Have you preordered an RF 200-800mm?

[Dragon]

OK, I can do something even simpler after poring over the dynamic range plots for the most useful way to summarise in practical terms, which is what we want in the end. If shutter speed is not important, then the 800mm at f/9 set at the same iso as the 500mm at f/7.1 but 2/3rds stop slower shutter speed (1.6x slower in real numbers), the image from the 800mm will be 1.6x larger in width and height and 2/3rds stop better DR. If shutter speed is of the essence, set the iso for the 800 2/3rds stop higher (1.6x) and you will get the 1.x1.6x larger image with the same DR as the 500mm.

Did you mean "1.6x1.6x larger image" in that last line?

 

[AlanF]

Did you mean "1.6x1.6x larger image" in that last line?

Yes

 

[Quack]

OK, I can do something even simpler after poring over the dynamic range plots for the most useful way to summarise in practical terms, which is what we want in the end. If shutter speed is not important, then the 800mm at f/9 set at the same iso as the 500mm at f/7.1 but 2/3rds stop slower shutter speed (1.6x slower in real numbers), the image from the 800mm will be 1.6x larger in width and height and 2/3rds stop better DR. If shutter speed is of the essence, set the iso for the 800 2/3rds stop higher (1.6x) and you will get the 1.6x1.6x larger image with the same DR as the 500mm.

Thanks again Alan, but maybe I am just dense, but your 2 examples seem to give me different conclusions. I am also not understanding what image size means in your comparisons.

Images taken from both lenses, let's say, are 6000 x 4000 pixels. One has a field of view magnified 1.6 times, but the overall image size is still 6000 x 4000 pixels. I understand that when one crops, and then displays the image at the same size as the uncropped image, the noise is more pronounced because it is being enlarged. I would not think that the noise is being enlarged when shooting an 800mm focal compared to a 500mm focal length, but let me know if I am mistaken. If noise is not being enlarged, then I do not understand why the image size is being multiplied by 1.6. I know folks keep mentioning the pixels on duck. I am shooting an entire image. Noise, DR, effect the entire image, not just the subject.

In your first example, you are lowering the shutter speed of the 800mm shot by 2/3rds of a stop, allowing more light to reach the sensor, concluding that the DR is also 2/3rds of a stop better. To me this sounds like if the 800mm shutter speed was set 2/3rds of a stop faster - thus equal to the 500mm shot - then the DR would be the same if they the ISO and shutter speed were the same.

In your second example, shutter speeds are the same, ISO is set 2/3rds of a sop higher on the 800mm shot, and the DR is the same. This tells me that if the ISO is set 2/3rds of a stop lower on the 800mm shot - thus equal, with the the 500mm shot - then the DR would be better on the 800mm shot (presuming higher ISO equals more noise and less DR).

Where am I going wrong?

Perhaps it is easier to ask it this way...I take a shot with the RF 200-800mm focal length lens, then the RF 100-500mm lens. First shot I take, f/9 on the 800, f/7.1 on the 500. ISO and shutter speed are the same. Does the larger entrance pupil of the 800mm lens give me a more or less noisy image? More or less DR?

 

[AlanF]

In your second example, shutter speeds are the same, ISO is set 2/3rds of a sop higher on the 800mm shot, and the DR is the same. This tells me that if the ISO is set 2/3rds of a stop lower on the 800mm shot - thus equal, with the the 500mm shot - then the DR would be better on the 800mm shot (presuming higher ISO equals more noise and less DR).

Where am I going wrong?

If you set the 100-500mm at 500mm f7.1, the 200-800mm at 800mm f/9, and and the same iso and shutter speed for both, then the 800mm shot will be underexposed by 2/3rds of a stop - that's the exposure triangle.

I also had explained that the noise is not caused by the iso but but by the number of photons hitting the image. You could push the underexposed through 2/3rds of a stop in post-processing and if the camera's amplifier was linear you would recover the correct exposure with the same noise and DR as if you had 2/3rd stop higher on the camera.

Importantly, it's not the iso that causes the noise, it's the low number of photons that usually accompanies the high iso that contains the noise. An image with a high iso but containing a higher number of photons because the image is large can be less noisy than a much smaller image with a lower iso, as we should know from an iPhone vs an APS-C vs FF vs MF.

