If you are here to understand (why) equivalence (is wrong) then read this: https://dtmateojr.wordpress.com/2014/09/28/debunking-equivalence/
Before you start reading this post, I highly recommend that you read my previous post because it explains the very basic concept of f-stop. Failure to understand this basic concept is what leads to nonsense.
It turned out that the myth I mentioned in the last paragraph of my previous post became a hot topic in one of the forums. This guy who owns a full frame, a D800 to be more specific, started arguing about the superiority of his camera vs the APS-C sensors. What’s even funnier is that this was a Pentax forum. You know that brand that doesn’t have a full frame camera yet (cough! cough!). I don’t even know why he was lurking in there.
Anyway, as the myth goes, a full frame sensor has less noise compared to a smaller sensor because it can gather more light. The basic premise is that all things being equal (lens, sensor, processor, incident light), the size of the sensor is the single most significant factor that affects noise because of its superior light gathering capacity. As I have mentioned in my previous post, this is not true. In fact, in reality, a full frame sensor gathers less light than a smaller sensor and I will expain this later.
First, let’s see what this myth really is in greater detail. Here’s the link that is supposedly the authority of this myth: http://www.josephjamesphotography.com/equivalence/#aperture
I would like to quote the following:
The concepts of, and connections between, total light, DOF, and noise, are much more easily understood in terms of aperture rather than f-ratio, especially when comparing different formats. While the same f-ratio will result in the same exposure (where exposure is the density of light that falls on the sensor — photons / mm²), regardless of the format, the aperture diameter, together with the shutter speed, determines the total amount of light that falls on the sensor, where the same total amount of light falling on the sensor results in the same noise in the photo, for equally efficient sensors.
I can already see where this is going. The argument is that since a larger sensor requires a longer focal length to cover the same FoV then it follows that for the same f-stop, it will have a larger aperture and therefore allow more light in. ROFL! Seriously?! You just can’t make that up you know 🙂
Here’s more:
” 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 · Effective Sensor Area)”.
You can read that from here: http://www.josephjamesphotography.com/equivalence/index.htm#noise
The first statement totally ignores the fact that distance plays a very important role in the intensity of light. Look at our sun for Pete’s sake. There are more stars way larger than our sun and yet it’s much much brighter than any other light source above our atmosphere because of distance. Even if you go out at night in the open (large aperture), it’s still way brighter inside a room with a single window (smaller aperture) during daytime. I have not even started arguing photographically yet.
Allow me to debunk this myth using a very simple example. Consider the simplest lens of all: a pinhole. Put a pinhole in front of a full frame and a crop sensor. Do you think that the pinhole is any brighter because it’s in front of a full frame? Of course not. Now here’s the problem: if the pinhole is too close to the sensor, it won’t illuminate the whole sensor. What do you do? You move the pinhole away from the sensor of course. Now anyone with half a brain will quickly realize that the larger the sensor the farther the pinhole should be. Refer to my crappy illustration below:
The distance D1 and D2 of the pinholes are proportional to the size of the sensor that needs to be illuminated below. Can you see how this relates to our sun and stars argument above? The farther pinhole is giving lesser light to the the sensor. Unfair to the full frame isn’t it? How do we make this fair to both sensors? Make the farther pinhole bigger. Obvious isn’t it?
How does this relate to a real lens? The size of the pinhole is your aperture, the distance from the sensor is your focal length and their ratio is your f-stop. That’s why a f-stop is a f-stop and gathers the same amount of light for any sensor size. Same amount of light means same SNR for the same type (not size) of sensor!
Now to counter the other statement on sensor area, refer to the illustration below. If we are to use the same lens at the same aperture then we will arrive at something like this for different sensor sizes:
So now we have the same distance and the same aperture therefore the same amount of light entering the sensor chamber. Does this mean the larger sensor is capturing more light? Of course not! It’s not like the individual sensels can share their signals with other sensels. Whatever light falls on each sensel is as is. The SNR for each sensel is (more or less) constant for a given incident light.
Look closely at the illustration above again because now I will go further into saying that in fact, the larger sensor is capturing less light in this situation. Now why is that? Because lenses aren’t perfect. Light diminishes as you move away from the center of the lens. You want proof? Wide open, the same lens will produce vignettes on a full frame while it looks perfect on a crop sensor. A vignette can be as bad as two stops down at the corners. That’s four times less light!!! If truly a full frame captures more light then there won’t be vignetting.
