light meter measurement: incident light and the inverse square law

[Q.G.]

If at 300 m we use a tele lens and we fill the image with the piano, a lesser quantity of light fills the entire image, so we have a different exposure.

Correct. The longer lens will change the size of the image, compared to, say, a shorter lens, but cannot change the amount of light it receives form the subject. So telelenses project a larger, but darker image, i.e. are slower, than shorter lenses.

(Assuming the same size light gathering area, of course. You can give a telelens a larger front lens/entrance pupil, and it will be able to gather more light than a smaller, shorter lens. It will need that - collect more light - to be able to project a larger image that is as bright as the smaller image. Make the front lens large enough, and it will be able to collect as much light as is needed to project a lager image that is also brighter than that of the shorter lens. But that shorter lens can also be made faster, etc., etc.)

If I place on the piano a grey card and read it with a spot reflected meter, which only reads light reflected from the card (that is, supposing the card "fills" the reading angle of the spot lightmeter), my understand of physics tells me that my spot meter will give me a different exposure than the incident light meter used near the piano.

Assuming that there are no other variables, it does not.
Why would it?

If I take a picture of a lit monument at night, the exposure for the monument isn't the same even if I am far from the monument.

I'm really puzzled.

The answer is in what is said before, and what was quoted from that encyclopedia.

 

[Ed Sukach]

The question is why, if you measure at your subject (which you certainly should with a light two feet away from the subject), will the reading stil be correct if you set the cmaera up two miles from your subject.
The light seen by the camera has to travel those two miles after being reflected off the subject.

So before anyone else gives the same non-answer you gave, tell us why you can use the same setting you should use with the camera not two miles, but six feet from your subject.

Incident metering measures the light falling on the object, and that does not change no matter where the camera, or eye, is located. A face, for example, properly exposed at one meter, will remain properly exposed at two hundred meters, due to the fact that its image size decreases - proportionately.

A star many light years away is just as bright, per unit area, as it is from a distance of one meter.

 

[Q.G.]

Incident metering measures the light falling on the object, and that does not change no matter where the camera, or eye, is located.

I don't understand why people feel the need to harp on that point.
It's not what is being discussed,. Not what the question is about.

I'll tell you one more time: that "light falling on the object" has to travel from that object to your camera for you to be able to capture that object on film. The question is why there is no light loss due to that.
The answer is that there is.

A face, for example, properly exposed at one meter, will remain properly exposed at two hundred meters, due to the fact that its image size decreases - proportionately.

Yes.
I have told you that a couple of times now...

A star many light years away is just as bright, per unit area, as it is from a distance of one meter.

Is it?

 

[tkamiya]

Guys?

You have been talking about sun as a source of light for this discussion. That's true SUN is far far far away making the little distance from subject to camera irrelavant. How do you explain, the same rule holds true when the light source is a studio flash/light? You, too, can take incident reading at the subject. Now, the light source, and point source at that, is only 10 feet away. You, the camera man/woman can be at 6 feet, 10 feet, or 20 feet, and incident reading is still valid.

The only rationale I can come up with is that the point source light falls on the subject. In this journey of the light, it spreads as it travels. Once the subject is illuminated, the reflected light doesn't spread. I don't know why and can't explain why.

Image size? OK, but the size of subject being 10% of the total frame or 100%, say just the cheek of the person or the whole face, at either magnifications, the spot in question, (let's say the cheek) is exposed identically. Does the size really matter??

Not arguing or disagreeing... I'm really confused now.

 

[Q.G.]

[...]

The only rationale I can come up with is that the point source light falls on the subject. In this journey of the light, it spreads as it travels. Once the subject is illuminated, the reflected light doesn't spread. I don't know why and can't explain why.

You will not have to search for an explanation why it would not spread, because it indeed does spread.
Read the thread. All is revealed. Several times, in fact.

 

[Diapositivo]

Just for clarity, I had written a sentence without a question mark, I corrected it immediately after posting but Q.G. had already answered to my post. My statement

If I take a picture of a lit monument at night, the exposure for the monument isn't the same even if I am far from the monument.

Should read:

If I take a picture of a lit monument at night, the exposure for the monument isn't the same even if I am far from the monument?

The explanation given is not clear to me. When I am far, the angle at which I see the piano is narrower than when I am near. The piano goes on spreading light in all directions, but when I am far I am reached by only one smaller quota of that light.

When I am near, I am reached by a larger quota.

So, the explanation confuses me even more.

When I am far, not just the light has "diverged more" than when I am near (it is more "spreaded" just like butter is thinner if spread on a larger slice of bread) but, in addiction to this, I even see a smaller "angle" of it.

I visualize light as if it were the surface of a rubber balloon. The more I inflate the balloon, the more the same rubber is spread over a larger surface, so the quantity of rubber per unit of surface is smaller.

If I consider the solid angle, well, even inflating the balloon, the amount of rubber for a certain solid angle never changes.

But when I take a picture of the piano from far, with a tele lens, I am using just a smaller "solid angle" of the light reflected by the piano.

Ever more confused

Fabrizio

 

[Q.G.]

