Home · Register · Search · View Winners · Software · Hosting · Software · Join Upload & Sell

Moderated by: Fred Miranda
Username   Password

  New fredmiranda.com Mobile Site
  

FM Forums | Lighting & Studio Techniques | Join Upload & Sell

1       2       3      
4
       5              9       10       end
  

Archive 2012 · silly inverse square law question.
  
 
curious80
Offline
• • •
Upload & Sell: Off
p.4 #1 · silly inverse square law question.


RDKirk wrote:
It does change brightness as you shift to the side, it is not reflecting light uniformly in different directions. Notice the phases of the moon. For that matter, move your reflected light meter 180 degrees around a subject. It's not reflecting light uniformly in all directions.



You are mistaking what I said. I am saying that if you look at the "same spot" on the subject from different directions then that spot seems identically bright from different directions. The phases of moon are indicating that "different parts" of the moon are reflecting differing amounts of light, which is a totally different issue. If you look at a part of the moon which is bright, then whether you look at it from the left of the moon or right, it will appear identically bright because that bright spot is sending out light rays in all directions.

You can do this experiment yourself. Go out in an open field on a sunny day. Put a cardboard box on the ground. Now move around the box and look at the top surface of the box from different directions. I hope you will agree that the top of the box will appear identically bright from all sides (of course assuming you don't cast your shadow on it in the process ). Now can you please tell me how can you watch the box top from various sides if it is not reflecting light in all directions. And also if it is appearing identically bright from all sides then how could it be that it is not reflecting light in a pretty much uniform way?

Edited on Nov 06, 2012 at 07:36 PM · View previous versions



Nov 06, 2012 at 07:22 PM
curious80
Offline
• • •
Upload & Sell: Off
p.4 #2 · silly inverse square law question.


And another little thing to ponder about. Lets switch the role of the moon here. Assume that we are photographing an object illuminated by moon light. Moon is now a light source for us. So you would agree that as the object goes closer to the moon it will get more light and as it move away from the moon, it will get less light as per the classic ISL. I hope there is no difference of opinion on that.

Now replace that object by a lens of equal size. Now can you explain why would it be that the amount of light falling on the lens would not change if you move the lens closer or farther from the moon? The lens is as good a subject as any other.

Edited on Nov 06, 2012 at 07:34 PM · View previous versions



Nov 06, 2012 at 07:28 PM
curious80
Offline
• • •
Upload & Sell: Off
p.4 #3 · silly inverse square law question.


PeterBerressem wrote:
A question to curious80, if I may :
an evenly lit flat e.g. board, metered with a 5 spotmeter at varying distances, achieves identical exposure values. How comes?


I will avoid answering that question for now, because we already have many threads of thought flowing and don't want to add another more confusion. But I will answer that once we close a couple of others mini-discussions.



Nov 06, 2012 at 07:31 PM
BrianO
Offline
• • • • •
Upload & Sell: On
p.4 #4 · silly inverse square law question.


PeterBerressem wrote:
A question to curious80, if I may : an evenly lit flat e.g. board, metered with a 5 spotmeter at varying distances, achieves identical exposure values. How comes?


curious80 wrote:
I will avoid answering that question for now, because we already have many threads of thought flowing and don't want to add another more confusion.


It seems to me that the question relates directly to the topic at hand.

By way of example, if I take spot readings of a gray target as I move back, and taking the light back with me, the readings change in accordance with the inverse square law; but if I leave the light stationary and move only the spot meter, the readings remain constant.

RDKirk's post explains the reason for this.



Nov 06, 2012 at 08:34 PM
curious80
Offline
• • •
Upload & Sell: Off
p.4 #5 · silly inverse square law question.


BrianO wrote:
It seems to me that the question relates directly to the topic at hand.

By way of example, if I take spot readings of a gray target as I move back, and taking the light back with me, the readings change in accordance with the inverse square law; but if I leave the light stationary and move only the spot meter, the readings remain constant.

RDKirk's post explains the reason for this.


Ok. So first tell me what will happen if you take a meter reading of the light source itself. Will that change if you move back? Will it go down since inverse square law is in effect? Let me tell you the answer that it will not. If you don't believe me then try it.



Nov 06, 2012 at 08:46 PM
BrianO
Offline
• • • • •
Upload & Sell: On
p.4 #6 · silly inverse square law question.


curious80 wrote:
...Ok. So first tell me what will happen if you take a meter reading of the light source itself. Will that change if you move back? Will it go down since inverse square law is in effect? Let me tell you the answer that it will not. If you don't believe me then try it.


Precisely! Because you're reading only the direct rays traveling from the light to the meter; the Inverse Square Law is NOT in effect because you're reading more-collimated-than-not light...exactly what you're reading when you measure the reflected light from the subject.