Consider this thought experiment:
Supposing that you have a D800 sensor under any lens. Keep the same lens but now cover one half of the sensor with a completely opaque material therefore exposing only half of it. Do not cover the lens; just the sensor. If the myth is true, then it follows that this sensor half (a crop) will now gather less light and therefore produce more noise. Now uncover this half and cover the other half. Again, if the myth is true then this second half would also gather less light and, just like the other half, will produce more noise. It follows that if you fully uncover the full frame sensor, it’s really just a combination of two more noisy halves! The myth is telling us that two noisy half-sensors will produce one clean full frame sensor?! This is absurd!
This thought experiment is not just a thought experiment. It actually happens every time you click your shutter. At fast shutter speeds, the shutter curtain does not actually fully open and close. Instead, the shutter curtain behaves like a very small slit that glides over the sensor. Therefore, the sensor is not exposed as a whole at the same time but in chunks defined by the curtain slit size. This slit is way smaller than half a sensor. Since this slit exposes only a part of the sensor at any given time, your full frame is really acting like a combination of multiple smaller sensors! If the myth is true then a full frame sensor is really just a combination of multiple very small noisy sensors!!!
We can continue this experiment by further subdividing the sensor until we come to a point where all we have left is just one sensel. If the myth holds then this sensel will be hopelessly noisy. It follows that a full frame sensor is composed of individual hopelessly noisy sensels! That is insane! That is beyond reason! This is just wrong, people!
The only conclusion is that sensor size does NOT matter! Note that half of a full frame is your APS-C sensor. Therefore full frame and APS-C have the same light gathering capacity and therefore exhibit the same noise profile!
What affects light gathering capacity (among other factors) is the size of the sensels. The K5/D7000 and D800 have practically the same sensel pitch and therefore have the same light gathering capacity. Again, it’s NOT the sensor size but SENSEL size! That is why the SNR of the D7000 and K5 are the same as the D800. That is why the SNR of the D700 and D4 are better than the D800. (https://dtmateojr.wordpress.com/2014/04/21/rain-can-teach-us-photography/)
Update:
Note: re-editing since my update disappeared mysteriously when I accessed it using the wordpress iPad app 😦
With the recent discussions that happened in the comments below I could not help but to conduct my own real experiment on this. According to the myth, as long as the same shutter speed and same physical aperture are used, the resulting images should have the same amount of noise and so I made an experiment.
Here are the parameters:
35mm, f5.6, ISO 200 VS 100mm, f16, ISO 1600
Why those settings? Because they produce the same aperture sizes:
35mm/5.6 = 6.25mm
100mm/16 = 6.25mm
Since f16 is 3 stops slower than f5.6, I boosted the brightness in-camera by using ISO 1600.
I used the low-light monster Nikon D700. My lens of choice was the Tamron 24-135/4-5.6. I chose this lens because of the following reasons:
1. It covers the focal range I required for the experiment without hitting the extreme zoom limits of the lens.
2. It has focal length markings at 24, 35, 50, 70, 100 and 135.
3. It locks to infinity focus.
Item #3 is very important because my subject and light source are very far. I used the clear night sky as my subject and my light source. The sky is very wide and uniformly lit that differences in AoV are immaterial. I shot the scene approximately 30 minutes after sunset.
Without further ado, let me present to you the results:
Above photo was taken at 35mm, f5.6, 30s, ISO 200.
Above photos was taken at 100mm, f16, 30s, ISO 1600.
As you can see, the first shot is very clean. The second shot has very visible luminance and chroma noise. Admittedly, I expected a lot worse from the second shot but I guess this is testament to how good the D700 is in low light situations. The blurry white lines are stars. The long exposure has created star trails. As expected, the longer focal length has produced longer trails.
I think that at this point, this myth has been truly busted. As always, comments are more than welcome.
Demo Mateo 
Totally agreed on this.
I think you need to go back and read the full article you have linked to. I haven’t had time to read it carefully enough to follow all of the detail to draw conclusions as to the accuracy of the article but it is definitely not the simplistic argument that you seem toy have assumed.
Which part of their article are you referring to?
I went back and could not find anything new in there. He keeps talking about equivalence where two photos are not “the same” unless they have the same AoV and DoF. Why must AoV and DoF be equivalent when measuring noise?
