Crop sensors need cropped lenes
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Pi said:BTW, the convention what EV means does not involve ISO at all. Not that it matters because EV is just a term. I am fine with you not knowing what it is, after all, I did not until today.
oh dear oh dear. Have you been reading up on EV when taking a break from digging this hole ?
I shoot at 250 / f5.6 / ISO 100 for 'correct' exposure. The EV is............ 13
I shoot at 250 / f11 / ISO 400 for the same 'correct' exposure. The EV is..............15
No it has nothing to do with ISO at all
<sarcasm>
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Since this discussion generated too much, I want to copy and paste a quote from Joe's essay, boldface mine:
http://www.josephjamesphotography.com/equivalence/#1
I only disagree with his mentioning of DR but I know that he knows about read noise, etc.
1) f/2 = f/2 = f/2
This is perhaps the single most misunderstood concept when comparing formats. Saying "f/2 = f/2 = f/2" is like saying "50mm = 50mm = 50mm". Just as the effect of 50mm is not the same on different formats, the effect of f/2 is not the same on different formats.
Everyone knows what the effect of the focal length is -- in combination with the sensor size, it tells us the AOV (diagonal angle-of-view). Many are also aware that f-ratio affects both DOF and exposure. It is important, however, to understand that the exposure (the density of light falling on the sensor -- photons / mm²) is merely a component of the total amount of light falling on the sensor (photons): Total Light = Exposure x Effective Sensor Area, and it is the total amount of light falling on the sensor, as opposed to the exposure, which is the relevant measure.
Within a format, the same exposure results in the same total light, so the two terms can be used interchangeably, much like mass and weight when measuring in the same acceleration field. For example, it makes no difference whether I say weigh 180 pounds or have a mass of 82 kg, as long as all comparisons are done on Earth. But if makes no sense at all to say that, since I weigh 180 lbs on Earth, that I'm more massive than an astronaut who weighs 30 lbs on the moon, since we both have a mass of 82 kg.
The reason that the total amount of light falling on the sensor, as opposed to the density of light falling on the sensor (exposure), is the relevant measure is because the total amount of light falling on the sensor, combined with the sensor efficiency, determines the amount of noise and DR (dynamic range) of the photo.
For a given scene, perspective (subject-camera distance), framing (AOV), and shutter speed, both the DOF and the total amount of light falling on the sensor are determined by the diameter of the aperture. For example, 80mm on FF, 50mm on 1.6x, and 40mm on 4/3 will have the same AOV (40mm x 2 = 50mm x 1.6 = 80mm). Likewise, 80mm f/4, 50mm f/2.5, and 40mm f/2 will have the same aperture diameter (80mm / 4 = 50mm / 2.5 = 40mm / 2 = 20mm). Thus, if we took a pic of the same scene from the same position with those settings, all three systems would produce a photo with the same perspective, framing, DOF, and put the same total amount of light on the sensor, which would result in the same total noise for equally efficient sensors (the role of the ISO in all this is simply to adjust the brightness of the LCD playback and/or OOC jpg).
Thus, settings that have the same AOV and aperture diameter are called "Equivalent" since they result in Equivalent photos. Hence, saying f/2 on one format is the same as f/2 on another format is just like saying that 50mm on one format is the same as 50mm on another format.
http://www.josephjamesphotography.com/equivalence/#1
I only disagree with his mentioning of DR but I know that he knows about read noise, etc.
1) f/2 = f/2 = f/2
This is perhaps the single most misunderstood concept when comparing formats. Saying "f/2 = f/2 = f/2" is like saying "50mm = 50mm = 50mm". Just as the effect of 50mm is not the same on different formats, the effect of f/2 is not the same on different formats.
Everyone knows what the effect of the focal length is -- in combination with the sensor size, it tells us the AOV (diagonal angle-of-view). Many are also aware that f-ratio affects both DOF and exposure. It is important, however, to understand that the exposure (the density of light falling on the sensor -- photons / mm²) is merely a component of the total amount of light falling on the sensor (photons): Total Light = Exposure x Effective Sensor Area, and it is the total amount of light falling on the sensor, as opposed to the exposure, which is the relevant measure.
Within a format, the same exposure results in the same total light, so the two terms can be used interchangeably, much like mass and weight when measuring in the same acceleration field. For example, it makes no difference whether I say weigh 180 pounds or have a mass of 82 kg, as long as all comparisons are done on Earth. But if makes no sense at all to say that, since I weigh 180 lbs on Earth, that I'm more massive than an astronaut who weighs 30 lbs on the moon, since we both have a mass of 82 kg.
The reason that the total amount of light falling on the sensor, as opposed to the density of light falling on the sensor (exposure), is the relevant measure is because the total amount of light falling on the sensor, combined with the sensor efficiency, determines the amount of noise and DR (dynamic range) of the photo.
For a given scene, perspective (subject-camera distance), framing (AOV), and shutter speed, both the DOF and the total amount of light falling on the sensor are determined by the diameter of the aperture. For example, 80mm on FF, 50mm on 1.6x, and 40mm on 4/3 will have the same AOV (40mm x 2 = 50mm x 1.6 = 80mm). Likewise, 80mm f/4, 50mm f/2.5, and 40mm f/2 will have the same aperture diameter (80mm / 4 = 50mm / 2.5 = 40mm / 2 = 20mm). Thus, if we took a pic of the same scene from the same position with those settings, all three systems would produce a photo with the same perspective, framing, DOF, and put the same total amount of light on the sensor, which would result in the same total noise for equally efficient sensors (the role of the ISO in all this is simply to adjust the brightness of the LCD playback and/or OOC jpg).
Thus, settings that have the same AOV and aperture diameter are called "Equivalent" since they result in Equivalent photos. Hence, saying f/2 on one format is the same as f/2 on another format is just like saying that 50mm on one format is the same as 50mm on another format.
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Sporgon said:Pi said:BTW, the convention what EV means does not involve ISO at all. Not that it matters because EV is just a term. I am fine with you not knowing what it is, after all, I did not until today.
oh dear oh dear. Have you been reading up on EV when taking a break from digging this hole ?