ISL only applies to the amount of light falling ON THE SUBJECT from a non-collimated light source.



Nov 06, 2012 at 09:02 PM
curious80
Offline
• • •
Upload & Sell: Off
p.4 #7 · silly inverse square law question.


BrianO wrote:
Precisely! Because you're reading only the direct rays traveling from the light to the meter; the Inverse Square Law is NOT in effect because you're reading more-collimated-than-not light...exactly what you're reading when you measure the reflected light from the subject.

ISL only applies to the amount of light falling ON THE SUBJECT from a non-collimated light source.


EDIT: I would prefer right now that we stick to the discussion which I already started. Don't want to again have too many discussion going on in too many directions. If we could continue the thread that I started with the box on a sunny day above and reach an agreement on that, then this light meter problem is just a corollary of that.



Nov 06, 2012 at 09:17 PM
ukphotographer
Offline
• • •
Upload & Sell: Off
p.4 #8 · silly inverse square law question.


Confusing??

This:
BrianO wrote:
Nope.


And then this:
BrianO wrote:
By way of example, if I take spot readings of a gray target as I move back, and taking the light back with me, the readings change in accordance with the inverse square law; but if I leave the light stationary and move only the spot meter, the readings remain constant.


More confusing..?

PeterBerressem wrote:
A question to curious80, if I may :
an evenly lit flat e.g. board, metered with a 5 spotmeter at varying distances, achieves identical exposure values. How comes?


A spotmeter meters reflected light from a specific (5) area using the same lens and using the same sensor area each time. It will give you the same exposure value of the subject based on its reflected light reading design disregarding distance. This reflected reading is not affected by ISL.

Without going very far from the subject, (and assuming the subject isn't very large compared to the measuring distance) the use of an incident meter pointed towards the subject (now a pseudo light source as curious80 seemed to be first inferring), at the same varying distances will illustrate to you that the ISL is still (only approximately) in effect.

Subjects and light sources are being mixed up it seems. Confusing matters in an under current is then the role of apertures, focal length and magnification !

Try this:
gpop wrote:
if you light a subject with off camera light source, but move the camera away from the light and subject will this effect exposure to the same degree as if you moved the light farther from the subject?

No.


pathetic illustration:
L2/S2 <-20 feet-> L1/S1 <-10 feet-> camera.

(light source and subject #2 is 30 feet from the camera, light source and subject #1 is 10 feet from camera.)


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.






Nov 06, 2012 at 09:58 PM
RustyBug
Offline
• • • • • •
Upload & Sell: On
p.4 #9 · silly inverse square law question.


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 be reflected in iaw with AI=AR and that remaining value of energy will continue until it is acted upon by another object that it is acted upon (i.e. the lens refraction / reflection) and subsequently reach the sensor where it its energy gets absorbed and changed into an electrical signal.

A point light source releasing energy does so in accordance with path of least resistance, which is theoretically none, and as such the distribution and transfer of the amount of energy will be distributed in a spherical manner resulting in a trigonometric distribution of that energy iaw with ISL.

Once that energy has been given a direction, it will continue along that path indefinitely until acted upon by another object, which includes absorption, refraction and reflection ... all of which are vector quantities that are responsible for the degree of form & direction change that occurs from the available energy.

The concept that you are trying to apply ISL to reflected energy via camera to subject distance is simply not true. You are wrongly applying the dispersion / distribution of a point light source that is omni-directional at it's origin to directional light that continues to follow AI=AR (absorption & refraction) to infinity, iaw with conservation of energy.

Take a laser pointer and you can more readily see that light does NOT travel iaw with ISL. It does NOT spread iaw with ISL. It does NOT diminish iaw ISL.

Sure, the pointer is not full spectrum, and has been given "direction" ... but every ray of light that is emanating from a point light source also has been given a "direction" (i.e. spokes on a wheel). The collection of rays emanating from a point light source is an omni-directional distribution of the energy released from the point light source ... but it continues in a directional manner iaw with AI=AR and the energy contained for a given ray remains unchanged until acted upon by another object. Diffuse light simply being a collection of incident angles and/or reflecting angles that collectively cover a wider angle, whereas specular light is limited to a more narrow coverage.

ISL is a trigonometric calculation of the distribution of the energy released. The absorption, refraction and reflection of light continues directional iaw AI=AR and vector forces ... iaw with conservation of energy ... not predicated upon ISL & offsetting camera to subject distance.

Edited on Nov 06, 2012 at 10:11 PM · View previous versions



Nov 06, 2012 at 10:02 PM
curious80
Offline
• • •
Upload & Sell: Off
p.4 #10 · silly inverse square law question.