You suggested that he was ignoring distance, which it didn’t, it was included and explained later and also included a much deeper explanation of the sources of noise http://www.josephjamesphotography.com/equivalence/#efficiency
If you are just looking at noise then there is a definite benefit to a larger sensor, I may not fully understand the reasons but I can see that it is there if you look at samples.
The AoV and DoF need to be the same if you are trying to compare image quality (which is what his article is about, noise is only one part of the whole picture). Incidentally his overall conclusion is similar to yours in many ways – there is no best format, there are pros and cons to all of them. There will only be a best format for a particular set of circumstances.
If you read that block I quoted in my post, there was no mention of distance in there. It’s all about aperture which is kinda short-sighted (pun intended). Efficiency is affected by the sensor itself but NOT the size of the sensor. We expect the D7000 and D800 to have the same efficiency because they are exactly the same sensor. Size, in that case does not matter.
AoV and DoF matters in photography but has no effect on noise measurements assuming that the light source or subject is evenly lit. I still can’t see how aperture alone affects noise. F-stop affects noise but we know that this incorporates distance.
I have updated the post with more examples.
Your aperture diagram is flawed because you have confused intensity and amount of light – the intensity of light on the full frame sensor will be less because of the distance but the amount of light captured will be the same because the full frame sensor can capture the light over a larger area. The amount of light does not diminish with distance, it is just spread over a larger area.
Assuming that the sensors have the same number of pixels and the rest of the sensor systems have the same efficiency, they will have captured the same signal even though the full frame sensor has been technically underexposed (less intensity). If you adjust the size of the aperture to give the same f-number you will let in more light, increase the intensity and the sensor will capture more light.
The same is true of your diagram with the two sensors at the same intensity, the larger sensels capture more light because more photons hit them, therefore more electrons produced. The images will look the same brightness because it is processed differently in the analogue to digital conversion.
Isn’t this the whole argument you are using in your Noise performance comparisons article, to quote:-
“It’s common knowledge that one of the biggest factors affecting high ISO noise performance is the size of the sensels (sensor pixels). Ultimately it’s the individual sensels that capture light. Bigger sensels mean higher signal-to-noise ratio (SNR) assuming of course that everything else is equal.”
Yes, bigger sensels mean higher SNR. The myth says something more though. It says that everything else being equal (i.e. D800 and D7000 sensors) the bigger sensor size guarantees more light gathering capacity. That’s wrong.
Here’s another quote from the same author: ” 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 · Effective Sensor Area)”. You can read that from here: http://www.josephjamesphotography.com/equivalence/index.htm#noise
So according to him noise is affected not just be light intensity (f-stop and shutter speed) but by sensor area as well. If you read my post update, my explanation on dividing the sensor debunks that statement. Sensor area does NOT affect noise. Sensel size and exposure do.
And if you read the rest of the article you will see that photon noise is not a major contributor to the noise we see in the shadow area of our photos, what we are predominantly seeing there is read noise caused from the system. This becomes more obvious with lower signal.
If you also read the original experiment to which the sensor area quote refers (cropping the sensor to produce the same image taken with different lenses) it makes sense. You are arguing a different point to what he is discussing.
There are different types of noises inherent to the imaging system but shot noise (photon noise) is what’s affected by incident light.
Why would you crop? Cropping artificially magnifies everything including noise. If you crop then we are back to my example on dividing the sensor. Crop indefinitely and you arrive at a single pixel which according to this myth will have no signal at all but just pure noise. His logic is flawed.
As I said in my original comment, I think you should read the full article carefully. You have assumed because an idiot in a forum has misunderstood the article that the article is backing up the idiot. Reading through the whole article shows that the writer is not pushing the myth that you are arguing against. The article covers a whole range of issues to do with image quality in, by my reading, a balanced and logical way. It actually has detailed information that backs a good deal of what you have argued here but pulling parts of the article and placing them out of context can certainly give the opposite impression.
With reference to the cropping, I’m not sure why that was included in the article. The only idea I have is probably to dispel the myth of shooting with a full frame sensor and cropping down because you don’t have a long enough lens. The article points out that this doesn’t give you an advantage, you agreed cropping makes the noise more noticeable. Using a single pixel analogy as a reductio ad absurdum to prove that the idea is flawed argues against your agreement of the fact. At this level you have no noise since noise is a variation of signal and you can’t have a variation with one data point.
I would have to agree with your comments except the single pixel analogy. That analogy still holds. In fact it proves that noise (shot noise, read noise, etc…) happens at the pixel level. That is indeed a fact.
