Members: 68,266   Posts: 1,468,427   Online: 977

# Thread: light meter measurement: incident light and the inverse square law

1. 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.

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

2. Originally Posted by Diapositivo
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.

Originally Posted by Diapositivo
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.

Originally Posted by Diapositivo
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.

3. 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.

4. Originally Posted by Q.G.
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?

5. 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.

6. Originally Posted by 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.
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?

7. 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.

8. Originally Posted by Q.G.
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.

9. 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.

10. 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.

Page 4 of 7 First 1234567 Last

 APUG PARTNERS EQUALLY FUNDING OUR COMMUNITY: