expected.
> 
> It's hard to imagine, because we always assume that light is packaged into nice little bundles that are easily determined from one another. This isn't how light behaves - we've developed lenses so that we can determine objects. We think of a cameras lens as a sort of artificial aid to make light behave as it "truly is". But light in it's natural form is just ambient radiation, scattering out in all directions. It's not focused into neat little packets and projected into distributions relative to distance, simply because we need it to.
> 
> ...



Interesting. Ok, thank you. That's whole another thing to consider now, I wish you mentioned it before.


----------



## unpopular (Nov 14, 2012)

*blink* it was THIS thread that explained it to me!


----------



## Fred Berg (Nov 14, 2012)

Oh lord, this inverse square law is difficult. What I learnt last time this came up (and is useful and relevant to me) is, the larger an object is the more light (or time) you need to properly expose it. Is this why when you zoom out in A mode the shutter time speeds up? Zoom in to 70mm on an ornament from 3 meters away: 1/60 , zoom out to 35mm: 1/250. This didn't come directly out of the last discussion, but this is what I boiled things down to after scratching my head and digging around in various books. Otherwise, the only real use for this law seems to be when working out how much drop off there will be when using flash.

If I wasn't tea-total, I'd pour myself a stiff drink whilst trying to get to grips with this.


----------



## christop (Nov 14, 2012)

Fred Berg said:


> What I learnt last time this came up (and is useful and relevant to me) is, the larger an object is the more light (or time) you need to properly expose it.


I don't know how you came up with that idea. It doesn't make any sense.



> Is this why when you zoom out in A mode the shutter time speeds up? Zoom in to 70mm on an ornament from 3 meters away: 1/60 , zoom out to 35mm: 1/250.


More likely you included brighter objects in the frame when you zoom out (such as lights) which causes the auto metering to speed up the shutter speed to compensate. Or the auto metering chooses a wider aperture at 35mm than at 70mm. Inexpensive zoom lenses have variable maximum aperture: my 18-50mm lens ranges from f/3.5 at 18mm to f/5.6 at 50mm, for example. This is about 1.3 stops difference.


----------



## runnah (Nov 14, 2012)

Nice try photography...you almost made me learn physics.


----------



## tris_d (Nov 14, 2012)

unpopular said:


> *blink* it was THIS thread that explained it to me!



All I see is they told you, like they told me, that things appear four times smaller at double the distance and that is supposed to explain inverse square law, which it does not. Now, while your explanation with focusing makes sense I don't see how focus could compensate for a light that is supposed to appear four times dimmer at double the distance or sixteen times dimmer at quadruple the distance. Also I could get that you can focus one light source, but how in the world could you focus on many light sources at once, like on that image with street lights.

Looking back at christop's light bulbs I could say now the further one does not look dimmer because the light source is not spherical, but the filament is cylindrical, so inverse square law was not to be expected. However, if we were to look at ambient (walls) around that light bulb we would expect to see inverse square law drop off as you explained, so I still can't make sense of it. -- I wish we could summon Isaac to shed some more light on the subject. Ysarex, you there?


----------



## amolitor (Nov 14, 2012)

tris_d said:


> unpopular said:
> 
> 
> > *blink* it was THIS thread that explained it to me!
> ...



You are precisely and exactly wrong, here. Lest someone stumble across this and become confused, let me be quite clear: Tris is wrong on this point. Things appear 4 times smaller at double the distance, and that is exactly the explanation for the inverse square law.


----------



## fjrabon (Nov 14, 2012)

amolitor said:
			
		

> You are precisely and exactly wrong, here. Lest someone stumble across this and become confused, let me be quite clear: Tris is wrong on this point. Things appear 4 times smaller at double the distance, and that is exactly the explanation for the inverse square law.



It's like you think he finally gets it, and then all of the sudden we are back at square one.


----------



## unpopular (Nov 14, 2012)

fjrabon said:


> amolitor said:
> 
> 
> 
> ...



and 1^2 always equals 1.


----------



## amolitor (Nov 14, 2012)

fjrabon said:


> amolitor said:
> 
> 
> 
> ...



Tris will never get it. That's kind of the point of these threads, is to not get it. I neither know nor care whether Tris actually understands this or not, it doesn't make any difference from our perspective. The behavior is 100% predictable, it's boring, and it wastes everyone's time.


----------



## fjrabon (Nov 14, 2012)

amolitor said:
			
		

> Tris will never get it. That's kind of the point of these threads, is to not get it. I neither know nor care whether Tris actually understands this or not, it doesn't make any difference from our perspective. The behavior is 100% predictable, it's boring, and it wastes everyone's time.



