Light Metering & Changing Distance

<p>Hi friends,<br /> I was looking for info about photographing a concert, arrived trough google to what seems to be an experienced photographer doing the job for years, but suddenly he touched a question of photography I am asking myself for more than two decades.<br /> He said that he meters with a spot meter Caucassian skin, (kindly excuss me) and opens one customary stop. Fine no problem, he is not the first nor will be the last.<br /> But since a concert needs very different shooting distances, and if you do not agree it doesn't matter since the best angles to shoot a concert it is not my question here, my old question I carry on my mind was aroused:<br>

it doesn't sound me "logical" or realistic, that an f/stop under light stress conditions will remain the same from a distance of one meter (your own hand metered at the same lighting like the orquestra) and stay the same if you move 100 meters away from the orquestra. It simply doesn't make sense for me. The f/stop to shoot the orquestra must fall as you walk farther. In addition we all know a certain law for artificial lighting saying the light decreases by some factor related to the distance between your studio light and your subject.<br /> However, on the other hand and against my sense and logics, we all know that the average aperture for a sunny day is f/11 and 1/250 and it doesn't change with distance.<br /> If you can, kindly extract me from the chaos.</p>

<p> </p>

<p>This involves two different questions that are easily confused.</p>

<p>The first is luminance, that being the light intensity per unit area. This illustrates the "inverse square" law of light, where the amount falling on the subject decreases as the distance between the source and the subject increases.</p>

<p>Let's take for example a 100-watt light bulb enclosed in a sphere.<br /> If the radius of the sphere is chosen such that its surface area is 1 square meter, then the luminance is 100 watts per square meter, by definition.<br /> If we enlarge the sphere so its surface are is now 10 square meters, the luminance drops to 10 watts per square meter. Similarly, if we enlarge the sphere to 100 square meters, the luminance drops to 1 watt per square meter.</p>

<p>Since in photography the size of the subject doesn't change (usually), we can use the distance from the light to the subject to control the brightness.</p>

<p>The other half of the question is what happens to light between the subject and the camera. The key to this answer is the size of the receptor, i.e. the camera lens.</p>

<p>This can be illustrated using a newspaper and your nose. Hold the paper up touching your nose, and notice its brightness. Then move the paper out to arm's length. The brightness doesn't change (unless it was shadowed by your head) even though the distance changed by a ratio of 100:1 or more.</p>

<p>When the receptor (your eye or your lens) is sufficiently small, the apparent brightness of the subject does not change significantly over a very wide range of distances.</p>

<p>Hope this helps.</p>

<p>- Leigh</p>

<p> </p>

<p>The light rays reaching your lens from a point in space are effectively parallel, irrespective of the distance; thus the exposure does not change. The identical scenario in lighting would be if you had a spotlight/dedolight on your subject. There would be no light falloff with distance. The inverse square law does not hold for parallel rays. As Leigh said, the light bulb needs to be enclosed in a sphere (whose surface area changes with square of radius).</p>

<p>I have never tried it, but by my own logic, exposure should change between an object shot as macro (like from a centimeter away) and from farther away (since the change in surface area covered by the lens here is probably huge).</p>

<p>some more split hairs about spotlights <a href="../photography-lighting-equipment-techniques-forum/00XIRH">here</a>.</p>

<p>To eliminate the mental chaos you've indicated, I'ld suggest you stop thinking about all these exposure rules and do a quick test shot of the scene in question and adjust accordingly to how the histogram appears on your camera's LCD while shooting manual.</p>

<p>It's quick and doesn't take a lot of thought. Of course this is assuming you're using a digital camera.</p>

<p>Ruben : think of it this way : The face of your subject is lit by a spot at,let's say,10 feet.<br>

The 'fall off',related to distance,is only from subject to light source. The effective lighting intensity,on the subject,will not change whether you photograph from the front row,or the back of the stadium.<br>

That's why Sunny 16 works : the difference,subject to light source,is almost infinite. So,in bright sun,it hardly matters if the camera is 2 feet, or 2000 yards,from the subject.<br>

Just Physics,really.</p>

<p>Imagine someone's face, evenly lit, filling the frame. Now imagine the same face but in the distance so the size in the frame is now a tenth of it's previous size.</p>

<p>Due to the inverse square law, the light being received is now 1/100 of the light from the close up but it is also now only covering 1/100 of the area of the film or sensor which it previously covered. </p>

<p>Therefore, the light at any single point of this image of a face is the same in both cases. </p>

<p>I find none of the answers particularly illumining (lots of laughs).</p>

<p>As you known, light plays on the subject and the intensity of the light at the subject plane decreases as the distance to light source increases. Doubling the distance lamp-to-subject decreases the light intensity to 1/4 of the original, tripling the distance decreases it to 1/9.</p>

<p>Consider, 100 lumens plays on the subject from a lamp 16 meters from the subject, ncreasing the distance lamp-to-subject to 22 meters reduce the light energy at the subject plane to 50 lumens (1/2). Going the other way, a lamp delivering 100 lumens at a distance of 16 feet, when moved closer to 11 meters delivers 200 lumens (2x). The key here is the law of inverse square. If we multiply, or divide lamp-to-subject distance by the square root of 2, we get a revised distance that brings about a 2x change. That factor is 1.414 however its best if we round to 1.4. Thus: 16 x 1.4 = 22 and 16 ÷ 1.4 = 11.</p>

<p>I t is no coincidence that the f/number set we use is based on 2x incremental steps and it too uses the 1.4 factor. Note the f/numbers set: 1.4 - 2 - 2.8 - 4 - 5.6 - 8 - 11 - 16 - 22 -32 each number going right is its neighbor on the left multiplied by 1.4.</p>

<p>Now as to reflected light: We see and photograph objects by reflected light (reflect from Latin to bounce back). Reflected light follows rules similar to rebounding billiard balls. The angle or arrival (incidence) equals the angle of departure (I = r) (reflection).</p>

<p>The caveat is, this rule applies only to flat surfaces. Now it is rare to find truly flat surfaces in nature. Our subject reflects light over a wide arc. This is true because you can see the people on the stage equally well from the center or the sides of the auditorium. This is because bounce back from the subjects is diffused reflections. Diffused reflections do not sticky follow the law of the inverse square. We capitalize on this when be we use umbrella light or soft-box. Also note that only a point source like a bare bulb will strictly follow the law of the inverse square. However, some small areas of the subject are quite flat or polished. If your orientation is just right, from these areas your camera will receive virtually all parallel rays. These will be the "highlights" and "cetch-lights". <br /> So it is safe to say, camera-to-subject distance is not a major contributor to exposure modifications.</p>

<p>OK enough gobbledygook from me!</p>

<p>"So it is safe to say, camera-to-subject distance is not a major contributor to exposure modifications"</p>

<p>Hi fellows,<br /> I thank you all for the attempt, including the bottom lines, but perhaps what appears to be a simple question requires a rather complex (for me) mathematical level. I have read the answers and the most I think about I go nuts by inverse square proportion. Therefore I will spare from you the paradoxes in my mind about this issue.<br>

Perhaps I am not taking into account the difference between the effect of the lenses when processing light, and our nude eyes. Meaning that different lenses can give .... No. I am going nuts again, lenses cannot change the constant value of light.<br>

So I think that I will leave the chaos inside a closet, go practical and pick up the above quotation. And make myself a coffee.<br /> Cheers.</p>

<p> </p>