[Ian Stuart Forsyth]

I must admit that I indeed didn't watch the video.

I trusted that despite their reputation for being pretty terrible when it comes to comparisons and technical explanations (they seriously argue that the Profoto B10 is as powerful as a Godox AD400 which anyone with a brain and access to both devices could convince themselves is a boatload of nonsense) that they would manage to regurgitate some facts about sensor sizes that they read somewhere.

Alas, Lee managed in his typical style in which confidence trumps knowledge to misinform his viewers.


If you keep everything constant (except sensor size, presumably) then you get two wildly different images which cannot be reasonably compared to each other.

At the 6:56 minute mark, Lee attempts to start the one experiment that makes sense:

• have both cameras at the same position (otherwise perspective would change and with it DOF).
• compensate for the larger sensor by using a longer focal length (otherwise the FOV would change and with it the light gathered, DOF upon cropping, etc.)
• keep the shutter speed the same.

His result is that the larger sensor produces an image with a shallower DOF.

This apparently contradicts the title of the video but he offers an explanation: The longer lens is responsible for creating the shallower DOF.

This explanation is wrong. While it is true that using a longer focal length reduces DOF when not changing the sensor size at the same time, the reason for that latter phenomenon is not the longer focal length as such, but keeping the f-stop constant.

You'll only see the reduction in DOF when going from a shorter to a longer focal length, if you keep the f-ratio constant, say, use f/2.8 for both a 50mm and 100mm lens.

You could achieve exactly the same DOF if you stopped down the 100mm to f/5.6. Why f/5.6 exactly? Well, DOF is determined by the aperture diameter. In the case of a 50mm lens at f/2.8, the aperture diameter is 50mm/2.8 = 17.86mm. Now if you only change the focal length from 50mm to 100mm but keep the f-stop at f/2.8 then you increase the aperture diameter to 100mm/2.8 = 35.71mm. This larger light-passing port is responsible for the shallower DOF, not the the increase in focal length as such. One can prove this by stopping down the 100mm lens to f/5.6, resulting in the same 17.86mm aperture diameter we had for the 50mm f/2.8 lens.

I hope you can agree so far.

Note that f-stoppers use some nice diagrams to show why distance, for instance, affects DOF.
Have you noticed they they don't show you any diagrams to illustrate why increasing focal length changes DOF?
The reason why they don't show you respective diagrams with light rays, etc., is because there is no causal relationship between focal length and DOF.
I suspect that the actual effect -- the increase of the aperture diameter when the f-stop is kept constant -- eludes them and that they don't understand that DOF is essentially the result of a parallax effect, i.e., the larger the aperture, the more angles/positions exist to view the same point in the subject space.

To summarise: Contrary to what f-stoppers purport, DOF is not affected by focal length but by changes to the aperture diameter.

F-stoppers were correct in claiming (mainly through the video title) that a larger sensor does not produce shallower DOF per se.
They were also correct in attributing the change in DOF to the lens.
They were wrong in attributing the change in DOF to focal length.

The reason why the second image shot with the larger sensor showed shallower DOF was not because he used a focal length that was twice as long, but because he used an aperture diameter that was twice as large. Had he used the equivalent f-stop of f/5.6 for the longer lens, he would have achieved exactly the same DOF.



The difference between "myths" and "facts" is that the former can be disproven by experiments while the latter are confirmed by experiments.

I hope you can agree that 100mm is the FF-equivalent focal length compared to 50mm on a 4/3 sensor.
It is not the same focal length (100 <> 50) but it is equivalent because the longer focal length exactly cancels out the wider FOV of the larger sensor.

For all practical intents and purposes, the combination of 50mm on 4/3 and 100mm on FF are indistinguishable. There are some real world practical implications that would allow you to experimentally determine which combination you are dealing with, but theoretically, you cannot distinguish one combination from the other.

I think you know that already. Where it gets slightly trickier is to understand that in order to keep combinations indistinguishable, one also has to change the f-stop. The reason is simple, though. We need to keep the aperture diameter the same, as the latter is responsible for determining the DOF. Hence, the f-stop on the 100mm lens must be twice as large as the one on the 50mm lens.

This tells us that the FF-equivalent f-stop for f/2.8 on 4/3 is f/5.6.

I hope you can see that this is exactly the same conversion we had to do for the focal length.
All photographers seem happy with the idea of equivalent focal lengths, i.e., the notion that you have to adapt the focal length when you change the sensor size.
Far fewer photographers understand the idea of an equivalent f-stop, i.e., the notion that you have to adapt the f-stop when you change the sensor size.



In the comparisons I make or assume, I never ever change the camera to subject distance.

With equivalent shooting parameters, you'll find that you cannot distinguish the FF image from the APS-C image.

You'll find that only one eye is in focus for the FF image, if you choose to use non-equivalent f-stops.
Choosing, say f/2.8, for both shots, make the f-stops look the same numerically, but we know that the equivalent (i.e., the one that achieves the same effect) f-stop for FF is ~f/4.

I hope this helps.

Good post

---------- Post added 11-15-2019 at 10:44 PM ----------

You could achieve exactly the same DOF if you stopped down the 100mm to f/5.6. Why f/5.6 exactly? Well, DOF is determined by the aperture diameter. In the case of a 50mm lens at f/2.8, the aperture diameter is 50mm/2.8 = 17.86mm.

The only change that I would make is that it is the entrance pupil that is 17.86mm. I know this is nitpicking but when people see this they think the aperture is 17.86mm and somehow they think that this is the size of the iris.

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Last edited by Ian Stuart Forsyth; 11-15-2019 at 09:46 PM.

[Class A]

The only change that I would make is that it is the entrance pupil that is 17.86mm. I know this is nitpicking but when people see this they think the aperture is 17.86mm and somehow they think that this is the size of the iris.

Very good point.

The entrance pupil can only be equated with the size of the iris if the lens elements in front of the iris achieve a magnification of factor one, or don't exist at all.

[Ian Stuart Forsyth]

The entrance pupil can only be equated with the size of the iris if the lens elements in front of the iris achieve a magnification of factor one, or don't exist at all.

Been following this thread watching you fight an uphill battle, I know how difficult it can be over something that is very basic
lens light-gathering abilities - Page 15 - PentaxForums.com

[Serkevan]

Been following this thread watching you fight an uphill battle, I know how difficult it can be over something that is very basic
lens light-gathering abilities - Page 15 - PentaxForums.com

It is indeed very basic. A pixel is either empty (no-information shadow), saturated (clipped highlight) or in between (yay, proper exposure). Dynamic range depends exclusively on, well, how much distance between "recording begins" and "saturates" is. Since an image is made of pixels, the maximum "image DR" is equal to the maximum DR of the pixels.

On another note, Class A says to keep shutter speed the same to take "the same photo". Of course, everyone knows that going from 1/2000 to 1/1000 in a landscape is going to make the mountains look different.

I am fascinated by the insistence on "total light", when that is irrelevant to the image (what we see is the exposure. No one cares or thinks "oh wow, I really like how the light captured in this striking landscape was whatever Cd at whatever Lumens).

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Last edited by Serkevan; 11-16-2019 at 04:23 AM.

[Class A]

Of course, everyone knows that going from 1/2000 to 1/1000 in a landscape is going to make the mountains look different.

1. Images with static content exist. Images with dynamic content exist as well. In the latter case one wants the motion blur to be the same. Hope that makes sense. However, the is a more fundamental reason than desiring the same level of motion blur which is:
2. The shot taken at 1/1000s will use more photons than the shot taken at 1/2000s (everything else being equal). The mountains may look the same in terms of their shape but when you want to push shadow areas, you'll be more successful with the 1/1000s shot.

I am fascinated by the insistence on "total light", when that is irrelevant to the image (what we see is the exposure....

Whether you call it "exposure" or "total light" doesn't matter for what you "see".

With any given image you look at, you can convert "exposure" into "total light" and vice versa.

Also, there is no "insistence" on total light. It is just useful to realise that it is the total photon count that matters for image noise, whereas "exposure" is only useful to know when you also know the sensor size. In other words, if you only know the exposure but not the sensor size, you cannot make any prediction about the image noise. In contrast, if you know the total number of photons used to produce the image, you can predict the image noise without even having to know how big the sensor is.

That's why "total light" is a useful concept (in addition to making it easy to understand the relationship between shooting parameters and image properties).

[photoptimist]

It is indeed very basic. A pixel is either empty (no-information shadow), saturated (clipped highlight) or in between (yay, proper exposure). Dynamic range depends exclusively on, well, how much distance between "recording begins" and "saturates" is. Since an image is made of pixels, the maximum "image DR" is equal to the maximum DR of the pixels.

On another note, Class A says to keep shutter speed the same to take "the same photo". Of course, everyone knows that going from 1/2000 to 1/1000 in a landscape is going to make the mountains look different.

I am fascinated by the insistence on "total light", when that is irrelevant to the image (what we see is the exposure. No one cares or thinks "oh wow, I really like how the light captured in this striking landscape was whatever Cd at whatever Lumens).

Total light does matter because it is differences in the recorded numbers of photons that let us perceive differences in the relative brightness of parts of the scene.

The central issue is that light from a scene is ALWAYS noisy even if the sensor is perfectly noise-free. The exact number of photons recorded in any given pixel in any given image is a Poisson-distributed random variable. The higher the expected or average number of photons, the better the ability to resolve adjacent differences in shading.

Bigger sensors with bigger pixels behind bigger lenses with bigger physical entrance pupils gather more photons which enable us to resolve more subtle differences in shading better than can smaller sensors with smaller pixels behind smaller lenses with smaller physical entrance pupils.

[Kunzite]

"Total light" is nothing more than a way to obscure useful information. It is something trivial to derive once you know the exposure and sensor area; but weirdly there are people wanting to replace the two factors with the derived one.
Weirder, people are still claiming it's making things "easy to understand".

[Rondec]

Let's just leave it that bigger sensors tend to have better high iso performance and better dynamic range at base iso. Obviously this is dependent on age of the sensor as well.

I don't really care why it is that what, but it is the reason that I use my full frame cameras more than the APS-C ones. Others who don't need that sort of performance or who shoot out of camera jpegs will scoff at this as unimportant. This is a photo of my daughter shot at iso 8000 with the DA *55 1/80 second, f2.5. It was brightened 1.55 EVs in Lightroom and had a touch of noise reduction added. Is it a perfect photo? Of course not. It has a bit of noise in the shadows and the noise reduction has softened it a bit, but I guarantee that if I had shot this with most APS-C cameras with similar light and then brightened the exposure a stop and a half, the image would have been terrible, even at web sizes.



Is it total light? Is it magic? Is it my imagination? Maybe some of all of those things, but I feel like Galileo talking about the earth going around the sun. "And yet it moves..."

[Ian Stuart Forsyth]

Dynamic range depends exclusively on, well, how much distance between "recording begins" and "saturates" is. Since an image is made of pixels, the maximum "image DR" is equal to the maximum DR of the pixels.

Your forgetting about noise and what is the threshold of the cutoff point of what you determine as an acceptable amount. Then how we view that noise will also determine the amount of DR we have.

Here we have 2 images taken using the same exposure the only difference is that I have change the size of the medium capturing the image. The sensor is identical in between the 2 images the only thing that has changed is that for 1 image a 18mm lens was used and for the other 53mm lens both shot at iso 100 ƒ8 8sec. Both images are pushed 3 stops in the raw converter to see how much noise starts to appear


The only difference other than the focal length used was the enlargement factor needed to display the images at this output. For this there was a scale factor of 1.7

Because of this the one image is made up of more light around 3 stops

[Dartmoor Dave]

Your forgetting about noise and what is the threshold of the cutoff point of what you determine as an acceptable amount. Then how we view that noise will also determine the amount of DR we have.

Here we have 2 images taken using the same exposure the only difference is that I have change the size of the medium capturing the image. The sensor is identical in between the 2 images the only thing that has changed is that for 1 image a 18mm lens was used and for the other 53mm lens both shot at iso 100 ƒ8 8sec. Both images are pushed 3 stops in the raw converter to see how much noise starts to appear


The only difference other than the focal length used was the enlargement factor needed to display the images at this output. For this there was a scale factor of 1.7

Because of this the one image is made up of more light around 3 stops

I'm assuming that the one on the right is a 100% crop, and the one on the left is a crop of far fewer pixels that you've upsampled to appear the same size on screen. So the appearance of more noise in the one on the left is a by-product of the upsampling, and nothing at all to do with "more light" being used to make the one on the right. Upsample the one on the right by the same amount that you upsampled the one on the left and you'll see the same amount of noise in it.

And of course, in the case of digital photography we are almost never looking at an upsampled image. Even if you use a 4K monitor, a photograph taken with any camera with more than 8 megapixels has to be downsampled so that you can see it all on screen. Upsampling only becomes an issue in digital photography if you decide to print bigger than the size that your actual pixel resolution would give you at the standard 300dpi.

The way that the sensor data is processed, and the pixel resolution that we choose to view it at, plays a large role in the appearance of noise in digital photography. It's the refusal to include those factors and instead to rely on the notion of "total light" that's at the heart of my own objections in threads like this.

Perhaps in the interests of fair play, you could now post a 100% crop of the 18mm image, then downsample a crop from the 53mm to appear the same size on screen, so that we can all consider the difference in noise when you do that.


Edit: Just to be clear, I'm absolutely not disputing the fact that bigger pixels result in less noise. Of course they do, and yes, the reason for that is that more photons are collected by a bigger pixel. So a bigger sensor with bigger pixels will give you less noise than a smaller sensor with smaller pixels (at least if they are from the same generation of sensor technology). But it's the pixels that matter, not the "total light" hitting the whole sensor, and the actual appearance of noise in the final image also depends on any resampling required to view it at the pixel resolution that we decide to view it at. As long as your lens throws an image circle that covers your sensor size, you've got all the "total light" you need.

[Serkevan]

So you change the magnification factor of the one taken with the "smaller" sensor and then claim it's more noisy. Ok.
As I said, I have done the tests myself: the clipping starts at the same time, and noise is virtually the same. You simply cannot enlarge as much as you have less than half the pixels to work with.

[swanlefitte]

A point here. Noise is only seen or at least known from pixel variations.
Show 1 pixel there is no way to know.
Yet if a piece of a scene that looks like a dot is captured on a 4x4 pixel area or a 6x6 point area and each pixel has a random noise die roll thats like rolling 16 dice and and averaging them vs 36 dice. Then the variation between 2 of these dots will on average be significantly more with the 2 16 rolls vs the 2 36 rolls.

[StiffLegged]

This is fantastic! There’s been pages of this; invoking DxO, depth of pixel focus, dynamic light and partial frames and I still don’t know what y’all are arguing about. Brilliant!!

[Serkevan]

This is fantastic! There’s been pages of this; invoking DxO, depth of pixel focus, dynamic light and partial frames and I still don’t know what y’all are arguing about. Brilliant!!

I hear it has something to do with the raffle (and, on my side, I like senseless banter )

[photoptimist]

The way that the sensor data is processed, and the pixel resolution that we choose to view it at, plays a large role in the appearance of noise in digital photography. It's the refusal to include those factors and instead to rely on the notion of "total light" that's at the heart of my own objections in threads like this.

Agreed! And each of those steps has an effect on the number of photons that went into making an output pixel and the number of photons in the overall output screen or print image.

Perhaps in the interests of fair play, you could now post a 100% crop of the 18mm image, then downsample a crop from the 53mm to appear the same size on screen, so that we can all consider the difference in noise when you do that.

Ian will correct me if I'm wrong, but that seems to be exactly what he did. The left image is a 100% crop from the 18mm shot and the right one is about a 3:1 (=53:18) downsample of the shot taken with the 53mm lens. About 9X the number of photons were collected and averaged to make each screen pixel of the image on the right relative to the image on the left.

Edit: Just to be clear, I'm absolutely not disputing the fact that bigger pixels result in less noise. Of course they do, and yes, the reason for that is that more photons are collected by a bigger pixel. So a bigger sensor with bigger pixels will give you less noise than a smaller sensor with smaller pixels (at least if they are from the same generation of sensor technology). But it's the pixels that matter, not the "total light" hitting the whole sensor, and the actual appearance of noise in the final image also depends on any resampling required to view it at the pixel resolution that we decide to view it at. As long as your lens throws an image circle that covers your sensor size, you've got all the "total light" you need.

So we can agree that more photons -> less noise at the sensor pixel level (for the same generation of sensor technology). That rule of thumb of more photons -> less noise also extends to overall images.

What if I have a 2400 x 3600 pixel image taken with 5 micron pixels and a 4800 x 7200 pixel image taken with 5 micron pixels with both taken with identical exposure settings and of the identical subject? Sure, if we look at 100% crops of both images -- the pixel-peeping scenario -- noise levels appear similar (and the total light in each crop is the same). Next, if we downsample both images make 300 PPI prints of the same size, the amount of light per output pixel will be 4X higher for the 4800 x 7200 pixel image and the noise will be half that of the 2400 x 3600 pixel image.

Total light goes by other names in astronomy and optical engineering: etendue, AΩ, light-collecting power, light grasp, throughput, and others. It is essentially the product of the physical diameter of the lens entrance pupil times the diameter of the sensor's image circle.