Or maybe tris really DOES get it and the inverse square law is the key to a unified field theory and we are all just sheeple falling under the ills of general relativity, quantum mechanics and string theory.


----------



## unpopular (Nov 14, 2012)

It's actually kind of interesting. I think there is a sort of status frustration in scientific discovery. Average people of average resources and education can no longer participate in breakthrough science. We're human, we're naturally curious, we have a need to make discoveries and contribute to scientific endeavors, but we can't. So some of us convince ourselves that we've made some important breakthrough by flawed observations of everyday objects. The need itself to make the discovery is so vastly important, that we look over the obvious - people have been observing candles and other arrangements of lights for tens of thousands of years, and that if candles arranged in rows did behave in some way to uproot fundamental understanding of cosmology, you'd think we would have observed this behavior by this point.

I can understand the desire to explore, discover and understand. I can understand the frustration in the realization that simply thinking outside the box is no longer enough to make significant headway in experimental science. But none of this changes the fact that human knowledge has surpassed the ability of any one person's mind and resources, and in a way, that's an amazing realization in and of itself.


----------



## unpopular (Nov 14, 2012)

fjrabon said:


> amolitor said:
> 
> 
> 
> ...



And energy industry cronyism lead by the reptilian overlords who seek to destroy earth and know that if they keep us away from Captret technology (seriously, youtube it) we'll never develop the energy necessary to defend ourselves, and ultimate destroy their planet with warp-capable starships with phase canons!


----------



## tris_d (Nov 14, 2012)

amolitor said:


> tris_d said:
> 
> 
> > unpopular said:
> ...



Just please point some reference. Surely if what you say is true it should be mentioned in some text books, there would be some article on the internet about it, right? It's not some kind of secret knowledge, I suppose, so please just give me some link where I can confirm what you are saying.


----------



## amolitor (Nov 14, 2012)

tris_d said:


> Just please point some reference. Surely if what you say is true it should be mentioned in some text books, there would be some article on the internet about it, right? It's not some kind of secret knowledge, I suppose, so please just give me some link where I can confirm what you are saying.



No. I will not be baited.


----------



## tris_d (Nov 14, 2012)

fjrabon said:


> amolitor said:
> 
> 
> 
> ...




Take a piece of paper and put lit candle 10cm away from it, you see a bright blob on the paper. Move the candle to 20cm away from the paper and you see not only four times smaller blob, but is also four times less bright. Correct? So, when we replace the paper with a photo, why would it be different?


----------



## unpopular (Nov 14, 2012)

NO!!

Because the blob becomes smaller at the same rate which the energy is dispersed, the blob's relative intensity stays the same.

Yes, absolute energy is dispersed, if you take a photovoltaic cell and read the current at 1cm from the source it will be greater than at 10m, but brightness stays the same. _Brightness_ is determined by energy distribution, not quantity. This regardless if it is focussed onto our retna, film or a sensor.

See http://en.wikipedia.org/wiki/Lumen_(unit) and by extension http://en.wikipedia.org/wiki/Lux 

and for god sake, follow every single link that you don't understand.


----------



## tirediron (Nov 14, 2012)

Why?  Because Physics said so, that's why!


----------



## tris_d (Nov 14, 2012)

amolitor said:


> tris_d said:
> 
> 
> > Just please point some reference. Surely if what you say is true it should be mentioned in some text books, there would be some article on the internet about it, right? It's not some kind of secret knowledge, I suppose, so please just give me some link where I can confirm what you are saying.
> ...



I googled it for days, and I found hundreds of articles talking about light and inverse square law, yet I could not find anyone anywhere mentions anything like what you said.


----------



## tris_d (Nov 14, 2012)

tirediron said:


> Why?  Because Physics said so, that's why!



What physics, what equation? Where did you read about it? Can you point some reference? Where exactly is the difference between a paper and a photo?


----------



## amolitor (Nov 14, 2012)

tris_d said:


> Take a piece of paper and put lit candle 10cm away from it, you see a bright blob on the paper. Move the candle to 20cm away from the paper and you see not only four times smaller blob, but is also four times less bright. Correct?



Actually, incorrect.


----------



## unpopular (Nov 14, 2012)

tris_d said:


> tirediron said:
> 
> 
> > Why? Because Physics said so, that's why!
> ...



maybe you should start with Pythagorean Theorem.

If you don't know how that applies, then maybe you should start here:

Lens (optics) - Wikipedia, the free encyclopedia


----------



## fjrabon (Nov 14, 2012)

tris_d said:


> amolitor said:
> 
> 
> > tris_d said:
> ...



