Basic question on Light Meter (I should know this)

Okay,

I'm gonna suck up a little pride and ask a basic question, that quite frankly I should know the answer to.

Our camera's use a reflective meter, which means they measure the light reflected off of a subject to determine proper exposure. Light meters measure the amount of light hitting the subject, to determine proper exposure. So, theoretically, I use my light meter, measure how much light is hitting my subject, adjust my camera to the values on the meter and I get proper exposure.

But I got to thinking about how distance plays into this. The Inverse Square Law of Light (http://en.wikipedia.or​g/wiki/Inverse-square_law) says that the intensity of light is inversely proportional to the square of the distance of the light from the subject. In essence this says that light has a predictable falloff in intensity based on the distance from the source.

Now let's take this and apply it back to my example. If I use a light meter, this measures the amount of light hitting my subject. But light hitting my subject does me no good. That light must bounce off the subject and hit my sensor. The Invers Square Law says that how much light reaches my sensor is determined by the distance I am away from the subject. If I take a picture of the subject at 10 ft, and then take another picture of the subject at 20ft, the Inverse Square law says I will receive only 1/4 of the amount of light. Therefore, my exposure would have to compensate for the drop in the light. But yet the light meter gives the same reading, and doesn't know/care at what distance I'm shooting from. It only records what is hitting the subject.

So the question I have is how do you compensate for distance, when taking readings with a light meter. I will give you a real world case. I shoot theater productions for my daughters high school. I can use my Sekonic 358 and take a meter reading of the lights directly from the stage. However, I can't take pictures on the stage. I am forced to shoot from 70-80 feet back, from the auditorium. If the sekonic says my exposure is f/2.8, 1/100 and ISO 1600 when I take the reading on the stage, I assume that applies only for the stage. Since I know light falls off proportional to the distance, how do I interpret what the correct exposure is at the back of the auditorium, where there is no ambient light.

Reflective meters (aka built in camera meters) would not appear to be affected by this issue, as they measure the light actually hitting the metering sensors. As such, they have already accounted for light falloff (because the light has already "fell off" before it reached the meter), where as incident meters only tell me what is hitting the subject, and not how much of that will actually make it back to me. Is there some rules/guidelines I can apply that helps me understand how to adjust the exposure readings suggested by the light meter, when I'm shooting at a large distance away? Or do I even need to ?

Am I over complicating this? What am I missing?

Subscribing... Interesting question.

BigBlueDodge wrote in post #8609552
The Inverse Square Law of Light (http://en.wikipedia.or​g/wiki/Inverse-square_law) says that the intensity of light is inversely proportional to the square of the distance of the light from the subject. In essence this says that light has a predictable falloff in intensity based on the distance from the source.

Am I over complicating this? What am I missing?

The inverse square law only applies for point sources.

If you had a hypothetical infinite plane with constant radiance, then the intensity does not vary with distance. (Wikipedia states this as: "For an infinite planar-source the equation is: E = I (E is invariant with d).")

The front of the stage more closely approximates an infinite plane than a point source, hence the measured intensity does not vary significantly with distance.

I think the only problem you'll encounter is when the relative permittivity of air to light waves changes...I mean, is the Moon any less bright through a telescope than it is at the same physical distance required for the same magnification (a few hundred miles) (presuming they both 'fill the frame').

It appears to be rationic - fill your sensor with 85mm lens at 10 feet, step back to 100 and you'll need more FL to fill the frame but will there be less light?
I'd say not - as your 'beam width' (FOV) has narrowed to a few degrees.

Now, regard Q. Q being a 'quality factor' of center-frequency / bandwidth(-3dB). Or for light :

Brightest part of the image/area of illumination (50% darker than the brightest part).

If you have a sensor full of lit face at 10 feet and 85mm lens, Q=100 (for argument's sake).
If you have a sensor NOT!!! full of lit face at 100 feet and 85mm lens, Q=1 (faaaaar less light reaching the sensor due a tiny little lit subject and vast blackness).
If you have a sensor full of lit face at 100 feet and 1200mm lens, Q still =100. Comprende?


Inverse-square-law is an interesting thing, and certainly more prevalent at RF frequencies. But then we have 'telescopes' (high-gain antennae)...

BigBlueDodge wrote in post #8609552
Okay,

I'm gonna suck up a little pride and ask a basic question, that quite frankly I should know the answer to.

Our camera's use a reflective meter, which means they measure the light reflected off of a subject to determine proper exposure. Light meters measure the amount of light hitting the subject, to determine proper exposure. So, theoretically, I use my light meter, measure how much light is hitting my subject, adjust my camera to the values on the meter and I get proper exposure.