Rusty, we have both expressed our views on this multiple times and it will probably not convince either of us to listen the other person reiterate his view/theory one more time. If you are willing to go with my thought experiment then we can take this one small step at a time and try to reach a consensus. If you agree to that then as a first step I would like to know what you think about what I said about the box lying outside scenario above. Do you agree or not that the box in that scenario is reflecting light in all directions? (and right now lets not worry about why is that exactly the case)


Nov 06, 2012 at 10:10 PM
 

Search in Used Dept. 



RustyBug
Offline
• • • • • •
Upload & Sell: On
p.4 #11 · silly inverse square law question.


Simple ... the box is being illuminated from a multitude of angles that reflect light iaw AI=AR.

It's that simple.



Nov 06, 2012 at 10:27 PM
curious80
Offline
• • •
Upload & Sell: Off
p.4 #12 · silly inverse square law question.


RustyBug wrote:
Simple ... the box is being illuminated from a multitude of angles that reflect light iaw AI=AR.

It's that simple.


As I said I am not arguing about WHY is that the case. For the time being I am just requesting you to bear with me and just tell me if you agree that the box is reflecting light in all directions and that it is reflecting light more or less uniformly in a semisphere. Depending on whether you agree to that or not, I will proceed with the rest of my thought experiment.



Nov 06, 2012 at 10:31 PM
ukphotographer
Offline
• • •
Upload & Sell: Off
p.4 #13 · silly inverse square law question.


curious80 wrote:
You can do this experiment yourself. Go out in an open field on a sunny day. Put a cardboard box on the ground. Now move around the box and look at the top surface of the box from different directions. I hope you will agree that the top of the box will appear identically bright from all sides (of course assuming you don't cast your shadow on it in the process ). Now can you please tell me how can you watch the box top from various sides if it is not reflecting light in all directions. And also if
...Show more

Light gets reflected off subjects in all directions its true.

The surface of the box does not conform to AI=AR, but you will find a 'brighter bit' at the AR. The surface of the box is matt and scatters light it is not a flat mirror like the analogy of a snooker ball hitting a bumper. In that analogy the bumper would be the flattest, most polished mirror not yet created if the snooker ball was a photon. Maybe throwing a ping-pong at a ball pool would work better?

Care needs to be taken not to take things too literally.



Nov 06, 2012 at 10:35 PM
RustyBug
Offline
• • • • • •
Upload & Sell: On
p.4 #14 · silly inverse square law question.


Okay, I'll bite ... I'll agree that the box is reflecting light in "essentially" all directions, more or less "uniformly" in a "semi-sphere" (although I already know why).

@ UK ... actually all surfaces conform to reflection @ AI=AR (absorption, and refraction) ... the more non-uniform the surface angles, the greater the number of subsequent angular variation. Thus, while a more matte surface does not yield as specular a reflection as a more uniform surface ... it does not violate AI=AR.



Nov 06, 2012 at 10:37 PM
curious80
Offline
• • •
Upload & Sell: Off
p.4 #15 · silly inverse square law question.


RustyBug wrote:
Okay, I'll bite ... I agree that the box is reflecting light in "essentially" all directions, more or less "uniformly" in a semi-sphere (although I already know why).


Thanks! The why part is not what I am worried about, as long as we agree to this fact that you just agreed to.

Let me also extend this by saying that every spot on the box surface is sending light in all directions (for example you can cover the box surface by a black cloth with a small hole in it, and the part of the box showing through the hole will still have the same brightness as before and again will seem uniformly bright from all directions).

Now. The following diagram shows how the reflected light is leaving various points in the box surface. And then how that light interacts with the lens. From each point on the box surface, some of the rays hit the lens surface and the lens focuses them to a single point on the sensor (assuming the box surfaces is in focus). Not that only a subset of the rays is captured from each point. The other rays from each point miss the lens. Note also that in this configuration the sensor is capturing the image of a part of the box surface. There will be rays which strike the lens from points outside this imaged area, however they will be focused at a point which is outside the sensor:







Now what happens when we move the lens + sensor combination back? Note that the lens will now capture a smaller set of rays from each point on the surface, because some of the original rays will miss the lens now. And the decrease in number of rays captured from each point will more or less have an inverse square relationship with the distance.

However, at the same time the lens will now be focusing rays from a larger number of points on the subject to the sensor. The following diagram shows the larger area images by the sensor as you move the lens + sensor back (This time I have only drawn one ray from each point as it makes it easier. And sorry for my sloppy drawing):







Note that the rays from the original part of the box are now concentrated on a smaller area on the sensor. So while the number of rays that are getting to the lens from the original part of the box to the sensor are less than before, they are also now concentrated on a smaller area which is why the light intensity i.e. light per units area remains same as before and we get the same exposure as before. Another way to say this would be that while the sensor is getting less light from each point on the box, it is overall getting light from a larger portion of the box so the total light delivered to the sensor remains the same and thus the exposure remains the same (this second explanation is a less desirable way to describe this because it doesn't handle all the cases).