Your issue is this, which we have said multiple times: THE INVERSE SQUARE LAW ONLY APPLIES TO POINT SOURCES, ie sources of light that do not have any dimensionality and radiate light equally from that point in all directions.  You can then generalize that to other forms, based on a serious of more complicated equations, or just fudge it, if the non point object is small enough.  

Because spheres are literally formed by the inverse square law (ie gravity), they also follow the inverse square law. Flat panels of light, or groups of light that are visually in line with one another or any shape that isn't a tiny point or a sphere, DO NOT FOLLOW THE INVERSE SQUARE LAW.  You can use a giant softbox, which essentially turns a small source of light into a large panel of light, and meter the light coming from it, and it clearly doesn't follow the inverse square law, which is the whole point of the thing in the first place.  Under your justification, all objects would have to follow the inverse square law, regardless of size or shape.  WHy would a sphere dim differently than a flat panel?  

However, when you understand the justification for the inverse square law, we've laboriously been trying to give you, you'll understand why the inverse square law works for points and spheres and extremely small objects in the visual frame.  

I think the biggest issue is you keep confusing light emitted by a subject on a viewer with apparent brightness of a point.  ie think of a large floodlight, closer, it takes up more of your visual frame, it's emitting more total light on you.  However, the intensity of the vectors of light isn't less further away, it's that fewer vectors are hitting you.  any single vector of light is the same intensity as before.  The light didn't lose energy.  

Additionally, if you consider your diagrams, the lights you are drawing aren't following the inverse square law, their brightness is decreasing by a power raised to the 4th.  If the diagrams were as you drew them, the power of my speedlight would, for example 10,000 at 1 foot.  At 2 feet the power would only be 100.  If we followed the diagrams you are showing.


----------



## fjrabon (Nov 14, 2012)

amolitor said:


> tris_d said:
> 
> 
> > Take a piece of paper and put lit candle 10cm away from it, you see a bright blob on the paper. Move the candle to 20cm away from the paper and you see not only four times smaller blob, but is also four times less bright. Correct?
> ...



haha, actually in his example the blob gets bigger, not smaller.  Like hes seriously never flashed a flashlight at a sheet?


----------



## unpopular (Nov 14, 2012)

He's probably a physics grad student at Cornell just trying to **** with us.

Physicist troll....


----------



## Helen B (Nov 14, 2012)

tris_d said:


> amolitor said:
> 
> 
> > tris_d said:
> ...



It is mentioned in _many_ textbooks, and in many places on the internet, but I guess that you can't recognize it for what it is. Here's one version from _Optics in Photography_ by Rudolf Kingslake, who used to be the Director of Optical Design at Eastman Kodak, and Emeritus Professor at the University of Rochester. He also received the Gold Medal from The International Society for Optical Engineering. I don't think he has a blog, or even a Facebook page, so I'm not sure if he can really be trusted, to be honest. 

E = t &#960; B / (4 N^2)

Where 
E is the image illumination,
t is the lens transmittance (dimensionless),
&#960; is Pi (dimensionless),
B is the object brightness (at the object), and 
N is the f-number (dimensionless).

Note the absence of any term relating to the distance to the object.


----------



## tris_d (Nov 14, 2012)

unpopular said:


> NO!!
> 
> Because the blob becomes smaller at the same rate which the energy is dispersed, the blob's relative intensity stays the same.



Light has intensity, blob has brightness. I'm talking about blob's brightness. 

Previously you said: 
- "What that light illuminates does not focus it, this is why inverse square appears to apply to ambient illumination.."




> _Brightness_ is determined by energy distribution, not quantity.



But quantity is determined by distribution. That's what inverse square law is about. Distribution gets more sparse and thus quantity per surface area becomes less. Less dense distribution means less quantity, which means less brightness. 




> See Lumen (unit) - Wikipedia, the free encyclopedia and by extension Lux - Wikipedia, the free encyclopedia
> 
> and for god sake, follow every single link that you don't understand.



What exactly should I look at there, what parts are you referring to?


----------



## unpopular (Nov 14, 2012)

helen's equation is more concise.

but seriously, if you _can't_ get this from what has been said, you have a LONG ways to go.


----------



## tris_d (Nov 14, 2012)

fjrabon said:


> haha, actually in his example the blob gets bigger, not smaller.  Like hes seriously never flashed a flashlight at a sheet?



You are right. I was thinking of the blob I get as a reflection on my LCD.


----------



## christop (Nov 14, 2012)

Helen B said:


> I don't think he has a blog, or even a Facebook page, so I'm not sure if he can really be trusted, to be honest.