@Surab and @Dragon have attempted to explain the size issue.

 

[Surab]

if the camera's amplifier was linear you would recover the correct exposure with the same noise and DR as if you had 2/3rd stop higher on the camera.

Is linear amplifier response what defines ISO invariance? I believe the R6II has an ISO invariant sensor, but I might misremember?

 

[AlanF]

Is linear amplifier response what defines ISO invariance? I believe the R6II has an ISO invariant sensor, but I might misremember?

Both the R5 and R6II are isoinvariant above iso 800 and pretty much so from 400 - https://www.photonstophotos.net/Charts/PDR_Shadow.htm#Canon EOS R5,Canon EOS R6 Mark II(EFCS)

 

[Kit.]

You haven‘t defined what x is. So what you have written is completely meaningless, let alone simpler.

x is x, a generic name for a free variable in a math formula.

sin(x) ≈ x is the common name for the approximation you were implicitly relying on. In this particular case, this approximation lets you use areas instead of solid angles.

I learned this approximation (and this name) in the high school, as a part of basic trigonometry. So, it should be common knowledge, or...?

And you don’t explain what it leads to even if we knew what x is.

It leads to more photons per duck, obviously.

If it's still unclear where it leads then, I could reuse your phrase "Signal/noise in low light is proportional to the square root of the total number of photons hitting the whole image per unit time", except that I need to add one modification to make it more relevant: for the main use of this lens, it's not "the whole image", but "the subject (duck) image" what counts.

I'm pretty sure the readers of CR can fill the rest by themselves.

Anyway, my point was that the calculations involving focal length and f-number were redundant here and were detracting from the fact that, physically, assuming your implied assumptions, it's just the area of the entrance pupil that matters.

 

[AlanF]

x is x, a generic name for a free variable in a math formula.

sin(x) ≈ x is the common name for the approximation you were implicitly relying on. In this particular case, this approximation lets you use areas instead of solid angles.

I learned this approximation (and this name) in the high school, as a part of basic trigonometry. So, it should be common knowledge, or...?


It leads to more photons per duck, obviously.

If it's still unclear where it leads then, I could reuse your phrase "Signal/noise in low light is proportional to the square root of the total number of photons hitting the whole image per unit time", except that I need to add one modification to make it more relevant: for the main use of this lens, it's not "the whole image", but "the subject (duck) image" what counts.

I'm pretty sure the readers of CR can fill the rest by themselves.

Anyway, my point was that the calculations involving focal length and f-number were redundant here and were detracting from the fact that, physically, assuming your implied assumptions, it's just the area of the entrance pupil that matters.

Most know that x is a generic term, but what that term is in a specific problem is crucial - it has to be defined as without its definition it is meaningless. If you don't say what x is we don't know what you are talking about. I have a degree that included a term of optics and I don't know what you are talking about as well as your use of high school trigonometry and I don't know how the average member of CR would.

Your statement that I was implicitly relying on the "this approximation lets you use areas instead of solid angles" baffles me because I wasn't relying on it at all, and further this approximation is based on very, very small angles and with lenses of f-numbers in the range of f/7 - f/9 or we are dealing with large angles.

So, would you please help me through my problems by 1), explaining explicitly: what x and sin(x) refer to in these calculations; and 2) where the approximation x ~ sin(x) comes in.

 

[neuroanatomist]

So, would you please help me through my problems by 1), explaining explicitly: what x and sin(x) refer to in these calculations; and 2) where the approximation x ~ sin(x) comes in.

Apparently @Kit. is referring to the ‘small angle approximation’ (though typically for angles one uses θ and not x to represent a generic variable), whereby at sufficiently small enough angles, sin θ ≈ θ, tan θ ≈ θ, and cos θ ≈ 1.

In astronomy, for example, that approximation is used in calculations involving stellar diameters, because stars are sufficiently far that they cover only a few arcseconds of the FoV.

The main relevance to lens optics is for angles essentially nearly on the optical axis. When applied to ray tracing, the approximation means there’s no difference between sagittal and meridional rays. Empirically, you can get a sense of where in the image that approximation holds by looking at an MTF chart – basically, just at the left side, before the solid lines (sagittal contrast/resolution) and dashed lines (meridional contrast/resolution) diverge.



