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Archive 2012 · silly inverse square law question.
  
 
RustyBug
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p.5 #1 · p.5 #1 · silly inverse square law question.




ukphotographer wrote:
...It doesn't matter how far you are from your subject with your camera, only the subject to light source distance and the exposure value it provides is important.


+1

Energy sent - (energy absorbed + energy refracted) = energy reflected iaw AI=AR.



Nov 07, 2012 at 05:19 AM
BrianO
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p.5 #2 · p.5 #2 · silly inverse square law question.


RustyBug wrote:
Energy sent - (energy absorbed + energy refracted) = energy reflected iaw AI=AR.


Sorry, Rusty, but the first half of your equation has to do with intensity, and the second has to do with direction. "Energy [intensity] reflected" does NOT mean the same as "angle of incidence" or "angle of reflection."



Nov 07, 2012 at 05:27 AM
curious80
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p.5 #3 · p.5 #3 · silly inverse square law question.


RustyBug wrote:
WRONG !!!

We get the same number of light rays reflecting off the black (unless the black is a black hole that will not allow any light to escape/reflect).

We get the same number of light rays, but the additional amount of energy being absorbed by the black cloth, renders less energy being reflected. The magnitude of the energy is reduced, but the direction remains iaw AI=AR. Thus the amount of energy reaching the sensor is now less and and cannot be converted into as strong of a signal (or is insufficient energy to effect a change to the negative), and
...Show more

Sure, to be completely pedantic we have to say that the black part does reflect rays which are very low energy and these form the very dark image on the sensor (though it doesn't effect the end result). So to summarize with this assumption: Initially we set the camera such that we capture just the original uncovered part of the box as I drew above. Now we move the lens+sensor back. We loose some rays from the original part of the box, but as you said we gain some more "low energy rays" from the black part of the image. However these extra low energy rays are just used to form the dark image of the black cloth. As far as the image of the original box is concerned, it only depends on the rays from come from that original part of the box. And we have lost some of those rays. So the total energy that is now reaching from original part of the box to the image of that original part on the sensor has gone down. Are we in agreement now?


Edited on Nov 07, 2012 at 05:40 AM · View previous versions



Nov 07, 2012 at 05:37 AM
curious80
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p.5 #4 · p.5 #4 · silly inverse square law question.


BrianO wrote:
I think so, too.

Thank you Ian! That sums it up in one succinct sentence. (Although I'm sure at least one person here still won't accept it.)


We are all in perfect agreement that the exposure does not change with subject to camera distance. The only question that we are arguing is WHY that is the case.



Nov 07, 2012 at 05:39 AM
RustyBug
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p.5 #5 · p.5 #5 · silly inverse square law question.


Sorry ... I'll recant rather than explain vector forces (direction & magnitude). We've already gone way too far into the physics of light and the conservation of energy laws.

Light Science & Magic is sufficient for practical understanding of light for photographic purposes.

To all the lurkers ... get the book, it's a good read.

@ curious80 ...

We are not in agreement. I'll forego any further attempts to explain why ... we've gone around this mountain too many times already.



Edited on Nov 07, 2012 at 06:09 AM · View previous versions



Nov 07, 2012 at 05:58 AM
curious80
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p.5 #6 · p.5 #6 · silly inverse square law question.


Rusty, answer me this. You are claiming that as we move the camera back the energy that gets delivered from the original part of the box to the image of that part on the sensor does not change. Right?

You also agree that as we move the camera back the image of that original part becomes smaller on the sensor. Right?

So if we go by your assumption then we now have the same amount of energy now concentrated on a smaller area of the sensor. But that would mean that the energy per unit area i.e. light intensity would go up. Why then the image of that part of the box doesn't become brighter?



Nov 07, 2012 at 06:03 AM
curious80
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p.5 #7 · p.5 #7 · silly inverse square law question.


RustyBug wrote:
Sorry ... I'll recant rather than explain vector forces (direction & magnitude). We've already gone way too far into the physics of light and the conservation of energy laws.