Lol.


----------



## fjrabon (Nov 14, 2012)

here's my last try:

I got PICTURES!

First, to sort of disprove your idea earlier of the blob getting smaller and dimmer:

First a piece of paper resting on a lamp, with none of the image clipped:




DSC_4405 by franklinrabon, on Flickr

Next, picked off of the lamp by 1/4"




DSC_4406 by franklinrabon, on Flickr

Finally about an inch above the lamp:




DSC_4407 by franklinrabon, on Flickr

As you can see, the 'blob' gets bigger, as the light gets dimmer.

What this doesn't mean was that if I was to look at the lamp directly that the actual lamp itself would appear dimmer.  It's that the total light falling on my eye would be less, because it's coming from a smaller part of my visual frame, due to how our eyes focus.  Once you take focus out of the equation, further light sources are actually in some sense bigger the further they are.  

here is a sort of graphical explanation with some writing and diagrams as well:




DSC_4404 by franklinrabon, on Flickr


----------



## tris_d (Nov 14, 2012)

unpopular said:


> tris_d said:
> 
> 
> > tirediron said:
> ...



Where exactly is the difference between a paper and a photo?


----------



## unpopular (Nov 14, 2012)

the lack of a lens in front of the paper.


----------



## tris_d (Nov 14, 2012)

fjrabon said:


> Your issue is this, which we have said multiple times: THE INVERSE SQUARE LAW ONLY APPLIES TO POINT SOURCES, ie sources of light that do not have any dimensionality and radiate light equally from that point in all directions.  You can then generalize that to other forms, based on a serious of more complicated equations, or just fudge it, if the non point object is small enough.



Finally we can agree on something, so let's take it from here. Consider what Isaac said:
- "_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._"


Does what he said not mean if we photograph such spherical light source it will produce less bright blob on the image proportionally to the square of the distance, as if it was a point light source?


----------



## fjrabon (Nov 14, 2012)

tris_d said:


> fjrabon said:
> 
> 
> > Your issue is this, which we have said multiple times: THE INVERSE SQUARE LAW ONLY APPLIES TO POINT SOURCES, ie sources of light that do not have any dimensionality and radiate light equally from that point in all directions.  You can then generalize that to other forms, based on a serious of more complicated equations, or just fudge it, if the non point object is small enough.
> ...



No, because you're optically altering the light due to having a lens in front of it.  You're taking all that dispersed light and then recombining it into a smaller light, of equal brightness.  Which is why we kept telling you all along that you can represent the falloff as the light sources being dimmer, or smaller, BUT NOT BOTH.


----------



## Bitter Jeweler (Nov 14, 2012)




----------



## christop (Nov 14, 2012)

Think about this....

You have two identical pieces of paper, one at 1m and the other at 2m from a spherical light source (such as a paper lantern). The paper at 2m receives one fourth as much light as the paper at 1m, so it appears dimmer. Correct?

Ok, so now we replace the square papers with square lenses of the same size. The total amount of light energy that passes through the lenses is the same as was originally illuminating the papers. We then put a flat surface behind the lenses so we can view the real image of the light sphere that is projected by each lens. The surfaces are placed at the same distance behind each lens. This surface could be a sensor or film if we wish to record the image.

The lens at 2m projects an image onto the surface that is half the size (one quarter of the area) as the lens at 1m. Since the total amount of light projected into that image is also one quarter as much as the image at 1m, the image appears just as bright on both images.


----------



## fjrabon (Nov 14, 2012)

FInally, I was able to find a lamp that gave reasonable exposures and could do your 'test' from earlier:

lamp, 1 ft away




DSC_4410 by franklinrabon, on Flickr

Lamp, 2 ft away




DSC_4411 by franklinrabon, on Flickr

In the first image, does the lamp look 4 times brighter?  It makes up a small enough portion of the visual frame that the inverse square law would only be off by about 5%.  So I guess you'd be asking does it look 395% brighter, technically.


----------



## amolitor (Nov 14, 2012)

It doesn't matter what you say, or what pictures you post, or what references you cite, guys. Tris is going to either ignore your post, or dismiss it on some trivial quibble (which might be real but will more likely be nonsensical), and then request that you "please just do one more little thing."

It doesn't end. It's about attention, not learning.


----------



## unpopular (Nov 14, 2012)

Bitter Jeweler said:


>



this is the only solution.


----------



## amolitor (Nov 14, 2012)

The question is, really, who's head to we bang here?