I’ve enjoyed the ‘discovery’ of this seminal equivalence article on this blog by you two and have welcomed the discussion. Andrew, you said “idiot in a forum who misunderstood the article”, and you’re probably referring to me – however, you’re actually referring to Mr Mateo’s characterization of me, in which he was doing the same things with my responses as he was with Joseph James’ equivalence article – not fully reading what was being written, or the significance of it. I fully understand the article and all it’s implications, have for years, and I originally brought it to Mr Mateo’s attention to highlight how a statement he made to the effect of “an F stop is an F stop on any format” is true, but highly misleading and ‘missing the point’ in the same way “focal length is focal length” on any format. (see: http://www.josephjamesphotography.com/equivalence/#1)
‘Equivalence’ provides the necessary background framework for understanding why keeping the F stop and the AOV the same for any given shot is going to provide more light on the sensor with a larger format – because the linear aperture (entrance pupil) will have to be larger at the FL that would provide the same AOV. For example, FF: 70mm / 2.8 == 25mm linear aperture (entrance pupil)
aps-c: 46mm / 2.8 == 16mm linear aperture. Same AOV, same shutter speed allowed, would result in about one stop less noise on the larger sensor (photon shot noise, assuming sensor QE an read noise of similar generations,) directly determined by the larger linear aperture. That’s what I was arguing (in part,) and Mr Mateo seemed to think that everything I was writing was a defense of full frame, or a purchase justification, or something like that. (wrapped up here: http://www.pentaxforums.com/forums/172-pentax-k-3/253508-please-give-me-reason-buy-pentax-k3-not-nikon-d3-6.html#post2732906). Not in any way the case – just like the Equivalence article shouldn’t be characterized that way.
(Now, we’ll see if this gets posted here! 🙂 )
I welcome any sane discussion to my blog. Welcome btw.
As I have stated numerous times in my arguments with you, aperture alone does not guarantee that the same incident light is falling on the sensor. Distance (i.e. focal length) is another factor. I have demonstrated this in my pinhole example and sun-stars analogy. You keep bringing this up as if you have not read my post.
May I request that you perform this quick experiment for me: Shoot at 35mm, f5.6, ISO 100 on a crop sensor. Take note of the shutter speed. Now on a full frame shoot at 70mm, f11, ISO 100 at the same shutter speed. According to your myth, since both settings result in the same aperture diameter of approx 6.25mm, both will have the same amount of noise. Boost the full frame shot in photoshop by +2 or shoot the full frame at ISO 400. Use similar sensors like a D7000 or K5 and a D800.
I don’t have cameras available to run the test but I’m very interested in the results. Good luck!
Sorry for the idiot comment, without having read the forum. I had no idea what was stated and I could easily see how the original article could be misquoted or misinterpreted. It’s too easy to make assumptions online so I hope there is no offence.
OK, I should have said you can’t measure noise on a single pixel because you have no standard to measure against.
You could measure SNR for different incident light intensities. This would then naturally lead to my follow up post on noise performance which compares sensors with different sensel sizes.
I can’t reply 2 levels deep here, there’s no ‘reply’ button under your reply, so I’m posting at the bottom. You wrote “May I request that you perform this quick experiment for me: Shoot at 35mm, f5.6, ISO 100 on a crop sensor. Take note of the shutter speed. Now on a full frame shoot at 70mm, f11, ISO 100 at the same shutter speed.”
That’s not an equivalent image in FOV, though – for one thing you’d be comparing an image with different subject enlargement and then comparing noise if you were to try to match the FOV later with a crop.
Equivalent images would be 35mm f/5.6 on aps-c, and about 52mm, f/8.5 to get the same linear aperture, and the same noise, DOF and FOV. If you were to keep the F-stop the same there (35mm f/5.6 and 52mm f/5.6), to get the same shutter speed and keep the same FOV (which is what most photographers would strive to do – frame the subject the same and worry about adequate shutter speed,) then you can take advantage of the larger sensor gathering more total light (exposure x sensor area) and you get better noise performance – but you’d have to accept about a stop less DOF.