Light Science & Magic is sufficient for practical understanding of light for photographic purposes.

To all the lurkers ... get the book, it's a good read.


Sure, if you feel that way. I have explained to you in perfect detail the exact science of how everything works here. And the explanation that I have given you handles all the cases comprehensively. However you are choosing to stick with an explanation which doesn't explain various example cases. If you ever get a chance to talk to that physics friend of yours then please do show him the diagrams and description that I gave you and ask him what he thinks. You might be surprised.



Nov 07, 2012 at 06:08 AM
Guari
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p.5 #8 · p.5 #8 · silly inverse square law question.


RustyBug wrote:
1. Light is energy
2. Conservation of energy states that energy can be neither created nor destroyed (i.e. it only changes form)
3. Unless something acts upon that energy to change its form, it will retain its form and magnitude

When light is traveling it will continue to have the same amount of energy until it interacts with another object. Upon striking an object, a portion of that energy will be absorbed (giving us subtractive color) and a portion of the energy will be reflected (or refracted to continue through a transparent or translucent object).

The energy that is neither absorbed, nor refracted will
...Show more

This is true.

Really, there is not much to say but poorly understood science...

I hope everyone has a nice day



Nov 07, 2012 at 11:36 AM
Chiefdog72
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p.5 #9 · p.5 #9 · silly inverse square law question.


What a great discussion……..very respectful. It’s how everything should be discussed on all forums.

RustyBug is right about the “Light, Science, and Magic” book. It is not a physics book, but it is probably the best book written for photographers to understand light. I have a couple editions myself and regularly send a copy of it as a gift to friends and family that are interested in photography.

I’m not looking to start a whole new argument……..but trying to recall my college physics, it seems this might be a particle vs. wave discussion which is way past most of us.



Nov 07, 2012 at 10:09 PM
tedwca
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p.5 #10 · p.5 #10 · silly inverse square law question.


So, to recap, if I understand correctly, regardless of the science behind it, the exposure does not change as you move towards or away from the subject.


Nov 08, 2012 at 01:34 AM
 

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BrianO
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p.5 #11 · p.5 #11 · silly inverse square law question.


tedwca wrote:
So, to recap, if I understand correctly, regardless of the science behind it, the exposure does not change as you move towards or away from the subject.


Correct...as long as you're not moving the light(s) with you.



Nov 08, 2012 at 05:30 AM
RustyBug
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p.5 #12 · p.5 #12 · silly inverse square law question.


BrianO wrote:


Even though your camera's average/weighted/matrix reflective metering reading might suggest otherwise with subsequent changes in camera orientation / placement. Take an incident meter reading and it doesn't care if you put your camera on the moon or in the toilet, it will still give you the same meter reading of the amount of light falling onto your subject.

Correct...as long as you're not moving the light(s) with you.

Yup.



Nov 08, 2012 at 08:40 AM
Peter Figen
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p.5 #13 · p.5 #13 · silly inverse square law question.


Incident or reflective. It doesn't matter. The exposure is the same, unless you're outside and you happen to have enough distance for atmospheric absorption to become noticeable - y'know, like a Stage 3 smog alert day here in L.A. and the difference between ten feet and a mile. That this subject has taken, what, five pages of posts only to have the same crap vomited over and over is hilarious. All you have to do is go outside and take a couple of photos to prove the idea one way, but not the other. This is not even a question to anyone with a smidgeon of practical experience.


Nov 09, 2012 at 06:36 AM
Cr VI
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p.5 #14 · p.5 #14 · silly inverse square law question.


The confusion, in simple terms, is the difference between brightness and illumination. It is roughly analogous to the difference between temperature and heat.

No matter how far you get from a fire, (or a hot object), its temperature will always be the same. However, the amount of radiant heat you will get from it varies as the inverse square.



Nov 09, 2012 at 03:29 PM
tedwca
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p.5 #15 · p.5 #15 · silly inverse square law question.


, that was actually the point of my last post. They probably should have gone to PM to work it out.