For extra fun, google "tris_d inverse square law". Tris has been fighting this one across the internet for a few days now, and the battle still rages. It's about attention, and Tris is winning.


----------



## christop (Nov 14, 2012)

Here's a version of my previous post with ASCII art illustration!


Papers:
.....A.........................B
.....|................o........|
......<-----.2m.----->.<-.1m.->

Lenses:
.....A.........................B
|....|................o........|....|
......<-----.2m.----->.<-.1m.->


(Sorry, this forum doesn't like to preserve spaces, and it doesn't have a "code" tag for some reason. Just ignore the periods.)

Paper A and lens A receive the same amount of light. Lens A passes that energy onto the surface behind it. Image A is the real image of the spherical light that lens A projects. Image A consists of 1/4 as much light energy as image B, but image A also has 1/4 the area as image B.


----------



## tris_d (Nov 14, 2012)

Helen B said:


> It is mentioned in _many_ textbooks, and in many places on the internet, but I guess that you can't recognize it for what it is. Here's one version from _Optics in Photography_ by Rudolf Kingslake, who used to be the Director of Optical Design at Eastman Kodak, and Emeritus Professor at the University of Rochester. He also received the Gold Medal from The International Society for Optical Engineering. I don't think he has a blog, or even a Facebook page, so I'm not sure if he can really be trusted, to be honest.
> 
> E = t &#960; B / (4 N^2)
> 
> ...



Great, thank you. It's just that astronomers use brightness to calculate distance to stars, and if distance is supposed to be irrelevant then you should see where my questions are coming from and why I can not settle with such answer. Ok, so the answer to my confusion is that stars are point light sources and thus their brightness is actually related to distance, right?


----------



## fjrabon (Nov 14, 2012)

tris_d said:


> Helen B said:
> 
> 
> > It is mentioned in _many_ textbooks, and in many places on the internet, but I guess that you can't recognize it for what it is. Here's one version from _Optics in Photography_ by Rudolf Kingslake, who used to be the Director of Optical Design at Eastman Kodak, and Emeritus Professor at the University of Rochester. He also received the Gold Medal from The International Society for Optical Engineering. I don't think he has a blog, or even a Facebook page, so I'm not sure if he can really be trusted, to be honest.
> ...



The answer is that stars are indiscernibly small from the distance they are, and thus we only perceive the drop in brightness, and cannot perceive their decrease in size.  However, they still emit the same amount of light no matter how far they are, until they are no longer in our hubble sphere, and thus, are invisible to us, because the distance between us and them is expanding faster than the speed of light, and thus their light never reaches us.


----------



## tris_d (Nov 14, 2012)

amolitor said:


> It doesn't matter what you say, or what pictures you post, or what references you cite, guys. Tris is going to either ignore your post, or dismiss it on some trivial quibble (which might be real but will more likely be nonsensical), and then request that you "please just do one more little thing."
> 
> It doesn't end. It's about attention, not learning.



Stop trolling me, please. You are not contributing and your comments are unnecessary. If you don't care just mind your own business. I think we are now close to wrap this thing up, so again, please don't pull my tail. We are so close, don't get me banned now.


----------



## fjrabon (Nov 14, 2012)

tris_d said:


> amolitor said:
> 
> 
> > It doesn't matter what you say, or what pictures you post, or what references you cite, guys. Tris is going to either ignore your post, or dismiss it on some trivial quibble (which might be real but will more likely be nonsensical), and then request that you "please just do one more little thing."
> ...



You've been provided with pictures disproving your theory.  How are we 'close to wrapping this up'?  It's over, just like we all told you from the beginning.


----------



## unpopular (Nov 14, 2012)

I don't even know why you even thought this would be a good forum to discuss this. Helen is the only one qualified to concisely answer your question, and she has. If she weren't here, then what could you expect?

Either post some photos, or GTFO.


----------



## Helen B (Nov 14, 2012)

tris_d said:


> Helen B said:
> 
> 
> > It is mentioned in _many_ textbooks, and in many places on the internet, but I guess that you can't recognize it for what it is. Here's one version from _Optics in Photography_ by Rudolf Kingslake, who used to be the Director of Optical Design at Eastman Kodak, and Emeritus Professor at the University of Rochester. He also received the Gold Medal from The International Society for Optical Engineering. I don't think he has a blog, or even a Facebook page, so I'm not sure if he can really be trusted, to be honest.
> ...



Aaagh. We're going round in circles. The above equation does not contradict the inverse square law at all.


----------



## tirediron (Nov 14, 2012)

*I think we've pretty much done this to death...  perhaps the OP could start a new thread to explain why it's possible to divide by zero?*


----------