I shoot at 250 / f5.6 / ISO 100 for 'correct' exposure. The EV is............ 13
I shoot at 250 / f11 / ISO 400 for the same 'correct' exposure. The EV is..............15
No it has nothing to do with ISO at all
<sarcasm>
What if you shoot at 250 / f11 / ISO 200?
Hint: now is the time to google it.
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Pi said:
Alright. I understand what this is saying, and it makes sense, but it is somewhat of an incomplete explanation. Exposure, which in his article is the same as Exposure Value (the amount of light allowed onto the sensor for a given aperture and shutter speed), IS identical across formats. Assuming you ignore ISO (which is what actual exposure does...ISO does not change exposure, it simply amplifies the result), an f/2 lens will always produce the same exposure.
What this article is explaining is that total noise is the same when photographing the same subject, at the same distance, with the same AoV. Total noise is the same because the total amount of light on the sensor is the same. Even though it is spread out more (<-- important point!!), and ISO is used to compensate!
What this article does not fully explain is that an 80mm lens at f/4 is not equivalent to a 50mm f/2.5 lens because of the inverse square falloff of light. Light will spread out more and lose energy more over the 80mm distance to the sensor than over the 50mm distance to the sensor, despite an identical entrance pupil diameter. The use of an f/4 aperture will result in an EXPOSURE that one and a third stops darker than on the APS-C sensor. In order to produce the same final outcome, ISO must be increased by an equivalent amount, one and a third stops, to compensate for the LOWER exposure.
So, assuming the following:
A) APS-C, 50mm, f/2.5, ISO 100
B) FF, 80mm, f/4, ISO 320
The images from both will have the same final appearance. Same DoF, same AoV, same amount of image noise. True point. I think that is a point that is a little different than everyone else has been trying to make...equivalence requires complete equalization of EVERY SINGLE factor, including an increase in ISO, so therefor it is not actually discussing exposure (as in exposure value, which is usually what arguments about aperture and/or shutter speed "equivalence" pertain to).
The point Pi is trying to make is that assuming equivalence (as explained in his linked article), then the final outcome of any sensor size would be identical.
I think the point everyone else has been trying to make is assuming an identical subject distance and lens, an f/2 aperture is an f/2 aperture regardless of sensor size.
Technically speaking, both points are correct. Everyone has been debating oblique points.
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I would also point out that if you read the entire site Pi has linked to, his arguments (not just in this thread) make more sense. The core, critical point that Joseph James makes is that the only truly fair way to compare camera systems is to do so at the same entrance pupil diameter and AoV. I think in a general sense, that is probably quite true, and will indeed produce more useful comparisons in any situation that is not focal length limited.
I guess I would note that normalizing entrance pupil and AoV is not always possible in all forms of photography. I tend to shoot subjects that are difficult to get close to, so my "normal mode" of thinking about such problems is from a focal length limited standpoint, which relies on a few different rules than Joseph James's article assumes.
I guess I would note that normalizing entrance pupil and AoV is not always possible in all forms of photography. I tend to shoot subjects that are difficult to get close to, so my "normal mode" of thinking about such problems is from a focal length limited standpoint, which relies on a few different rules than Joseph James's article assumes.
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I just read few last posts and understood that all you guys are so smart and that I need to learn a lot I thought that I know photography quite well, however, I was not able to fully understand some of posts. I am wondering whether everybody in this forum are so smart or this just me alone 
I thought that I know photography quite well, however, I was not able to fully understand some of posts. I am wondering whether everybody in this forum are so smart or this just me alone
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OK I hold my hand up. Unlike some I didn't learn my photography from wiki.
Obviously when you alter the ISO you alter the shutter/aperture combo - hence EV.
The EV scale was developed years before digital, and I believe ( not from wiki ) that it was primarily for cine film. In those days you couldn't chop and change your ISO ( or even ASA
) like you can now, so in practice ISO effects EV, though I agree not in the definition of the term.
For anyone who argues the technicalities of photography but believes 'exposure value' has just been made up is quite funny.
The only photographic technicalities I do use wiki for is to decipher some of the forum slang such as in this case:
ROFLMFAO
Obviously when you alter the ISO you alter the shutter/aperture combo - hence EV.
The EV scale was developed years before digital, and I believe ( not from wiki ) that it was primarily for cine film. In those days you couldn't chop and change your ISO ( or even ASA
) like you can now, so in practice ISO effects EV, though I agree not in the definition of the term. For anyone who argues the technicalities of photography but believes 'exposure value' has just been made up is quite funny.
The only photographic technicalities I do use wiki for is to decipher some of the forum slang such as in this case:
ROFLMFAO
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jrista said:
Right, but, as Joe says, ... exposure ... is [not] the relevant measure. Total light is.
Too many mistakes in the rest you wrote. F/2 will result in the same intensity (assuming a uniformly lit and fixed scene) no matter what the FL is. The inverse square law does not hold for focused light (even if you do not believe me, do not hold you hand under a bare lens in bright sunshine!) Anyway there is no point arguing this, we all know how this works in reality, if not in theory. You do not need faster f-stops with longer lenses.
To summarize, f/2 on crop vs. f/2 on FF, regardless of FL, gives you same intensity, same EV (with the same SS) but this is an irrelevant measure of photon noise, and to a large extent of anything IQ related aside from resolution (a different discussion). Total light, in contrast, is relevant. In terms of total light, f/2 on crop "is not" f/2 on FF. Of course, this depends on what the meaning of the word "is" is.
If the outcome of this discussion is your reading of Joe's essay, I consider my job done. Joe writes well, and shoots even better - check out his galleries. BTW, he is in math, too...
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Pi said:
I still think your being a little incomplete. No, you do not need a faster "f-stop" (relative aperture) with longer lenses, but that is because it is a relative measure. Just to be complete and clear, a 400mm f/4 lens has a much larger entrance pupil diameter than a 40mm f/4 lens. The longer lens has a 100mm diameter entrance pupil, while the 40mm has a 10mm diameter entrance pupil. The reason for the larger diameter on the longer lens is because of the inverse square law...falloff is greater over the greater distance to sensor with the 400mm lens.