Now I would like to know that which part of this illustration do you disagree with.



Nov 06, 2012 at 11:41 PM
pjbuehner
Offline
• •
Upload & Sell: Off
p.4 #16 · silly inverse square law question.


I like pictures. Makes perfect sense to me....unfortunately I am not in the discussion but merely a curious onlooker.
I also like the moon analogy which doesn't seem to be refuted yet.

Cheers to all.

I really am enjoying this discourse...reminds me of my philosophy classes in college



Nov 07, 2012 at 12:51 AM
RustyBug
Offline
• • • • • •
Upload & Sell: On
p.4 #17 · silly inverse square law question.


The part that says it will now collect a smaller set of rays ... you are forgetting to include the rays that were previously excluded by too narrow an angle of reflection to reach the lens ... which are now able to do so. Oooops.

You can't "pick & choose" to only have the distance/angle change work to the part that you want it to work for.

Again, pick up a copy of Light Science & Magic. It does a good job of illustrating reflection control inclusion/exclusion iaw with AI=AR ... including changes in camera to subject distance / fov changes. BUT, those changes do not render exposure changes that are being offset by ISL as you're suggesting.

Excuse the rotten artwork (which explains why I use a camera ) ...

But when you move the camera lens position, you will be losing some rays, while picking up others, i.e. red/yellow vs. blue, but in your box scenario, the angular changes are going to be so miniscule that they are absolutely insignificant other than for academic minutia. This is not an ISL issue being offset by camera to subject distance.











Nov 07, 2012 at 03:43 AM
curious80
Offline
• • •
Upload & Sell: Off
p.4 #18 · silly inverse square law question.


RustyBug wrote:
The part that says it will now collect a smaller set of rays ... you are forgetting to include the rays that were previously excluded by too narrow an angle of reflection to reach the lens ... which are now able to do so. Oooops.

You can't "pick & choose" to only have the distance/angle change work to the part that you want it to work for.

Again, pick up a copy of Light Science & Magic. It does a good job of illustrating reflection control inclusion/exclusion iaw with AI=AR ... including changes in camera to subject distance / fov changes. BUT,
...Show more

Perfect! I for most part agree with your diagram!. In fact if you read my description you will see that I wrote that same thing at the very end. I quote:

"Another way to say this would be that while the sensor is getting less light from each point on the box, it is overall getting light from a larger portion of the box so the total light delivered to the sensor remains the same and thus the exposure remains the same"

Which is exactly what you have shown. In fact if you see closely I have shown these extra rays in my last diagram above. In that diagram, there are rays which go from the original part of the box to the image of that original part on the sensor. And then there are extra rays which are coming from beyond the original portion of the box and these are now forming an image on the sensor.

However I also wrote that this description is less complete. And let me tell you the first reason why. Consider the case that we cover everything around our original part of the box with a black cloth. So when we go back we do not get any more light rays because everything outside the original area is black. Why then in this case the image of the original part of the box remains as bright. I am sure you will agree that with everything black outside the original box area, there are no more rays to be gained and thus total light entering the lens has to go down.



Nov 07, 2012 at 04:28 AM
RustyBug
Offline
• • • • • •
Upload & Sell: On
p.4 #19 · silly inverse square law question.


curious80 wrote:
So when we go back we do not get any more light rays because everything outside the original area is black.


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 that area of the scene is then recorded as black. The original area still yields a same exposure value, because it is the same reflection of energy with or without a black cloth on surrounding areas.

AI=AR (absorption & refraction).

Conservation of energy requires that the total amount of energy remain constant and is divided among that which is absorbed (color), refracted (transmitted through translucent/transparent material) and reflected. As the refractive index, angle of incidence and color absorption characteristics change, so does the amount of energy remaining available for reflection.










Edited on Nov 07, 2012 at 05:13 AM · View previous versions



Nov 07, 2012 at 04:45 AM
BrianO
Offline
• • • • •
Upload & Sell: On
p.4 #20 · silly inverse square law question.


ukphotographer wrote:
...Subjects and light sources are being mixed up it seems.


I think so, too.

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.


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.)



Nov 07, 2012 at 05:11 AM
1       2       3      
4
       5              9       10       end




FM Forums | Lighting & Studio Techniques | Join Upload & Sell

1       2       3      
4
       5              9       10       end
    
 

You are not logged in. Login or Register

Username   Password    Retrive password