But I got to thinking about how distance plays into this. The Inverse Square Law of Light (http://en.wikipedia.or​g/wiki/Inverse-square_law) says that the intensity of light is inversely proportional to the square of the distance of the light from the subject. In essence this says that light has a predictable falloff in intensity based on the distance from the source.

Now let's take this and apply it back to my example. If I use a light meter, this measures the amount of light hitting my subject. But light hitting my subject does me no good. That light must bounce off the subject and hit my sensor. The Invers Square Law says that how much light reaches my sensor is determined by the distance I am away from the subject. If I take a picture of the subject at 10 ft, and then take another picture of the subject at 20ft, the Inverse Square law says I will receive only 1/4 of the amount of light. Therefore, my exposure would have to compensate for the drop in the light. But yet the light meter gives the same reading, and doesn't know/care at what distance I'm shooting from. It only records what is hitting the subject.

So the question I have is how do you compensate for distance, when taking readings with a light meter. I will give you a real world case. I shoot theater productions for my daughters high school. I can use my Sekonic 358 and take a meter reading of the lights directly from the stage. However, I can't take pictures on the stage. I am forced to shoot from 70-80 feet back, from the auditorium. If the sekonic says my exposure is f/2.8, 1/100 and ISO 1600 when I take the reading on the stage, I assume that applies only for the stage. Since I know light falls off proportional to the distance, how do I interpret what the correct exposure is at the back of the auditorium, where there is no ambient light.

Reflective meters (aka built in camera meters) would not appear to be affected by this issue, as they measure the light actually hitting the metering sensors. As such, they have already accounted for light falloff (because the light has already "fell off" before it reached the meter), where as incident meters only tell me what is hitting the subject, and not how much of that will actually make it back to me. Is there some rules/guidelines I can apply that helps me understand how to adjust the exposure readings suggested by the light meter, when I'm shooting at a large distance away? Or do I even need to ?

Am I over complicating this? What am I missing?

Hi David,

Yes, you are overcomplicating this.

You're getting confused and mixing and matching concepts. The reason we take incident readings is because the light that falls on the subject is the same no matter where the camera position is and that's what allows the photographer to move around once the metering is done. When we take incident readings we're no longer concerned with reflected light coming back to the camera's sensor.

Intensity does not change based on camera position or distance from the subject. Once you know how much light is falling on your subject and have the exposure for the part of the subject you want then no matter where you stand, that part of the subject will still be properly exposed.

This series of images I shot demonstrates how intensity does not change with distance.

Thanks Robert. This is very illustrative and educational.

Mufutau

TMR Design wrote in post #8611289
This series of images I shot demonstrates how intensity does not change with distance.

IMG NOTICE: [NOT AN IMAGE URL, NOT RENDERED INLINE]

Precisely the point I was attempting to get to - the subject is still lit as well as it ever was, no matter your distance. Luminosity stays the same, just there's less subject to fill the frame with and some may say less 'percieved' brightness (see the Q thing in my previous post), but the subject is still just as bright.
If your entire sensor metered uniformly all over, then...you can see what'd happen . Fortunately they dont

First of all, I see no advantage to using a handheld meter for reflective readings in your situation. Your camera does reflective metering and gives you more options for control (zoom the lens, evaluative, partial, spot, etc.).

Now with regard to the relationship between distance and exposure:
With a given focal length and aperture, as you move further away from your subject, your sensor will collect fewer photons, but these photons will be concentrated in a smaller area of the sensor. The result is the same number of photons per unit area on the sensor, and the same number of photons per photocell, thus exposure stays the same.

Suppose you're shooting a 4x6 foot piece of foamcore with a 5D and 48mm lens from a distance of 8 feet. The light from the foamcore will expose all 12 million pixels.

Now backup to 16 feet. With the same lens, the camera will receive 1/4 as many pixels from the foamcore, but those photons will expose only 1/4 of the total sensor area (half the width and half the length), or about 3 million pixels. The exposure is the same because each pixel receives the same number of photons.

But what if you zoom the lens? At 96mm focal length, the foamcore now covers the entire sensor at 16 feet. But to maintain the same aperture setting, the lens iris has to double its diameter when focal length doubles (f/ number = iris diameter as a fraction of focal length). When you double the iris diameter, you quadruple the area, allowing 4 times as much light to reach your sensor. By keeping the aperture (f/ number) constant as you zoom to a longer focal length, you are increasing the actual size of the hole the light travels through, which compensates for the light lost through distance.

Thanks Curtis, your explanation cleared it up the best (that's not to imply that the other responses didn't help, but Curtis's "clicked" for me)