On a FF sensor, you could use a 400mm f/4, where as on APS-C, to achieve the same IQ at an equivalent AoV as you put it, you would need a 250mm f/2.5 lens. I don't disagree about that point, however if we conform to Joe's approach of entrance pupil diameter matters here, the statement "You do not need faster f-stops with longer lenses." is incomplete. Same f-stop, but longer lenses still do initially let through more light (considerably more, actually), which then falls off to produce the same intensity at the sensor as a shorter lens at the same f-stop. A 400mm f/4 lens has an entrance pupil area of 7850mm^2, where as the 40mm f/4 lens has a 78.5mm^2 entrance pupil area. Applying inverse square just to the aperture areas, we get the same result for both:
Code:
7850 * (1/400^2) = 0.05
78.5 * (1/40^2) = 0.05This may be entirely irrelevant in terms of measuring photon noise. That doesn't really matter either, though. Not everyone is solely concerned with noise. Sometimes sharpness is what matters, and in focal length limited situations, comparing equipment at the same AoV and entrance pupil diameter is not only irrelevant, it is also often impossible. ;P
There are different goals in photography, and not all of them can be discussed or compared in the same way. I will happily concede that the point you were arguing from the angle you were arguing it was valid. I came at the problem from a different, and still wholly valid, angle, as I believe most of the other members of the discussion did. Different relevance for different situations...a point you might try to recognize sometime.
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Joseph James here (Great Bustard on DPR). Been following the discussion. You may find the following links useful, as they discuss some of the points being raised:
http://www.josephjamesphotography.com/equivalence/#introduction
A 50mm f/1.4 lens is a 50mm f/1.4 lens regardless of the sensor that sits behind it. However, the effect of 50mm f/1.4, in terms of the visual properties of the recorded photo, depend very much on the sensor that sits behind the lens:
25mm f/1.4 on mFT (4/3) is equivalent to 31mm f/1.8 on 1.6x (Canon APS-C), 33mm f/1.9 on 1.5x (APS-C for everyone else), and 50mm f/2.8 on FF (FX), where "equivalent to" means:
• The photos all have the same AOV (diagonal angle of view) and aperture (entrance pupil) diameter: 25mm / 1.4 = 31mm / 1.8 = 33mm / 1.9 = 50mm / 2.8 = 18mm.
• The photos all have the same DOF (as well as diffraction softening) when they have same perspective (subject-camera distance), AOV, aperture diameter, and display size.
• The photos all have the same motion blur and the same total amount of light falls on the sensor when the aperture diameter and shutter speed are the same. which means the larger the sensor, the lower the exposure (same total light over a larger area) and thus a higher ISO setting for a given brightness.
• The photos all have the same same noise when the same total amount of light falls on the sensor if the sensors are equally efficient (less noise if the sensor is more efficient, more noise if the sensor is less efficient).
• Other elements of IQ, such as resolution, bokeh, color, etc., as well as elements of operation, such as AF speed/accuracy, size, weight, etc., are not covered in this use of the term "equivalent".
http://www.josephjamesphotography.com/equivalence/#aperture
The term "aperture", by itself, is vague -- we need a qualifying adjective to be clear. There are three different terms using "aperture":
1. The physical aperture (iris) is the smallest opening within a lens.
2. The virtual aperture (entrance pupil) is the image of the physical aperture when looking through the FE (front element).
3. The relative aperture (f-ratio) is the quotient of the focal length and the virtual aperture.
Thus, the "f" in an f-ratio stands for focal length. For example f/2 on a 50mm lens means the diameter of the virtual aperture (entrance pupil) is 50mm / 2 = 25mm. Likewise, a 50mm lens with a 25mm virtual aperture has an f-ratio of 50mm / 25mm = 2.
http://www.josephjamesphotography.com/equivalence/#exposure
The exposure is the density of light (total light per area -- photons / mm²) that falls on the sensor during the exposure, which is usually expressed as the product of the illuminance of the sensor and the time the shutter is open (lux · seconds, where 1 lux · second = 4.1 billion photons / mm² for green light -- 555 nm). The only factors in the exposure are the scene luminance, t-stop (where the f-ratio is often a good approximation for the t-stop), and the shutter speed (note that neither sensor size nor ISO are factors in exposure).
For example, two pics of the same scene, one at f/2.8 1/200 ISO 100 and another at f/2.8 1/200 ISO 400 (on any system, regardless of format) will both have the same exposure, since the same number of photons per unit area will fall on the sensor, but the ISO 400 photo will appear 4x (2 stops) brighter than the ISO 100 photo since the signal is amplified by a factor of four due to the higher ISO setting.
The brightness, then, is the brightness of the final image after an amplification is applied to the exposure either by adjusting the ISO and/or a push/pull in the RAW conversion, and is often what people mean when they say "exposure". For example, pics of the same scene at f/2.8 1/200 ISO 100 and f/5.6 1/200 ISO 400 will be processed to have the same brightness, even though the f/2.8 photo has 4x (two stops greater) exposure than the f/5.6 photo.
The role of the ISO setting in exposure is in how the setting indirectly results in the camera choosing a different f-ratio, shutter speed, and/or flash power, any and all of which will change the exposure. For example, changing the ISO from 100 to 400 may result in the camera choosing f/5.6 instead of f/2.8, 1/200 instead of 1/50, f/4 1/100 instead of f/2.8 1/50, etc. Aside from that, the ISO control on the camera will apply an analog gain (which results in less read noise for higher ISOs with cameras that use non-ISOless sensors) and/or a digital push/pull (usually for intermediate ISO settings).