The explanation given is not clear to me. When I am far, the angle at which I see the piano is narrower than when I am near. The piano goes on spreading light in all directions, but when I am far I am reached by only one smaller quota of that light.

When I am near, I am reached by a larger quota.

That's it.
When you're far a smaller amount of light reaches you. And the image will be small.
When you're near a larger amount of light reaches you, and the image is porportionally larger.
Divide the amount of light over the area it has to fill, and the result is the same, a constant.

So, the explanation confuses me even more.

When I am far, not just the light has "diverged more" than when I am near (it is more "spreaded" just like butter is thinner if spread on a larger slice of bread) but, in addiction to this, I even see a smaller "angle" of it.

That's not in addition: you're saying the same thing, three times.

I visualize light as if it were the surface of a rubber balloon. The more I inflate the balloon, the more the same rubber is spread over a larger surface, so the quantity of rubber per unit of surface is smaller.

If I consider the solid angle, well, even inflating the balloon, the amount of rubber for a certain solid angle never changes.

But when I take a picture of the piano from far, with a tele lens, I am using just a smaller "solid angle" of the light reflected by the piano.

Ever more confused

But it's already there in what you wrote!

The amount of rubber per solid angle does not change, but the area it covers does.
So that's the same as "the more the same rubber is spread over a larger surface, so the quantity of rubber per unit of surface is smaller".

I.e. the further away from the source of light (in this case the subject reflecting the light that falls on it), the 'thinner' it gets spread out over any area unit you put in it's way.
The thing the inverse square law quantifies as a function of distance.

So two things next. First the telelens thing.
Assume the same light gathering area as a standard lens, i.e. the same amount of light available to form an image with, a telelens will produce a larger image of the source of light (the subject).
That image, by force, then must be less bright. And so it is.

Next the equally bright image/same setting thing.
With increasing distance, not only does the amount of light per area unit get less, but by the same geometry, i.e. by the same proportion the size of the image any given lens will produce of the source gets smaller.
The result: less light spread / a proportionally less large area = equal brightness per area unit.
So no matter how far or close, the image will be of the same brightness, and you can use the same settings for exposure.

 

[Hikari]

Exposure is in reference to the f-number system. No matter the focal length, the image plane receives the same illumination at a given f-number. Although you are increasing magnification when using a telephoto lens, you are also increasing the area of the entrance pupil--At f/2, a 50mm lens, 100mm lens, and 200mm lens have entrance pupils diameters of 25mm, 50mm, and 100mm, respectively. You will note the change in entrance pupil area compensates for the increase in magnification. So you are actually intersecting a greater arc of light.

 

[moose10101]

First the telelens thing.
Assume the same light gathering area as a standard lens, i.e. the same amount of light available to form an image with, a telelens will produce a larger image of the source of light (the subject).
That image, by force, then must be less bright. And so it is.

How would the difference in brightness be calculated? How much less bright would an image be in a 100mm lens than it would be in a 50mm lens when both are used at the same distance? How much exposure adjustment would be necessary?

 

[Q.G.]

The ratio of the angular field of views. Half the angular field of view is twice the size linear, is four times the area, is 2 stops. Along those lines

But, as Hikari already mentioned, it's not something we need to concern ourselves with. It's already taken care of in the f-numbers of a lens.


Don't forget, despite this by now rather lengthy thread, that it's a very simple and most of all easy matter: incident light metering. There's very little to do and/or worry about to get it right. Very, very little.

 

[moose10101]

The ratio of the angular field of views. Half the angular field of view is twice the size linear, is four times the area, is 2 stops. Along those lines

But, as Hikari already mentioned, it's not something we need to concern ourselves with. It's already taken care of in the f-numbers of a lens.

Yes, I know it's taken care of, but in Reply #26 you insist that the exposure changes if a "tele" lens is used. Specifically, to this question:

"If at 300 m we use a tele lens and we fill the image with the piano, a lesser quantity of light fills the entire image, so we have a different exposure."

you replied:

"Correct. The longer lens will change the size of the image, compared to, say, a shorter lens, but cannot change the amount of light it receives form the subject. So telelenses project a larger, but darker image, i.e. are slower, than shorter lenses."

Since neither the shutter speed nor the aperture (a ratio) has to be changed, exactly how will the exposure be different?

 

[Q.G.]

I see what you mean.
What is should have stressed more is that i was comparing lenses, short and tele, with the same size light gathering area.

So let's say that you use both lenses wide open, the short one will be, say, f/4, the longer one f/8. So while you can shoot at, say, 1/125 at f/4 with the one, you have to change the speed to match the darker image and use 1/30 for the telelens.

The thing it was illustrating is that the decrease in image size corresponding with the reduction of the amount of available light is responsible for the fact that you do not have to compensate for the increased distance and the 'inverse square' reduction of light it brings along. But when you smear the available amount of light out over an area that has not decreased proportionally with the increase in distance, illumination at the film will indeed change.
Perhaps, for clarity's sake, i should not have mentioned that at all.

 

[Ed Sukach]

I'll tell you one more time: that "light falling on the object" has to travel from that object to your camera for you to be able to capture that object on film. The question is why there is no light loss due to that.
The answer is that there is.