This is what I was originally trying to show when I was refuting your ‘F-stop is F-stop on any format’ comment, and that’s why I linked Joseph James’ equivalence article. It’s held up, by the way, and aside from some language quibbles I have never heard anyone, any physicist or engineer try to refute it’s validity. There’s nothing there to debunk, and I think if you read it carefully you’d find that you actually agree with everything there. I will admit that when I first encountered it years ago I went through a “that’s not right” phase, followed quickly by an “OK its right but it doesn’t describe real-world situations because read noise and sensor QE vary too much, and the constant ‘keeping things equivalent’ just bugs me.” Later I started to really appreciate finding it. 🙂
My mistake on the suggested experiment. Change the full frame settings as per your suggestion. 52mm at f8 then bump the ISO to 200 instead of 400.
Better yet, let’s create more noise;
aps-c: 35mm/5.6/800
ff: 52/8/1600
FF will have a “slight advantage” because I don’t know how you can get f8.3.
Anyway I expect the crop sensor will leave the FF in the dust of it’s own noise 🙂
You’d have to lend me your K5, because my two aps-c cameras (D90 and K20D) don’t have the same read noise/QE as my D800 – my D800 would probably look much better in that comparison, depending on how dominant read noise was there (if it was very low-light.) K5 (or D7000) & D800 would show about exactly the same noise there, assuming no in-camera raw NR or ‘smoothing’ was going on with either. If you shot both at f/5.6 though, ISO 800, the D800 would show about a stop better noise performance, and about a stop less DOF.
I’ll test later with one camera and zoom lens. I’ll use the clear sky as the light source so AoV should not matter. Just have to wait for sunset so that I can introduce noise.
Well you know what there is an easier way if you can find a uniformly lit white wall. It’s just uniform white so AoV should not matter. A huge softbox would work as well because that allows you to control the light (we want it to be dim enough to introduce noise). With that as your target you only need one camera and one zoom lens. We should be able to prove if 35/f5.6/800 is indeed as noisy as 70/f11/1600 at the same shutter speed.
Your example there (35mm f/5.6 vs. 100mm f/16) is showing more noise with the 100mm f/16 shot because the *reduced AOV* is bringing less total light projected onto the sensor. In your first FF/apsc example (aps-c: 35mm/5.6 FF: 52mm f/8), the AOV was the *same*, and the entrance pupil is the same, so the total noise would be the same (assuming sensors with equal pixel QE/read noise.)
This is one reason why understanding the equivalence framework is important as a starting point – if you don’t keep AOV/FOV the same, you’re not starting with the same amount of light being projected in the first place. Not understanding this leads to a lot of newbies confusion about full frame, they think using the same lenses on both formats automatically brings some advantages that don’t really appear (like DOF control and noise.) You have to keep the AOV the same, and use the same F-stops as before to see the greater DOF control and noise advantage. And when you do that, the thing that’s varying is the virtual aperture – the entrance pupil, as I was saying.
Another benefit of using the same FOV/F-stop in general is that it describes what photographers actually *do* – their priorities usually start with keeping the framing and shutter speed the same. You don’t shoot a couple at a wedding with FF showing a bunch of extra stuff in the frame, you frame it the same as you would with aps-c – for example, a tight head-torso shot of the couple. You’d be framing the same way you would with aps-c, but using 1.5x longer FL with FF, and if you use the same F-stop to get the same shutter speeds, your entrance pupil is larger and you get your stop less noise and a stop less DOF. This is why, as they say, ‘an F-stop is an F-stop on any format’ is misleading, to say the least. Please refer to this when you get a chance: http://www.josephjamesphotography.com/equivalence/#1
LOL! Please read my explanation on dividing the sensor AGAIN. Dividing the sensor crops it and therefore changes the AoV. As I have proven in that example, AoV does NOT change the amount of light hitting the sensels. I really feel like talking to a coconut here.
One last few words with you.
If we did follow exactly your methodology I doubt if we can see any difference in noise. It does not prove anything. It will only show that one stop difference is immaterial. I could barely see any difference in three stops immediately after sunset. I had to wait until the exposure drops to 30 seconds to notice any difference. One stop will not visibly prove anything. And that’s why I chose extreme settings of three stops difference.
Anyway, people have been ignoring your nonsense in forums. I have been ignoring your nonsense in forums. You leave me no choice but to ignore you here as well. I hope you enjoyed your less than 24 hours of visit to my blog.
Good bye.
‘normhead’ has ignored me, and you, That makes two of you 😉 Id suggest following his posts over there on the subject to see what type of company you’re in first.
Why are you getting so angry and defensive about this? You don’t need to resort to name-calling, if your argument is strong it stands on it’s own without needing a boost of abuse.
For one thing, in that sensor-covering example if you’re projecting the original image on the half that’s left, you’re cutting the sensor area in half. Remember, the image noise is determined by the total light gathering, which is exposure times sensor area. Your ‘covering half the sensor area example’ would be something closer to taking two micro-4/3 shots and stitching them – The resulting image would be noisier than an FF image taken with the equivalent AOV, with the image projected on the whole sensor. If you’re simply cutting the FF-projected image in half (if that’s what you meant to show,) it’s just exactly the same as cropping half the image away in post and then showing the half that’s left un-enlarged – that half wouldn’t be any noisier. I’m not sure what you think that example proves either way, or how it possibly refutes anything I’ve said, either way.
You don’t need to take my word for this if you don’t want to 🙂 Email Mr James, Bob Newman (sensorsgen,) there are a host of other folks who can explain it very well, and there are other resources available.
But if you want me to continue contributing to your blog, please keep it civil.
I’ll approve this not because it makes sense but because it is so incoherent it’s too funny not to share 🙂
Just for the record, you were the first to throw the first punch in the forums by insulting my degree. Don’t be surprised if people ignore you.
That’s the last word … for real this time.
I’ve read through this post a few times and I am not quite sure I understand what you are saying. I will comment on your two hand drawn illustrations.
The first. When you say ‘The farther pinhole is giving lesser light to the sensor’ we need to keep in mind that a pinhole’s size is supposed to be infinitesimally small and N, normally approximated by the f-stop, is instead equal to 1/[2sin(theta)] – theta representing half the angle of view as shown here http://www.dpreview.com/galleries/8617333752/photos/2748690/source-to-scene-to-photons-and-electrons-captured-2.
Your first illustration above is equivalent to adding a sensor closer to the lens along the AOV lines in my diagram, at a focal length equal to f/1.5 – and pretending that the lens is in fact a pinhole. If you place an APS-C sized sensor there, the earlier FF sensor at focal length f and the new APS-C sensor at f/1.5 will see exactly the same ‘light’ (number of photons). So I am not sure what you mean by ‘The farther pinhole is giving lesser light to the sensor’. The pinhole delivers exactly the same ‘light’ (number of photons) to both sensors in this ideal situation.
In your second hand drawn illustration you have swapped a lens for the pinhole. You say that with the same lens, ‘distance and the same aperture therefore the same amount of light entering the sensor chamber.’ If by that you mean that the same number of photons (light if you prefer) are contained within the circle you have indicated as the ‘Lens Image Circ’, I agree. To simplify things let’s also assume that the photons are equally distributed within the ‘Circ’.
Therefore I am not sure I understand what you mean right after that when you ask ‘Does this mean the larger sensor is capturing more light?’ and respond in the negative. Of course the larger sensor is capturing more photons (light ) than the smaller one because if we sense a smaller and smaller area of the ‘Lens Image Circ’ we are going to count fewer and fewer photons (or light).
Cheers,
Jack
Good question. A very important thing to note is that light intensity drops as the square of the distance. This is basic physics. In photography, this distance is your focal length. Longer focal length means lesser light intensity hitting the sensor.
A f/2.8 lens may have more light compared to f/2 depending on focal length and aperture but the intensity of light being “produced” by the f/2.8 lens will always be lesser. Again, that is basic photography. That’s why rules such as sunny f/16 are applicable to any format. That rule isn’t just for full frame. It applies to huge 8×10 view cameras as well and to point-and-shoot.
‘A f/2.8 lens may have more light compared to f/2 depending on focal length and aperture but the intensity of light being “produced” by the f/2.8 lens will always be lesser. ‘
I believe a lot of the misunderstandings in this page come from not specifying clearly what situations are being discussed. Your statement above is both true and untrue depending on the setup so imho it is not helpful in furthering the cause implicit in the title..
Hint: f-stop is the ratio of focal length to aperture diameter.
To answer your last paragraph: Yes, smaller sensor will capture less photons. That is why the are called “crop sensors”. They crop, the eliminate, throw away, half of a full frame image circle…half of the total light entering the chamber. That does not affect the concept of photographic exposure for the crop sensor at all. It still managed to capture all the light it needed for a particular AoV.
A simpler proof here: https://dtmateojr.wordpress.com/2014/06/10/debunking-the-myth-of-full-frame-superiority-part-2/
You may also want to review the results of the experiment I conducted at the end of the article. Notice that although the same total amount of light enters the sensor chamber by virtue of the same shutter speed and aperture (not f-stop), the shot with a longer focal length has exhibited more visible noise.
I think your equivalence calculation is not correct. This is what I have observed in practice
On My 4/3 camera for a situation if my exposure reads
25mm, 1/125, f2.8, ISO 6400
on FF it reads
50mm, 1/125, f2.8, ISO 1600
The DOF would be different if you are shooting from the same position but I noticed that the FF ISO is always 4 times lower for the same shutter speed and aperture.
This seems to be like a triangle – you cannot adjust one without the other
You can get images of similar total light with
25mm, 1/125, f2.8, ISO 6400
25mm , 1/125, f5.6, ISO 1600
You can try this easily. Download the light meter app on iphone – You can compare the settings with m43 for the same scene.
Not sure if I understood what you are trying to say. In your first example, the exposures are the same but the brightness of the images are very different. Those are not equivalent images either. What light meter is giving you different readings for different sensor formats?
on the last comparison, It should be 50mm for FF
i.e. 50mm, 1/125, f5.6, ISO 6400
Since we adjusted aperture which brings in less light ISO can be the same across both
Check my previous reply.
Hello. Surely the only advantage FF can offer, once one understands equivalence of focal length, f-stop and ISO, is that given the same mega pixel count, a FF sensor can have larger arguably higher quality sensels (is that the right word). With latest technology, this is becoming less and less true else fuelling increases in megapixel counts. Yes the total light argument makes no sense. If i fill a bucket with 1 inch depth of water and same with an olympic size swimming pool, I still have the same depth of water in each. Yes the pool has more total water but so what. Are some people arguing that the performance of a sensel in one part of a sensor enhanced by that of another on a different area of the sensor??!!
Exactly!!! Finally, somebody who understands!
Too bad his understanding is wrong 🙂
More water = more signal, more signal = higher signal to noise ratio = less perceived noise.
Maybe this example image makes it easy to understand: https://aberration43mm.files.wordpress.com/2015/02/compared.png
The two images are taken with the same camera, but taken on different lenses. Thus they were cropped by different amount, thus different number of pixels are used to create each image.
I hope both he and you do the same test to see if the results match or not.
One can also easily use the DPReview’s new studio comparison tool – use Nikons D7000 and D800, ISO 25600, indoor light, and press the “comp” button on the top to compare same sized images.
Ignored. You need a bigger cluebat.
Full frame evangelists need to cut two words from their vocabulary: total and light. But only if used together.
if you want the math you may want to read this as well:
It’s good to read this simple and logical explanation. Looking at photography forums, I often feel that half the world has gone raving mad!
Hi Richard,
I see that you got here from a link in dpreview. This equivalence nonsense is still alive in that part of the world eh? 😊
I have written so much about the pseudoscience these people are spreading. Here are some other articles that might help:
If they are not scared of the math then this would explain everything:
And of course, film has proven that nothing has changed in photographic exposure for decades:
Cheers!
Thanks Mateo
I read those with interest (and a bemused grin when reading the comments!)
It really is a simple concept to understand. In fact it is totally intuitive.
For a given fov and aperture, the image area, the aperture area and the focal length will have a fixed proportional relationship.
Keep up the struggle 🙂
I meant f stop not aperture.
If you take a FF camera and an m4/3 camera with the exact same sensor technology and sensel size. Put as an example a 300/f2.8 lens on both and take a picture of the same object at the same distance. Isn´t most of the total light on the FF sensor used to create the parts of the picture that is missing from the m4/3 picture? Stupid question, perhaps?
Not sure about what you are asking here. For starters, the m43 will only capture 1/4 of the object. That’s 1/4 the total light.
But that’s not the point of photographic exposure. Photography is not about total light. It’s about total light over total area. In essence, a m43 only requires 1/4 of the total light to record the same exposure.
As a comparison, a 150hp car can beat a 300hp car if the former is much lighter/smaller. It’s all about power to weight ratio.
Photographic exposure is all about light to sensor area ratio. Same sensor technology produces the same SNR. Have a look at SNR data for D800 and D7000 in dxomark.
Well this was the answer I hoped for, and tried to explain in my question. English isn´t my first language, so what sounds like a good question in my language is hard to a explain in a foreingn language.
“For starters, the m43 will only capture 1/4 of the object. That’s 1/4 the total light.
But that’s not the point of photographic exposure. Photography is not about total light. It’s about total light over total area. In essence, a m43 only requires 1/4 of the total light to record the same exposure.”