Great Bustard
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This is a continuation of:
http://forums.dpreview.com/forums/read.asp?forum=1032&message=36740698
Why doesn't exposure change with focal length?
because I just know everyone needed more than 150 posts on the subject.
OK, here's my claim:
For a uniformly lit scene, the exposure will be the same for the same shutter speed and f-ratio regardless of framing or distance from the scene (so long as we are not at "high" magnification).
The explanation comes in two parts. The first situation is for the same framing, but different distances. Let's consider the case of 5 ft 50mm f/2 and 10 ft 100mm f/2.
Both will have the same framing. The aperture diameter at 50mm f/2 is 50mm / 2 = 25mm. The aperture diameter for 100mm f/2 is 100mm / 2 = 50mm. Thus, the aperture at 100mm f/2 has four times the area as 50mm f/2.
The amount of light reaching the lens from the scene at 10 ft is 1/4 the amount of light reaching the lens at 5 ft (inverse square relationship for light). The validity of using the inverse square relationship is discussed at the end of this post as an addendum.
Since there is 1/4 as much light reaching the lens at 10 ft than at 5 ft, but the aperture has four times the area at 100mm f/2 than 50mm f/2, the two effects cancel each other out, and the same amount of light falls on the sensor. For a given sensor size, this results in the same density of light, which means the same exposure.
The second case is for the same distance, but different framing. Let's again consider 50mm f/2 and 100mm f/2. The framing at 50mm f/2 is twice as wide as the framing at 100mm f/2, so we are collecting four times as much light (recall that we are assuming uniform lighting).
However, the aperture area at 50mm f/2 is 1/4 the aperture area at 100mm f/2, so only 1/4 of that light is getting to the sensor. Thus, once again, the two effects cancel each other out, and the same total amount of light reaches the sensor. And, again, for the same sensor size, that means the density of the light is also the same, and thus the same exposure.
Thus, for the same f-ratio and shutter speed, the exposure is the same, regardless of distance to the scene or the framing of the scene (for uniformly lit scenes). This also applies for different sensor sizes, but I'll skip those examples here.
OK, about the "inverse square law". Light radiates from each point in a spherical pattern. A sphere with twice the diameter will have four times the area. Thus, for a given amount of light, the amount of light per area will be 1/4 as much for twice the diameter.
However, for a scene that is not a point, when we double the distance from the scene, we are not doubling the distance from all the points in that scene -- we are only doubling our distance from the focal point.
Let's first consider the portions of the scene in the focal plane. As an example, we'll consider a distance of 10 ft at 50mm. The corners of the focal plane are 10.9 ft away, as opposed to 10 ft from the center. When we double our distance from the scene, the center is now 20 ft away, but the corners are 20.5 ft away. This, of course, is an insignificant difference.
However, if I instead get twice as close, 5ft at 25mm, the corners are now 6.6 feet away -- 2/3 their previous distance instead of 1/2 the distance. This still amounts to only around a 1/4 stop difference for the scene as a whole off from what the inverse square law would tell us. But, as we get closer, and wider, still, the difference will increase.
But what about the portions of the scene not on the focal plane? Objects far from the focal plane will barely change their relative distance to the camera at all, so there is basically no loss of light at all. However, as we step back and use a longer focal length, we capture less of the background. This effect cancels out the fact that the intensity of the light reaching the lens changes very little.
For those that doubt what I've claimed (and for those that care), I propose the following experiment:
Find a scene that is uniformly lit. That's the hard part -- how are you sure it's uniformly lit? I would say to meter a bunch of portions of the scene with high magnification from the same distance with the same focal length and f-ratio and see that the shutter speed stays the same.
That done, take a pic of the scene at any focal length and any f-ratio (if the zoom is a variable aperture zoom, use an f-ratio at least as large as the large end -- for example, if using a 70-300 / 4-5.6, use at least f/5.6).
Step back or forward, taking care that your shadow, however faint, does not fall on the scene, or that your body does not block any of the light sources. Frame the scene exactly the same, and take a pic with the same f-ratio again (and ISO, of course!). Note any difference in shutter speed.
I contend the shutter speed will not change (unless, perhaps, you are very close to the scene) and the pics will be identically exposed.
http://forums.dpreview.com/forums/read.asp?forum=1032&message=36740698
Why doesn't exposure change with focal length?
because I just know everyone needed more than 150 posts on the subject.
OK, here's my claim:
For a uniformly lit scene, the exposure will be the same for the same shutter speed and f-ratio regardless of framing or distance from the scene (so long as we are not at "high" magnification).
The explanation comes in two parts. The first situation is for the same framing, but different distances. Let's consider the case of 5 ft 50mm f/2 and 10 ft 100mm f/2.
Both will have the same framing. The aperture diameter at 50mm f/2 is 50mm / 2 = 25mm. The aperture diameter for 100mm f/2 is 100mm / 2 = 50mm. Thus, the aperture at 100mm f/2 has four times the area as 50mm f/2.
The amount of light reaching the lens from the scene at 10 ft is 1/4 the amount of light reaching the lens at 5 ft (inverse square relationship for light). The validity of using the inverse square relationship is discussed at the end of this post as an addendum.
Since there is 1/4 as much light reaching the lens at 10 ft than at 5 ft, but the aperture has four times the area at 100mm f/2 than 50mm f/2, the two effects cancel each other out, and the same amount of light falls on the sensor. For a given sensor size, this results in the same density of light, which means the same exposure.
The second case is for the same distance, but different framing. Let's again consider 50mm f/2 and 100mm f/2. The framing at 50mm f/2 is twice as wide as the framing at 100mm f/2, so we are collecting four times as much light (recall that we are assuming uniform lighting).
However, the aperture area at 50mm f/2 is 1/4 the aperture area at 100mm f/2, so only 1/4 of that light is getting to the sensor. Thus, once again, the two effects cancel each other out, and the same total amount of light reaches the sensor. And, again, for the same sensor size, that means the density of the light is also the same, and thus the same exposure.
Thus, for the same f-ratio and shutter speed, the exposure is the same, regardless of distance to the scene or the framing of the scene (for uniformly lit scenes). This also applies for different sensor sizes, but I'll skip those examples here.
OK, about the "inverse square law". Light radiates from each point in a spherical pattern. A sphere with twice the diameter will have four times the area. Thus, for a given amount of light, the amount of light per area will be 1/4 as much for twice the diameter.
However, for a scene that is not a point, when we double the distance from the scene, we are not doubling the distance from all the points in that scene -- we are only doubling our distance from the focal point.
Let's first consider the portions of the scene in the focal plane. As an example, we'll consider a distance of 10 ft at 50mm. The corners of the focal plane are 10.9 ft away, as opposed to 10 ft from the center. When we double our distance from the scene, the center is now 20 ft away, but the corners are 20.5 ft away. This, of course, is an insignificant difference.
However, if I instead get twice as close, 5ft at 25mm, the corners are now 6.6 feet away -- 2/3 their previous distance instead of 1/2 the distance. This still amounts to only around a 1/4 stop difference for the scene as a whole off from what the inverse square law would tell us. But, as we get closer, and wider, still, the difference will increase.
But what about the portions of the scene not on the focal plane? Objects far from the focal plane will barely change their relative distance to the camera at all, so there is basically no loss of light at all. However, as we step back and use a longer focal length, we capture less of the background. This effect cancels out the fact that the intensity of the light reaching the lens changes very little.
For those that doubt what I've claimed (and for those that care), I propose the following experiment:
Find a scene that is uniformly lit. That's the hard part -- how are you sure it's uniformly lit? I would say to meter a bunch of portions of the scene with high magnification from the same distance with the same focal length and f-ratio and see that the shutter speed stays the same.
That done, take a pic of the scene at any focal length and any f-ratio (if the zoom is a variable aperture zoom, use an f-ratio at least as large as the large end -- for example, if using a 70-300 / 4-5.6, use at least f/5.6).
Step back or forward, taking care that your shadow, however faint, does not fall on the scene, or that your body does not block any of the light sources. Frame the scene exactly the same, and take a pic with the same f-ratio again (and ISO, of course!). Note any difference in shutter speed.
I contend the shutter speed will not change (unless, perhaps, you are very close to the scene) and the pics will be identically exposed.