Peter Figen wrote:
That this subject has taken, what, five pages of posts only to have the same crap vomited over and over is hilarious. All you have to do is go outside and take a couple of photos to prove the idea one way, but not the other. This is not even a question to anyone with a smidgeon of practical experience.




Nov 09, 2012 at 07:16 PM
jeremymeier
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p.5 #16 · p.5 #16 · silly inverse square law question.


The inverse square law should effectively be the same from the source of the light or from the source of the lights reflection (i.e.: subject being lit). The subject, too is a light source to the sensor, so it will in effect require adjustment of exposure the further the sensor would get from the subject being lit. The real practical thing here, is that you're not normally going to be in the situation where it's going to matter as if you are lighting a person for a portrait, you would not likely back off far enough to make the light difference matter when framing them in a shot...

Light falls off over distance, but it does fall off less as it gets dimmer, the further out you go as per the inverse square law...



Nov 10, 2012 at 10:34 PM
ukphotographer
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p.5 #17 · p.5 #17 · silly inverse square law question.


jeremymeier wrote:
The subject, too is a light source to the sensor, so it will in effect require adjustment of exposure the further the sensor would get from the subject being lit.


No it won't.



Nov 10, 2012 at 10:48 PM
RustyBug
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p.5 #18 · p.5 #18 · silly inverse square law question.


Check out the interactive tutorial in the link. It does a better job than my inept artwork ever will.

http://micro.magnet.fsu.edu/primer/java/reflection/specular/index.html

Slide the color slider all the way to the left to get "white" light.
Set the surface roughness to the far left to emulate a "mirror" surface.

Then as you change the surface texture, the multitude of surface angles begin to change ... which in turn is responsible for the collection of angles of reflection that still follow AI=AR, producing the diffusion of the light spreading out, that people are wrongly trying to attribute to ISL.

If reflected light was spreading iaw ISL ... the light from a 'signalling mirror' ... by the time it reached it reached the pupil of the pilot, it would be too weak for him to see it. Good thing light travels straight, iaw AI=AR ... otherwise a signalling mirror (uniform surface) would be as worthless a sheet of paper (non-uniform surface) scattering light in a plethora of directions.

I hope I don't ever need to use a signaling mirror ... but if I do, I'm glad that reflected light travels straight line, iaw AI=AR and the conservation of energy ... and not ISL ... to help me get a pilot's attention.


Those angles that reflect light back in the direction of the lens will reflect the same amount of energy until something acts upon them ... no matter how far back/close you move the camera from the subject. Barring atmospheric conditions like smog, fog, etc. that acts upon the light, the light continues straight line predicated upon the direction it was reflected, iaw AI=AR.

I realize it is one thing to debate in a cyber-blog, but if you check the source @ the link, you'll see it is in conjunction with FSU. Hopefully they can add a bit of credibility to the dialogue.

Light Science & Magic ... get the book.
Minimal physics ... yet, great practical photographic application properly derived from the physics.

Edited on Nov 11, 2012 at 02:39 AM · View previous versions



Nov 10, 2012 at 11:45 PM
RDKirk
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p.5 #19 · p.5 #19 · silly inverse square law question.


The real practical thing here, is that you're not normally going to be in the situation where it's going to matter as if you are lighting a person for a portrait, you would not likely back off far enough to make the light difference matter when framing them in a shot...

I commonly double and triple the distance to my subject in my studio, and the exposure does not change.



Nov 10, 2012 at 11:53 PM
Guari
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p.5 #20 · p.5 #20 · silly inverse square law question.


jeremymeier wrote:
The inverse square law should effectively be the same from the source of the light or from the source of the lights reflection (i.e.: subject being lit). The subject, too is a light source to the sensor, so it will in effect require adjustment of exposure the further the sensor would get from the subject being lit.
Light falls off over distance, but it does fall off less as it gets dimmer, the further out you go as per the inverse square law...



Nope, not right at all



Nov 10, 2012 at 11:59 PM
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