Have you preordered an RF 200-800mm?

[Dragon]

Bring back @HarryFilm!

50 mm to 800 mm Fast all-Acrylic Super-Zoom Lens for Canon EF/EF-M, RF, Sony E-Mount and Fuji MF being beta-tested now!

HEADLINE: 50 mm to 800 mm Fast all-Acrylic Super-Zoom Lens for Canon EF/EF-M, RF, Sony E-Mount and Fuji MF being beta-tested now! YES! You heard that right! An under-the-radar Canadian Company based in Vancouver, British Columbia, Canada is now now beta-testing in the wild as of December 2020...

www.canonrumors.com

That was a funny post, but a bit extreme. From a practical perspective, only a few of the larger elements would need to be made of plastic to save a lot of weight, particularly if the design was in the vein of latest generation of big whites. The rear elements could be made of whatever glass or fluorite necessary to correct the CA of the acrylic (which based on the periphery of my eyeglasses is pretty nasty). Alternatively, large DO elements are much thinner than equivalent solid lenses. The 600mm f/11 uses a DO front element and the 800 f/11 is essentially the same lens with a relatively thin glass element in front of the DO group. the result in both cases is very light weight.

 

[AlanF]

That was a funny post, but a bit extreme. From a practical perspective, only a few of the larger elements would need to be made of plastic to save a lot of weight, particularly if the design was in the vein of latest generation of big whites. The rear elements could be made of whatever glass or fluorite necessary to correct the CA of the acrylic (which based on the periphery of my eyeglasses is pretty nasty). Alternatively, large DO elements are much thinner than equivalent solid lenses. The 600mm f/11 uses a DO front element and the 800 f/11 is essentially the same lens with a relatively thin glass element in front of the DO group. the result in both cases is very light weight.

The regulars know Harry - he has a wonderful imagination!

 

[Del Paso]

The regulars know Harry - he has a wonderful imagination!

Harry come back!
We miss you!

 

[Quack]

An 800/9 is 2/3 stop faster than a 500/7.1 when it comes to photons per duck - it has an 89mm front lens (entrance pupil) compared with 70mm. So, dial up 2/3 stop of iso with the f/9 to maintain the shutter speed and you'll actually get better signal to noise in your duck and 60% more resolution if the lens is just as sharp. A little bit of physics is actually useful when making choices!

f-number is what you use to get the right exposure in the exposure triangle. When it comes to diffraction, resolution, S/N and some other factors, all things being equal, it's the diameter of the entrance pupil/front element that is the crucial factor, the bigger the better.

Alan, I hope you can elaborate a little on what you have written in regards to S/N ratio, as the topic of some of these Canon lenses comes up frequently on various FB groups I belong to. I get that the larger front element let's in more light - and that is good - but how does one explain the practical advantage of that when the f stop (of the 200-800, in this case) is slower? Is it simple math ( highly unlikely?) that the 89mm entrance pupil lets in 27% more light (89/70)? If so, does that translate (all else being equal) to a 27% reduction in noise with the 200-800 at f/9 compared to the 100-500 at f/7.1? Or something totally different?

 

[Dragon]

Alan, I hope you can elaborate a little on what you have written in regards to S/N ratio, as the topic of some of these Canon lenses comes up frequently on various FB groups I belong to. I get that the larger front element let's in more light - and that is good - but how does one explain the practical advantage of that when the f stop (of the 200-800, in this case) is slower? Is it simple math ( highly unlikely?) that the 89mm entrance pupil lets in 27% more light (89/70)? If so, does that translate (all else being equal) to a 27% reduction in noise with the 200-800 at f/9 compared to the 100-500 at f/7.1? Or something totally different?

Actually, you are looking an area function, not a linear function, so 89sq/70sq =1.61 or 61% more light. The issue is that with the longer lens, the object of interest (duck in AlanF's terms) will occupy a much larger area on the sensor, so even though the total light to the sensor may be less with a smaller aperture, the light from the subject can be greater. A simple example. if you have a scene where the duck reaches 1/4 of the way across the sensor with a 400mm lens, then it will reach half way across the sensor with an 800mm lens. If the f-stop of the two lenses is the same, you will have 4 times as many photons from the duck hitting the sensor (area function again). Alternatively, the 800mm lens could be two stops slower and the duck would deliver the same number of photons to the sensor. Yes, you will have to kick up the ISO and at a pixel level the image will be noisier, but at the duck level, the image will be the same quality if the ISO behaviour (that "u" was for you Alan) of the camera is reasonably linear in the region of interest.

 

[EricN]

Harry come back!
We miss you!

love for Harry!

 

[Del Paso]

Actually, you are looking an area function, not a linear function, so 89sq/70sq =1.61 or 61% more light. The issue is that with the longer lens, the object of interest (duck in AlanF's terms) will occupy a much larger area on the sensor, so even though the total light to the sensor may be less with a smaller aperture, the light from the subject can be greater. A simple example. if you have a scene where the duck reaches 1/4 of the way across the sensor with a 400mm lens, then it will reach half way across the sensor with an 800mm lens. If the f-stop of the two lenses is the same, you will have 4 times as many photons from the duck hitting the sensor (area function again). Alternatively, the 800mm lens could be two stops slower and the duck would deliver the same number of photons to the sensor. Yes, you will have to kick up the ISO and at a pixel level the image will be noisier, but at the duck level, the image will be the same quality if the ISO behaviour (that "u" was for you Alan) of the camera is reasonably linear in the region of interest.