The total light is the total amount of light that falls on the portion of the sensor used to for the photo during the exposure: Total Light = Exposure · Effective Sensor Area. The same total amount of light will fall on the sensor for equivalent photos but, for different formats, this will necessarily result in a different exposure on each format, since the same total light distributed over sensors with different areas will result in a lower density of light on the larger sensor. Using the same example above, pics of the same scene at f/2.8 1/200 on mFT (4/3) and f/5.6 1/200 on FF will result in the same total light falling on each sensor, but the exposure will be 4x (2 stops) greater for the mFT photo, and thus the FF photographer would usually use a 4x (2 stops) higher ISO setting to get the same brightness for the LCD playback and/or OOC (out-of-the-camera) jpg.
Lastly, the Total Light Collected is the amount of light used to make the photo, which is the product of the Total Light that falls on the sensor during the exposure and the QE (Quantum Efficiency of the sensor -- the proportion of light falling on the sensor that is recorded). For example, if QE = 1, then all the light falling on the sensor is recorded. For reference, the Olympus EM5, Canon 5D3, and Nikon D800 all have a QE of approximately 0.5 (50%).
In terms of IQ, the total light collected is the relevant measure, because both the noise and DR (dynamic range) of a photo are a function of the total amount of light that falls on the sensor (along with the sensor efficiency, all discussed, in detail, in the next section). That is, noise is determined by the total amount of light falling on the sensor and the sensor efficiency, not the ISO setting on the camera, as is commonly believed (the ISO setting is simply a matter of processing the signal, discussed in more detail here). In other words, the less light that falls on the sensor, the more noisy and darker the photo will be. Increasing the ISO setting simply brightens the captured photo making the noise more visible.
For a given scene, perspective, and framing, the total light depends only on the aperture diameter and shutter speed (as opposed to the f-ratio and shutter speed for exposure). Fully equivalent images on different formats will have the same brightness and be created with the same total amount of light. Thus, the same total amount of light on sensors with different areas will necessarily result in different exposures on different formats, and it is for this reason that exposure is a meaningless measure in cross-format comparisons.
Mathematically, we can express these four quantities rather simply:
• Exposure (photons / mm²) = Sensor Illuminance (photons / mm² / s) · Time (s)
• Brightness (photons / mm²) = Exposure (photons / mm²) · Amplification (unitless)
• Total Light (photons) = Exposure (photons / mm²) · Effective Sensor Area (mm²)
• Total Light Collected (electrons) = Total Light (photons) · QE (electrons / photon)
The ISO setting on the camera determines the amplification applied to the signal. For example, at ISO 1600, the gain is 1600 / 100 = 16x for a camera with a base ISO of 100. The amplification can either be analog and/or digital, with analog gain resulting in less read noise for some cameras (see here for an example with the Canon 5D), but at the risk of more blown highlights (discussed in more detail here). It is important to note that not all cameras apply the same amount of amplification for the same ISO. For example, f/5.6 1/100 ISO 400 on one camera may not show the same brightness as f/5.6 1/100 ISO 400 on another camera. The ISO standards allow the manufacturers a lot of latitude in their definition of ISO.
http://www.josephjamesphotography.com/equivalence/#noise
When people refer to noise in an image, what they mean is the density of the noise in an image (NSR -- noise-to-signal ratio) and is often represented as a percent. Often, we hear the term "SNR" (signal-to-noise ratio), which is the reciprocal of the NSR (SNR = 1 / NSR). For example, if the SNR = 5:1, then the NSR = 1/5 = 20%. However, since photo with high noise has a high NSR, and a photo with low noise has a low NSR, whereas it is exactly opposite for SNR, it is less confusing to think in terms of NSR than SNR.
There are two principle sources of noise in a photo: luminance noise and chroma (color) noise. Luminance noise is a function of the total amount of light falling on the sensor, and the efficiency of the sensor. The photon noise (often referred to as "shot" noise) is determined by how much light the sensor records. This, in turn, is determined by the total amount of light falling on the sensor (Total Light = Exposure x Effective Sensor Area) and the QE (Quantum Efficiency) of the sensor, which is the proportion of light falling on the sensor that is recorded (for example, a QE of 50% means half the light falling on the sensor is recorded). The other factor in luminance noise is the read noise, which is the additional noise added by the sensor and supporting hardware.
The chroma noise is the noise is a result of the color filters both blocking light which they are supposed to admit (AE -- Absorption Error), and admitting light they are supposed to block (TE -- Transmission Error). For example, if a red photon fails to make it through the red filter, this results in an Absorption Error, and if a red photon makes it through a green filter, this results in a Transmission Error.
The more transmissive the color filters (weaker color filter), the greater the QE, and thus the lesser the luminance noise, but the greater the chroma noise. The less transmissive the color filter (stronger color filter), the less the chroma noise, but lesser QE, and thus greater luminance noise. Different manufacturers choose a different balance in the transmissivity of their color filters resulting in a different balance between the luminance and chroma noise, which results is a different quality of noise in the photos, even if the quantity of noise remains the same.
The photon noise is the primary source of noise in the midtones and highlights of a photo. It is an inherent characteristic of incoherent light (the kind of light in almost all situations -- see the diagrams at the bottom of this page), and unavoidable -- one of those "Laws of Physics" things, as opposed to "an engineering challenge". Light has the properties of both a particle (photon) and wave, and the noise is measured in terms of its particle characteristics. The photons are collected and focused by the lens onto the sensor, where they are converted into electrons, and the signal is processed and recorded. The only role the sensor plays in the photon noise is what proportion of the photons falling on the sensor are converted into electrons (QE), since the electrons are the source of the electrical current that is processed by the hardware. The read noise, discussed in more detail further down, is how much noise is added when collecting and processing the signal produced by the photons.
The photon noise is proportional to the square root of the signal (N = sqrt S) so the photon noise density is inversely proportional to the square root of the signal (NSR = 1/ sqrt S). Thus, if we quadruple the signal (the total amount of light recorded), we halve the noise density.
There are, of course, other sources of noise, such as thermal noise, which plays a central role in long exposures, PRNU (Pixel Response Non-Uniformity) noise, which plays an important role in the highlights of the image, as well as other sources of noise. So, noise, is, of course, even more complicated than this essay makes it appear, and for some specific forms of photography (such as astrophotography) we may find that the noise is very different for equivalent images in some situations, much in the same way that corner sharpness is very different for equivalent images in some situations.
http://www.josephjamesphotography.com/equivalence/#fratio
It is instructive to understand why the same f-ratio results in the same exposure for the same scene and framing, regardless of the focal length or format. There are six factors that determine how much light falls on the sensor:
• The luminance of the scene
• The amount of the scene that is recorded
• The distance from the scene
• The diameter of the aperture
• Transmissivity of the lens elements
• The shutter speed
The amount of light from the scene depends on how wide we frame -- the wider we frame, the more light we will capture, since we are gathering light from a larger scene. If we assume a scene with the same average luminance, framing twice as wide, for example, will result in collecting light from four times as much area, and thus four times as much light reaching the aperture.
The amount of light from the scene reaching the aperture also depends on how far we are from the scene -- the further away we are, the less of that light that reaches the lens. For example, if we are twice as far away, only 1/4 as much light will fall on the lens in any given time interval. It should be noted that the inverse square law is exact only for point sources, and becomes "less exact" the wider we frame. The reason is that the distance from the camera to the center of the focal plane is not the same as the distance to the other portions of the frame. So, when we increase the distance from the scene, the distance from the other portions of the frame do not increase in the same proportion as the center. However, for most situations, the difference is trivial.
Furthermore, the amount of light from the scene falling on the aperture is proportional to the area of the aperture. For example, if we double the diameter of the aperture, the area will quadruple, so four times as much light can pass through in any given time interval. However, some light is lost as it travels through the lens which depends greatly on the lens (click here for some examples). Lastly, the amount of the light passing through the aperture onto the sensor is proportional to the exposure time. That is, double the shutter speed, and you halve the amount of light falling on the sensor.
Let's work a few examples, ignoring the effects of light lost from the elements of the lens, keeping in mind that the exposure is the total light per area falling on the sensor, not the total amount of light. In other words, we can express the exposure as the quotient of the total amount of light falling on the sensor and the area of the sensor.
Say a photographer takes a "properly exposed" pic of a subject 10 ft away at 50mm f/2 1/100 (aperture diameter = 50mm / 2 = 25mm). If they step back to 20 ft away and use 100mm f/2 1/100, the aperture diameter has doubled (100mm / 2 = 50mm) and the aperture area has quadrupled (area is proportional to the square of the diameter). However, the amount of light from the scene reaching the lens is 1/4 as much since they're twice as far away. Since the aperture area is four times as much, it exactly compensates, and the same amount of light will pass through the aperture onto the sensor, and, since the sensor has not changed size, the exposure will also be the same.
Alternatively, let's say they don't step back, but instead remained at 10 ft and shot at 100mm f/2 1/100. At 100mm, the framing will be twice as tight, and thus record only 1/4 the light of the scene as they would at 50mm (assuming, of course, a uniformly lit scene). Thus, despite the fact that 1/4 as much light is reaching the lens, since the aperture area is four times as great, it exactly compensates once again. An excellent video on this can be seen here.
The above two examples demonstrate how the same f-ratio and shutter speed results in the same total light and exposure for a given scene and format regardless of focal length on the same format. However, for different formats, the same exposure does not result in the same total amount of light falling on the sensor.
Let's now consider a photographer using using 50mm f/2 1/100 on 4/3 and another photographer with FF shooting the same scene from the same distance at 100mm f/4 1/100. In both cases, the framing is the same (ignoring the minor differences in aspect ratio between the systems, which amounts to a mere 4% difference), the aperture diameters are the same (50mm / 2 = 100mm / 4 = 25mm), the distances from the scene are the same, and the shutter speeds are the same. Thus, the same amount of light will pass through the apertures onto the sensors.
On the other hand, if the FF photographer shot the same scene from the same position at 100mm f/2 1/100, the aperture diameter would now be 100mm / 2 = 50mm as opposed to 25mm. Since the aperture diameter is twice the size, the aperture area is four times as large, and four times as much light will fall on the sensor. But, since the sensor has four times the area, the density of the light on the sensor would be the same as the 4/3 sensor, so the exposures would be the same.
Thus, for a given scene, perspective, framing, aperture diameter, and shutters speed, the same amount of light will pass through the aperture onto the sensor for all systems. However, there are some exceptions. Some light is absorbed and/or scattered by the glass elements in the lens, and is quantified by noting the difference between the T-stop and the F-stop (f-ratio). Click here to see some examples of this differential for a few lenses for the Nikon system. It's helpful to note that a transmission of 79% represents a 1/3 stop loss of light, 63% represents a 2/3 stop loss of light, and 50% represents a one stop loss of light.
I hope these quotes make things more clear.
http://www.josephjamesphotography.com/equivalence/#introduction
A 50mm f/1.4 lens is a 50mm f/1.4 lens regardless of the sensor that sits behind it. However, the effect of 50mm f/1.4, in terms of the visual properties of the recorded photo, depend very much on the sensor that sits behind the lens:
25mm f/1.4 on mFT (4/3) is equivalent to 31mm f/1.8 on 1.6x (Canon APS-C), 33mm f/1.9 on 1.5x (APS-C for everyone else), and 50mm f/2.8 on FF (FX), where "equivalent to" means:
• The photos all have the same AOV (diagonal angle of view) and aperture (entrance pupil) diameter: 25mm / 1.4 = 31mm / 1.8 = 33mm / 1.9 = 50mm / 2.8 = 18mm.
• The photos all have the same DOF (as well as diffraction softening) when they have same perspective (subject-camera distance), AOV, aperture diameter, and display size.
• The photos all have the same motion blur and the same total amount of light falls on the sensor when the aperture diameter and shutter speed are the same. which means the larger the sensor, the lower the exposure (same total light over a larger area) and thus a higher ISO setting for a given brightness.
• The photos all have the same same noise when the same total amount of light falls on the sensor if the sensors are equally efficient (less noise if the sensor is more efficient, more noise if the sensor is less efficient).
• Other elements of IQ, such as resolution, bokeh, color, etc., as well as elements of operation, such as AF speed/accuracy, size, weight, etc., are not covered in this use of the term "equivalent".
http://www.josephjamesphotography.com/equivalence/#aperture
The term "aperture", by itself, is vague -- we need a qualifying adjective to be clear. There are three different terms using "aperture":
1. The physical aperture (iris) is the smallest opening within a lens.
2. The virtual aperture (entrance pupil) is the image of the physical aperture when looking through the FE (front element).
3. The relative aperture (f-ratio) is the quotient of the focal length and the virtual aperture.
Thus, the "f" in an f-ratio stands for focal length. For example f/2 on a 50mm lens means the diameter of the virtual aperture (entrance pupil) is 50mm / 2 = 25mm. Likewise, a 50mm lens with a 25mm virtual aperture has an f-ratio of 50mm / 25mm = 2.
http://www.josephjamesphotography.com/equivalence/#exposure
The exposure is the density of light (total light per area -- photons / mm²) that falls on the sensor during the exposure, which is usually expressed as the product of the illuminance of the sensor and the time the shutter is open (lux · seconds, where 1 lux · second = 4.1 billion photons / mm² for green light -- 555 nm). The only factors in the exposure are the scene luminance, t-stop (where the f-ratio is often a good approximation for the t-stop), and the shutter speed (note that neither sensor size nor ISO are factors in exposure).
For example, two pics of the same scene, one at f/2.8 1/200 ISO 100 and another at f/2.8 1/200 ISO 400 (on any system, regardless of format) will both have the same exposure, since the same number of photons per unit area will fall on the sensor, but the ISO 400 photo will appear 4x (2 stops) brighter than the ISO 100 photo since the signal is amplified by a factor of four due to the higher ISO setting.
The brightness, then, is the brightness of the final image after an amplification is applied to the exposure either by adjusting the ISO and/or a push/pull in the RAW conversion, and is often what people mean when they say "exposure". For example, pics of the same scene at f/2.8 1/200 ISO 100 and f/5.6 1/200 ISO 400 will be processed to have the same brightness, even though the f/2.8 photo has 4x (two stops greater) exposure than the f/5.6 photo.
The role of the ISO setting in exposure is in how the setting indirectly results in the camera choosing a different f-ratio, shutter speed, and/or flash power, any and all of which will change the exposure. For example, changing the ISO from 100 to 400 may result in the camera choosing f/5.6 instead of f/2.8, 1/200 instead of 1/50, f/4 1/100 instead of f/2.8 1/50, etc. Aside from that, the ISO control on the camera will apply an analog gain (which results in less read noise for higher ISOs with cameras that use non-ISOless sensors) and/or a digital push/pull (usually for intermediate ISO settings).
The total light is the total amount of light that falls on the portion of the sensor used to for the photo during the exposure: Total Light = Exposure · Effective Sensor Area. The same total amount of light will fall on the sensor for equivalent photos but, for different formats, this will necessarily result in a different exposure on each format, since the same total light distributed over sensors with different areas will result in a lower density of light on the larger sensor. Using the same example above, pics of the same scene at f/2.8 1/200 on mFT (4/3) and f/5.6 1/200 on FF will result in the same total light falling on each sensor, but the exposure will be 4x (2 stops) greater for the mFT photo, and thus the FF photographer would usually use a 4x (2 stops) higher ISO setting to get the same brightness for the LCD playback and/or OOC (out-of-the-camera) jpg.
Lastly, the Total Light Collected is the amount of light used to make the photo, which is the product of the Total Light that falls on the sensor during the exposure and the QE (Quantum Efficiency of the sensor -- the proportion of light falling on the sensor that is recorded). For example, if QE = 1, then all the light falling on the sensor is recorded. For reference, the Olympus EM5, Canon 5D3, and Nikon D800 all have a QE of approximately 0.5 (50%).
In terms of IQ, the total light collected is the relevant measure, because both the noise and DR (dynamic range) of a photo are a function of the total amount of light that falls on the sensor (along with the sensor efficiency, all discussed, in detail, in the next section). That is, noise is determined by the total amount of light falling on the sensor and the sensor efficiency, not the ISO setting on the camera, as is commonly believed (the ISO setting is simply a matter of processing the signal, discussed in more detail here). In other words, the less light that falls on the sensor, the more noisy and darker the photo will be. Increasing the ISO setting simply brightens the captured photo making the noise more visible.
For a given scene, perspective, and framing, the total light depends only on the aperture diameter and shutter speed (as opposed to the f-ratio and shutter speed for exposure). Fully equivalent images on different formats will have the same brightness and be created with the same total amount of light. Thus, the same total amount of light on sensors with different areas will necessarily result in different exposures on different formats, and it is for this reason that exposure is a meaningless measure in cross-format comparisons.
Mathematically, we can express these four quantities rather simply:
• Exposure (photons / mm²) = Sensor Illuminance (photons / mm² / s) · Time (s)
• Brightness (photons / mm²) = Exposure (photons / mm²) · Amplification (unitless)
• Total Light (photons) = Exposure (photons / mm²) · Effective Sensor Area (mm²)
• Total Light Collected (electrons) = Total Light (photons) · QE (electrons / photon)
The ISO setting on the camera determines the amplification applied to the signal. For example, at ISO 1600, the gain is 1600 / 100 = 16x for a camera with a base ISO of 100. The amplification can either be analog and/or digital, with analog gain resulting in less read noise for some cameras (see here for an example with the Canon 5D), but at the risk of more blown highlights (discussed in more detail here). It is important to note that not all cameras apply the same amount of amplification for the same ISO. For example, f/5.6 1/100 ISO 400 on one camera may not show the same brightness as f/5.6 1/100 ISO 400 on another camera. The ISO standards allow the manufacturers a lot of latitude in their definition of ISO.
http://www.josephjamesphotography.com/equivalence/#noise
When people refer to noise in an image, what they mean is the density of the noise in an image (NSR -- noise-to-signal ratio) and is often represented as a percent. Often, we hear the term "SNR" (signal-to-noise ratio), which is the reciprocal of the NSR (SNR = 1 / NSR). For example, if the SNR = 5:1, then the NSR = 1/5 = 20%. However, since photo with high noise has a high NSR, and a photo with low noise has a low NSR, whereas it is exactly opposite for SNR, it is less confusing to think in terms of NSR than SNR.
There are two principle sources of noise in a photo: luminance noise and chroma (color) noise. Luminance noise is a function of the total amount of light falling on the sensor, and the efficiency of the sensor. The photon noise (often referred to as "shot" noise) is determined by how much light the sensor records. This, in turn, is determined by the total amount of light falling on the sensor (Total Light = Exposure x Effective Sensor Area) and the QE (Quantum Efficiency) of the sensor, which is the proportion of light falling on the sensor that is recorded (for example, a QE of 50% means half the light falling on the sensor is recorded). The other factor in luminance noise is the read noise, which is the additional noise added by the sensor and supporting hardware.
The chroma noise is the noise is a result of the color filters both blocking light which they are supposed to admit (AE -- Absorption Error), and admitting light they are supposed to block (TE -- Transmission Error). For example, if a red photon fails to make it through the red filter, this results in an Absorption Error, and if a red photon makes it through a green filter, this results in a Transmission Error.
The more transmissive the color filters (weaker color filter), the greater the QE, and thus the lesser the luminance noise, but the greater the chroma noise. The less transmissive the color filter (stronger color filter), the less the chroma noise, but lesser QE, and thus greater luminance noise. Different manufacturers choose a different balance in the transmissivity of their color filters resulting in a different balance between the luminance and chroma noise, which results is a different quality of noise in the photos, even if the quantity of noise remains the same.
The photon noise is the primary source of noise in the midtones and highlights of a photo. It is an inherent characteristic of incoherent light (the kind of light in almost all situations -- see the diagrams at the bottom of this page), and unavoidable -- one of those "Laws of Physics" things, as opposed to "an engineering challenge". Light has the properties of both a particle (photon) and wave, and the noise is measured in terms of its particle characteristics. The photons are collected and focused by the lens onto the sensor, where they are converted into electrons, and the signal is processed and recorded. The only role the sensor plays in the photon noise is what proportion of the photons falling on the sensor are converted into electrons (QE), since the electrons are the source of the electrical current that is processed by the hardware. The read noise, discussed in more detail further down, is how much noise is added when collecting and processing the signal produced by the photons.
The photon noise is proportional to the square root of the signal (N = sqrt S) so the photon noise density is inversely proportional to the square root of the signal (NSR = 1/ sqrt S). Thus, if we quadruple the signal (the total amount of light recorded), we halve the noise density.
There are, of course, other sources of noise, such as thermal noise, which plays a central role in long exposures, PRNU (Pixel Response Non-Uniformity) noise, which plays an important role in the highlights of the image, as well as other sources of noise. So, noise, is, of course, even more complicated than this essay makes it appear, and for some specific forms of photography (such as astrophotography) we may find that the noise is very different for equivalent images in some situations, much in the same way that corner sharpness is very different for equivalent images in some situations.
http://www.josephjamesphotography.com/equivalence/#fratio
It is instructive to understand why the same f-ratio results in the same exposure for the same scene and framing, regardless of the focal length or format. There are six factors that determine how much light falls on the sensor:
• The luminance of the scene
• The amount of the scene that is recorded
• The distance from the scene
• The diameter of the aperture
• Transmissivity of the lens elements
• The shutter speed
The amount of light from the scene depends on how wide we frame -- the wider we frame, the more light we will capture, since we are gathering light from a larger scene. If we assume a scene with the same average luminance, framing twice as wide, for example, will result in collecting light from four times as much area, and thus four times as much light reaching the aperture.
The amount of light from the scene reaching the aperture also depends on how far we are from the scene -- the further away we are, the less of that light that reaches the lens. For example, if we are twice as far away, only 1/4 as much light will fall on the lens in any given time interval. It should be noted that the inverse square law is exact only for point sources, and becomes "less exact" the wider we frame. The reason is that the distance from the camera to the center of the focal plane is not the same as the distance to the other portions of the frame. So, when we increase the distance from the scene, the distance from the other portions of the frame do not increase in the same proportion as the center. However, for most situations, the difference is trivial.
Furthermore, the amount of light from the scene falling on the aperture is proportional to the area of the aperture. For example, if we double the diameter of the aperture, the area will quadruple, so four times as much light can pass through in any given time interval. However, some light is lost as it travels through the lens which depends greatly on the lens (click here for some examples). Lastly, the amount of the light passing through the aperture onto the sensor is proportional to the exposure time. That is, double the shutter speed, and you halve the amount of light falling on the sensor.
Let's work a few examples, ignoring the effects of light lost from the elements of the lens, keeping in mind that the exposure is the total light per area falling on the sensor, not the total amount of light. In other words, we can express the exposure as the quotient of the total amount of light falling on the sensor and the area of the sensor.
Say a photographer takes a "properly exposed" pic of a subject 10 ft away at 50mm f/2 1/100 (aperture diameter = 50mm / 2 = 25mm). If they step back to 20 ft away and use 100mm f/2 1/100, the aperture diameter has doubled (100mm / 2 = 50mm) and the aperture area has quadrupled (area is proportional to the square of the diameter). However, the amount of light from the scene reaching the lens is 1/4 as much since they're twice as far away. Since the aperture area is four times as much, it exactly compensates, and the same amount of light will pass through the aperture onto the sensor, and, since the sensor has not changed size, the exposure will also be the same.
Alternatively, let's say they don't step back, but instead remained at 10 ft and shot at 100mm f/2 1/100. At 100mm, the framing will be twice as tight, and thus record only 1/4 the light of the scene as they would at 50mm (assuming, of course, a uniformly lit scene). Thus, despite the fact that 1/4 as much light is reaching the lens, since the aperture area is four times as great, it exactly compensates once again. An excellent video on this can be seen here.
The above two examples demonstrate how the same f-ratio and shutter speed results in the same total light and exposure for a given scene and format regardless of focal length on the same format. However, for different formats, the same exposure does not result in the same total amount of light falling on the sensor.
Let's now consider a photographer using using 50mm f/2 1/100 on 4/3 and another photographer with FF shooting the same scene from the same distance at 100mm f/4 1/100. In both cases, the framing is the same (ignoring the minor differences in aspect ratio between the systems, which amounts to a mere 4% difference), the aperture diameters are the same (50mm / 2 = 100mm / 4 = 25mm), the distances from the scene are the same, and the shutter speeds are the same. Thus, the same amount of light will pass through the apertures onto the sensors.
On the other hand, if the FF photographer shot the same scene from the same position at 100mm f/2 1/100, the aperture diameter would now be 100mm / 2 = 50mm as opposed to 25mm. Since the aperture diameter is twice the size, the aperture area is four times as large, and four times as much light will fall on the sensor. But, since the sensor has four times the area, the density of the light on the sensor would be the same as the 4/3 sensor, so the exposures would be the same.
Thus, for a given scene, perspective, framing, aperture diameter, and shutters speed, the same amount of light will pass through the aperture onto the sensor for all systems. However, there are some exceptions. Some light is absorbed and/or scattered by the glass elements in the lens, and is quantified by noting the difference between the T-stop and the F-stop (f-ratio). Click here to see some examples of this differential for a few lenses for the Nikon system. It's helpful to note that a transmission of 79% represents a 1/3 stop loss of light, 63% represents a 2/3 stop loss of light, and 50% represents a one stop loss of light.
I hope these quotes make things more clear.
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:-\ :'(
joseph james said:
If you had indeed read this thread, you would know that I have already read the entire page on equivalence. You would also know that I do not dispute its relevance in AoV/entrance pupil normalized situations.
However, I do believe there is a specific real-world circumstance where such a comparison is not possible. I regularly photograph birds and wildlife, which are generally considered reach-limited, or focal-length limited, circumstances. I own the EF 600mm f/4 L IS II and the EF 1.4x TC III. If I were to head out and photograph shorebirds, for example, I will usually find a spot near enough to the birds to get good shots, without being too near that I scare them all off.
In this situation, I could bring along a 5D III and a 7D, and swap them out for each photograph on my 840mm f/5.6 lens with the same shutter speed and ISO setting on both. The entrance pupil will remain the same 150mm regardless of the camera. I either have the option of taking a less detailed photo with less noise (5D III), or a more detailed photo with more noise (7D). If I aim to achieve as much parity between the results as possible, I have the option of cropping the 5D III image to the same AoV as the 7D and scaling the 7D image to the same dimensions as the 5D III. The result of that normalization is a reduction in noise in the 7D image such that it is roughly the same as the 5D III image, while concurrently increasing detail and sharpness. The exposure for both images would be the same.
As a side note, one could also attach a 2x teleconverter to the lens when using the 5D III. That would largely normalize effective focal length (1200mm f/8 for the FF, 1344mm f/5.6 for the APS-C), maintain an identical entrance pupil, and your equivalence framework would then apply. Your article does state that you ignored any additional factors that might affect IQ, however the addition of a 2x TC would very likely reduce the MTF of the lens and exacerbate optical aberrations to some degree (despite equivalence, although frankly I'd take the 5D III+600+2x over the 7D +600+1.4x if I had the option, if for nothing other than the superior AF and chance to pack more pixels on subject with a bit of extra sneaky technique to get closer to my shorebird subjects.)
This is the world I usually live in when it comes to my photography. It is not the same situation that your article discusses, which quite clearly states that AoV must be the same. Nevertheless, it is a valid scenario that photographers DO encounter in the real world. Is it, then, invalid to compare cameras in such a scenario? Perhaps we are not talking about "equivalence" as you have defined it, but does that truly invalidate the comparison?
Your framework works beautifully to compare something like landscape photography or portraiture. As a landscape photographer myself, I have eagerly been awaiting Canon's "Big Megapixel" body, however in the mean time I know that the 5D III will still produce better results with less noise and better dynamic range than my 7D, which with its crop sensor will never really compare. I just stand by my argument that there are different frames of reference that change how one might need to compare two cameras, and that it is not always possible to normalize AoV and entrance pupil.
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jrista said:
In the case you are magnification or focal length limited, in terms of IQ, you are best served with the system that puts the greatest number of pixels on the subject and/or has the most efficient sensor.
Perhaps this section of the essay addresses that concern:
http://www.josephjamesphotography.com/equivalence/index.htm#purpose
We can compare systems in many different ways. The five parameters of Equivalence are simply guidelines to comparing systems on the basis of the most similar visual properties of the final photo, and are certainly not a mandate that systems must be compared in such a fashion. Therefore, it is important to specify the purpose of the comparison, and then not artificially handicap one or the other system with the conditions of the comparison. In addition, it is important to interpret the results of the comparison in the context of the circumstances where the conditions of the comparison are valid.
The point of photography is making photos. As such, one doesn't choose the particular system to get images which are equivalent to another system. A person chooses a particular system for the best balance of the factors that matter to the them, such as price, size, weight, IQ, DOF range, available lenses, and/or operation. By understanding which settings on which system create equivalent images, the difference in their capabilities is more easily understood.
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joseph james said:
You are implying that the guy who cannot afford to spend eleven grand on a 500mm lens or thirteen grand on a 600mm lens is "artificially handicapping" the comparison when all he can use is his 400mm lens on both a FF and APS-C camera. I would call it a literal handicap, imposed not artificially, but by a real-world lack of funds.
Your last paragraph there is a good one, and is entirely relevant in the case where someone has not yet already bought into a system, and has the option of determining up front which setup will best service their needs within their budget. I guess I am more concerned with the alternative case, where someone has already bought into a system, and is considering the most cost effective upgrade that will improve their results. If money is an object, reach is critical, and one cannot afford to buy the best telephoto lenses available for the system they already own, an upgrade from a 7D to a 7D II (assuming an increase in pixels and upgrade to AF performance and accuracy), for example, is probably better than an upgrade from a 7D to a 5D III. I am not sure your equivalence framework would help such an individual to make that determination.
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