That's about a 1 mm circle at the center of the sensor, approximately 0.09% of the area of the sensor. Put another way, the approximation holds with a bird sufficiently far away that it covers about 20% of the area of a the central spot imaging circle that is etched/shown in the viewfinder. So if he's talking about pixels-on-duck, the small angle approximation holds only for a very small (distant) duck...



From now on, I shall refer to it as the ‘small duck approximation’.

 

[AlanF]

Most likely @Kit. is referring to the ‘small angle approximation’ (though typically for angles one uses θ and not x to represent a generic variable), whereby at sufficiently small enough angles, sin θ ≈ θ, tan θ ≈ θ, and cos θ ≈ 1.

In astronomy, for example, that approximation is used in calculations involving stellar diameters, because stars are sufficiently far that the cover only a few arcseconds of the FoV.

The main relevance to lens optics is for angles essentially right in the optical axis. When applied to ray tracing, the approximation means there’s no difference between sagittal and meridional rays. Empirically, you can get a sense of where in the image that approximation holds by looking at an MTF chart – basically, just at the left side, before the solid (sagittal contrast/resolution) and dashed (meridional contrast/resolution) diverge.

View attachment 213359

That's about a 1 mm circle at the center of the sensor, approximately 0.09% of the area of the sensor. Put another way, the approximation holds with a bird sufficiently far away that it cover about 20% of the area of a the central spot imaging circle that was etched/shown. So if he's talking about pixels-on-duck, the small angle approximation holds only for a very small (distant) duck...

View attachment 213360

To calculate the relative S/N in the two cases, what we are doing is to calculate the relative number of photons hitting the image of a duck comparing a lens of f-number N1 with another of N2 and of focal lengths f1 and f2 respectively. To do this, we need to know just the relative photon fluxes per unit area of the duck times the relative areas of the duck. For the ratio of lens1/lens2, the relative flux is given by (N2/N1)^2 and we can approximate the relative areas by (f1xf1)/(f2xf2) for an object 10x or more further away from the lens than f1 or f2 or precisely calculate it from the lens makers equation. We are not using angles. As for notation, θ is commonly used for angles and in elementary optics u and v for distances.

 

[Surab]

I'm still a little confused about the sin(x) approximation and it's application here. I feel it'll give great insight if I could actually figure it out, but I have never been a huge fan of geometric optics.. :S

 

[Surab]

To calculate the relative S/N in the two cases, what we are doing is to calculate the relative number of photons hitting the image of a duck comparing a lens of f-number N1 with another of N2 and of focal lengths f1 and f2 respectively. To do this, we need to know just the relative photon fluxes per unit area of the duck times the relative areas of the duck. For the ratio of lens1/lens2, the relative flux is given by (N2/N1)^2 and we can approximate the relative areas by (f1xf1)/(f2xf2) for an object 10x or more further away from the lens than f1 or f2 or precisely calculate it from the lens makers equation. We are not using angles. As for notation, θ is commonly used for angles and in elementary optics u and v for distances.

I'm sure that the lens makers equation's derivativion will be based on angles. Hence the statement about your inherent reliance on that approximation.

 

[neuroanatomist]

To calculate the relative S/N in the two cases, what we are doing is to calculate the relative number of photons hitting the image of a duck

Ok, so now we’ve gone from the small duck approximation to the theory of duck relativity. The logical extension of this progression is the Muscovy. As a very large waterfowl, it can represent the Grand Unified Duck.

 

[AlanF]

I'm sure that the lens makers equation's derivativion will be based on angles. Hence the statement about your inherent reliance on that approximation.

I use the simplest form of the equation for thin lenses 1/u + 1/v = 1/f where u and v are the object and image distances (both +ve) and f the focal length. You can derive the more complex lens makers equation that uses radii of curvature and refractive indices by Snell's law and make simplifying assumptions about tan(theta). The simple equation 1/u + 1/v = 1/f is derived from geometrical optics of a thin lens using similar triangles with no assumptions of tan(theta) ~ theta eg
https://www.quora.com/What-is-the-proof-of-the-thin-lens-equation-dfrac-1-u-+-dfrac-1-v-dfrac-1-f
https://byjus.com/physics/derivation-of-lens-formula/

 

[Quack]

....

@Surab and @Dragon have attempted to explain the size issue.

Obviously we are missing each other's points. I thought I made it perfectly clear that I was talking about the entire image, not just the subject (duck or otherwise). But thanks anyway for your time.

 

[AlanF]

Ok, so now we’ve gone from the small duck approximation to the theory of duck relativity. The logical extension of this progression is the Muscovy. As a very large waterfowl, it can represent the Grand Unified Duck.

We started off with a large duck and it's the intervening comments that shrunk it. This is the original post, and we are now back to it. It's absolutely straightforward. I don't why all these sin(theta) ~ tan(theta) ~ (theta) were brought into it.

Here's the maths/physics behind why it is the diameter, d, of the lens (strictly the entrance pupil) that is important for signal/noise rather than the f-number, which I'll call N. For a subject that is some distance away, the size of the image (height or width) is directly proportional to the focal length f. (Double the focal length and you double the width and height).
The brightness of the image will depend on the amount of light getting through the lens, which will be proportional to its area, ie d^2 (=0.25*pi*d^2) and it will be dispersed over an area that is proportional to the height x width of the image, ie f^2. So, the brightness of the image varies as:

d^2/f^2.

f-number N = f/d, so the brightness varies as (1/N^2),

which is something nearly every photographer knows that the brightness drops by a factor of 2 when the f-number gets larger by 1.4x (the square root of 2). So, the f-number is what is used in the exposure triangle. 1/N is a measure of the number of photons hitting per unit area of the image per unit of time.

Signal/noise in low light is proportional to the square root of the total number of photons hitting the whole image per unit time. The total number of photons is proportional to the area of the image x d^2/f^2

No. of photons varies as f^2 x d^2/f^2, ie varies as d^2.

So the S/N varies as d.

Importantly, it's not the iso that causes the noise, it's the low number of photons that usually accompanies the high iso that contains the noise. An image with a high iso but containing a higher number of photons because the image is large can be less noisy than a much smaller image with a lower iso, as we should know from an iPhone vs an APS-C vs FF vs MF.

 

[AlanF]

Obviously we are missing each other's points. I thought I made it perfectly clear that I was talking about the entire image, not just the subject (duck or otherwise). But thanks anyway for your time.

If you are looking at the image with wider field of view from a 500mm on FF vs a narrower from 800mm on FF, then a wider f-number could be an advantage for s/n. If you crop the 500mm to the same fov as the 800, then a wider lens diameter could be an advantage. Usually when people are considering the merits of using a 500mm or 800mm, which I had thought the discussion was about, it's because they need a telephoto and they are limited by reach and are having to crop. There's a similar situation when comparing different sensor sizes. For example, a 500mm f/7.1 will give exactly the same size image on an APS-C as an 800mm f/9 on a FF. Now, there are some who think that the wider f/7.1 on crop will give better s/n but the opposite is true. I hope that resolves it.

 

[EricN]

I heard the formula for the duel fisheye is sinXXX=$

 

[Surab]

I heard the formula for the duel fisheye is sinXXX=$

What if I fly to the EU, will it become sinXXX=0.93€?

Also better wear fish-eye correcting glasses to that duel!

 

[Kit.]

Your statement that I was implicitly relying on the "this approximation lets you use areas instead of solid angles" baffles me because I wasn't relying on it at all,

You are relying on the assumption that the irradiance received from the subject surface by the flat imaginary surface of the entrance pupil is uniform, per unit of entrance pupil area. Which is not exactly so.

It is customary to assume that diffuse subject surfaces in photography exhibit Lambertian reflectance behavior, i.e. that their reflected radiance is uniform, per unit of solid angle.

Typically in photography this difference is negligible in practice, but it might be noticeable, for example, in microphotography using fast lenses and in near-field photography using lens arrays. Also, there is a similar (but not exactly the same) effect in the corners of ultrawide rectilinear lenses.

So, would you please help me through my problems by 1), explaining explicitly: what x and sin(x) refer to in these calculations;

You can replace x with a pair of letters of your choice denoting the angular dimensions (2d) of the entrance pupil as seen from the position of the subject.

and 2) where the approximation x ~ sin(x) comes in.

It comes in when one expects the linearity of mapping a portion of a sphere into a plane.