Quote from the French author Nicolas Boileau: "Ce que l'on concoit bien, s'enonce clairement".
In English: "What you understand well, you enunciate clearly".
Nothing more to add, apart from THANK YOU !

 

[AlanF]

Alan, I hope you can elaborate a little on what you have written in regards to S/N ratio, as the topic of some of these Canon lenses comes up frequently on various FB groups I belong to. I get that the larger front element let's in more light - and that is good - but how does one explain the practical advantage of that when the f stop (of the 200-800, in this case) is slower? Is it simple math ( highly unlikely?) that the 89mm entrance pupil lets in 27% more light (89/70)? If so, does that translate (all else being equal) to a 27% reduction in noise with the 200-800 at f/9 compared to the 100-500 at f/7.1? Or something totally different?

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.

 

[Del Paso]

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 photos 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.

I found the very last paragraph particularly interesting!

 

[AlanF]

Actually, I've written about this several times in the past. This thread is worth going back to as it is better. https://www.canonrumors.com/forum/t...g-and-600mm-f-6-3-vs-500mm-f-5-6-zooms.38178/

 

[Del Paso]

Actually, I've written about this several times in the past. This thread is worth going back to as it is better. https://www.canonrumors.com/forum/t...g-and-600mm-f-6-3-vs-500mm-f-5-6-zooms.38178/

I know, but scientific explanations are usually too much for my little literary brain...

 

[AlanF]

I know, but scientific explanations are usually too much for my little literary brain...

In that case, I have prepared a simple take home message in three literary forms:

Rhyming couplet
Take it on my say-so, never mind your iso
If your lens is wider, it will be quieter

Quatrain
Take it on my say-so,
If your lens is wider,
Never mind your iso,
It will be quieter.

And for a Japanese camera
Haiku
Wider lens whispers,
Silent vistas unfold wide,
Noise fades, peace abides.

 

[Kit.]

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.

Could be made simpler:

For far enough subjects (sin(x) ≈ x), the number of photons from the subject that cross the entrance pupil (and thus by definition reach the sensor, assuming 100% transparency of the lens) is directly proportional to the entrance pupil area.

 

[Del Paso]

In that case, I have prepared a simple take home message in three literary forms:

Rhyming couplet
Take it on my say-so, never mind your iso
If your lens is wider, it will be quieter

Quatrain
Take it on my say-so,
If your lens is wider,
Never mind your iso,
It will be quieter.

And for a Japanese camera
Haiku
Wider lens whispers,
Silent vistas unfold wide,
Noise fades, peace ab

Felix qui potuit rerum conoscere causas

 

[EricN]

I know, but scientific explanations are usually too much for my little literary brain...

Me too... Sadly, when I was a student I easily understood equations, but didn't find much value in them because in the classes they were taught, they felt like torture after solving the tenth version and usually there were between 30 or 60 to solve every class. Now that I know the value, it's difficult to grasp.

 

[AlanF]

Could be made simpler:

For far enough subjects (sin(x) ≈ x), the number of photons from the subject that cross the entrance pupil (and thus by definition reach the sensor, assuming 100% transparency of the lens) is directly proportional to the entrance pupil area.

You haven‘t defined what x is. So what you have written is completely meaningless, let alone simpler. And you don’t explain what it leads to even if we knew what x is.

 

[Quack]

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.

Alan, Thank you so much for this detailed explanation. Unfortunately, much of it is going over my head. You mention image area a few times, is that the sensor size? Please don't feel obligated, but I think I might understand your equations if you used real number examples. Since the discussion of late is about the 200-800 lens, if you could, would you mind using your equations filling in real numbers for a comparison between the 200-800 at 800 f/9 and the 100-500 at 500 f/7.1 in terms of image brightness and number of photons and/or S/N ratio on a full frame sensor.

 

[becceric]

love for Harry!

It seems about time for Harry to return with a New Year Prediction.

 

[AlanF]

Alan, Thank you so much for this detailed explanation. Unfortunately, much of it is going over my head. You mention image area a few times, is that the sensor size? Please don't feel obligated, but I think I might understand your equations if you used real number examples. Since the discussion of late is about the 200-800 lens, if you could, would you mind using your equations filling in real numbers for a comparison between the 200-800 at 800 f/9 and the 100-500 at 500 f/7.1 in terms of image brightness and number of photons and/or S/N ratio on a full frame sensor.

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.

 

[Surab]

Alan, Thank you so much for this detailed explanation. Unfortunately, much of it is going over my head. You mention image area a few times, is that the sensor size?

Not particularly the sensor size but really the physical size of the image on the sensor (hence the sensor size typically be the upper limit for the physical size of the image):

- Full frame sensors are 36x24mm^2, so that's the maximum size (36mm wide and 24m tall) for the physical image. (Technically the physical image can be larger than the sensor but that means that the object of interest is now too close/zoomed in and won't be fully visible on the picture later.)

- Let's assume that at a focal length f=400mm the duck is projected onto the sensor as an image of size 10x8mm^2 (10 x 8 / 36 / 24 ~ 0.0926, so approximately 9% of the sensor's area).

- Then at f=800mm the linear dimensions (width and height) will be doubled since the focal length was doubled and the area thus quadrupled: the physical image is now 20x16mm^2 and hence approximately 37% of the sensor.

Hope this helps.