You are correct: there *IS* a loss of light. The energy of the light decreases ... definitely, but as it does, the area illuminated decreases as well, compensating for the distance loss, therefore, the original image area remains as bright.

"Brightness" (I am trying to avoid confusion by not referring to illumination, illuminance, albedo, and a host of other anally accurate definitions) is the factor that decides exposure of any given area, not overall energy absorbtion.

I have told you that a couple of times now...

And I've read that a coupe of times as well.

You keep trying to separate the distance and area covered by the light. They are inseparable.

Is it?

The star example was not a good one - the light in question is not "falling on", but "eminating from". Even so, yes, it is true.

 

[Mainecoonmaniac]

Incident readings will only read light falling on the subject. If the reflected light that is falling on the subject is greater than the source of illumination, it will be taken into account in your incident reading. If you want greater accurate reading of reflected light and your source of illumination, use a reflected reading. I prefer to use reflected readings because I could figure out lighting ratios especially in contrasty scenes.

 

[Tim Gray]

Inverse square law applies for point sources.

Extended line sources (effectively infinite in length) would obey an inverse law - no square.

Plane sources, infinite in extent, would show no drop off.

It's due to the geometry of the surface the flux is passing through. Point sources radiate radially outward, and the appropriate surface is a sphere, which as a surface area of 4*pi*r^2. That's where your r^2 comes from. For line sources, you use cylindrical surfaces to calculate the flux per surface area. Cylinders have surface areas of 2*pi*r*length. There's where the single r comes from.

Real lighting is a mix of all these probably. If you are positioning a flash, you can more or less treat this as a point source (ignoring any flash zoom function). Lambertian surfaces and other sorts of reflections, collimated light, etc. are different. For example, a flash bounced off of a white card would be closer to a lambertian surface, while a flash bounced off a mirror would be similar to just the flash itself.

 

[Q.G.]

You are correct: there *IS* a loss of light. The energy of the light decreases ... definitely, but as it does, the area illuminated decreases as well, compensating for the distance loss, therefore, the original image area remains as bright.

"Brightness" (I am trying to avoid confusion by not referring to illumination, illuminance, albedo, and a host of other anally accurate definitions) is the factor that decides exposure of any given area, not overall energy absorbtion.

And I've read that a coupe of times as well.

So why did you think you needed to tell me what i told you?

You keep trying to separate the distance and area covered by the light. They are inseparable.

Oh dear me...
No, i'm not!

You say you have read the explanation i gave several times, and you still come up with that line?

It's like pulling teeth... It really is...
:w00t:

 

[moose10101]

Oh dear me...
No, i'm not!

You say you have read the explanation i gave several times, and you still come up with that line?

It's like pulling teeth... It really is...
:w00t:

All due respect, but my initial reading of your first half-dozen or so responses caused me to believe that you were. Your position only became clear to me in the last few posts.

 

[Diapositivo]

I finally got it.

I didn't take into account that f/number in exposure is not an "absolute aperture" but an aperture "relative to the focal lenght" (not by chance is named f/) and that the actual aperture of the diaphragm, for a 300m set at f/8, is higher than for a 50mm set at f/8. The higher absolute aperture of the tele lens compensates for the different angle of light capture, so that the exposure is the same.

Q.G. tried to put me in the right direction when mentioning different lens diameters, but I did not put this in relation to the same f/# and so the same exposure for the two cases. And when Q.G. answered to my question confirming that the light from distance was less, I was even more confused because practical experience shows that exposure does not change.

Lux facta est.

Fabrizio

 

[JBrunner]

QG, please explain why my wall reads an EV of six, regardless if I place my spot meter one foot away or twelve.

 

[Tim Gray]

Do you really want AN explanation, or an explanation from QG?

 

[JBrunner]

Do you really want AN explanation, or an explanation from QG?

Yeah, it should clear it up for everyone.

 

[Q.G.]

QG, please explain why my wall reads an EV of six, regardless if I place my spot meter one foot away or twelve.

Because you painted them a rather dark colour?

Because the spot sees more of it when it moves away.

 

[Nicholas Lindan]

Something more to argue about:

Point sources fall-off as distance ^ 2
Line sources as distance
Planar sources don't fall off ...

So what's the fall-off inside a spherical source?

 

[Q.G.]

Something more to argue about:

Point sources fall-off as distance ^ 2
Line sources as distance

Or a collection of adjoining point sources, the light of each falling off as distance ^ 2.

Planar sources don't fall off ...

They do, as a collection of adjoining line souces.

The principle remains the same. The math gets complicated.

So what's the fall-off inside a spherical source?

Does the source get brighter when its diameter gets bigger?
Else it's a complicated instance of the inverse square once again.

 

[Tim Gray]

They do, as a collection of adjoining line souces.

The principle remains the same. The math gets complicated.

You treat the line source as a line or a collection of points, there's no real difference. Actually the math is quite easy. I'm not sure what you are trying to prove with this kind of statement.

If you really insist on calling a line a collection of points, then go right ahead. Most of us, casually and mathematically, have a name for a collection of points: a line. As was stated before, inverse square law stems from essentially geometric arguments.