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If you keep the image scale constant in object space, a larger aperture collects more signal. I see a lot of misconceptions about how aperture, focal ratio, the sensor, and the type of object interact to produce signal at the sensor. So it looks like it might be worthwhile to do a little review of how radiometry works for those of you who are interested. Unfortunately, I'm super busy today rushing around to leave on a week long river trip so I don't have much time right now to produce a comprehensive post. Ultimately, I think that this subject really needs its own thread--and I really didn't mean to hijack the OPs question here. So I'll think about it while I'm on the river and I'll post something when I get back. In the meantime, here are a couple of quick comments. Bjorn, That's not quite what I said. If you use sensors with equal QE that sample the same patch of sky in object space, the signals will be identical regardless of the focal ratio or aperture size. Rouz, Thank your your images (which, BTW, look very nice) but comparing stretched images isn't how you compare signals. You have to look at the actual ADU values in the linear data. Here are some slides showing the basics. You guys can chew on this stuff until I get back and I have more time to go over this stuff in more detail. In the meantime make sure to do some research on the camera equation. That's the equation that describes the radiometry of imaging extended objects. It shows that the only thing that matters with regard to irradiance in the focal plane is focal ratio--not aperture. How you then sample that image determines the signal strength from the sensor. John         |
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John Hayes: Hi John, Let me try to elaborate my thinking, maybe you can show me at which point the error is: 1. Let's take an idealized universe/sky which is empty and we only have a tiny square patch of sky with an area of 1 square-arcsec that emitts some photons. 2. My telescope has a certian focal length that maps this tiny patch onto a single pixel. 3. This tiny patch has a given irradiance and by irradiance I mean quantity that's given by Energy pear Unit Area per Time Unit which is basically eqiuvalent saying a certain number of photons hits one square meters per second here on Earth. 4. Obvioulsy, if I manage to keep the focal length constant, my pixel will always see this patch in the sky. Now, by changing the aperture, I'm going to collect more photons as the irradiance "hitting" the aperture is constant but by multyplying/integrating over the aperture, I collect more photons. So I am not sure how the signal (which is the number of electrons?) should not change as I have changed the aperture here? (Otherwise I would violate the conservation of energy) Björn PS: I'm basing my calculations on the equation shown on the slide at timecode 42:32 (https://youtu.be/te4UVYi6n44?t=2549). Keeping everything (including focal length) constant except aperture does change the focal ratio ratio and thus the signal ratio. |
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John Hayes:If you keep the image scale constant in object space, a larger aperture collects more signal. John, my original question has been answered in the sense that the combination of small scope + tiny pixel won't work out. So feel free to use this thread, since we already have the discussion started here anyway. Clear skies Wolfgang |
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| I suppose it is meant in the sense that the signal is aperture invariant as the image scale is only defined by the focal length and hence the same imaged patch of the sky. In other words the signal is indipendent as it is defined only by the source itself. |
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If you use sensors with equal QE that sample the same patch of sky in object space, the signals will be identical regardless of the focal ratio or aperture size. There is something I am missing here because I don’t see how this could possibly be true. For example, keeping the camera the same, a 35mm f/1.4 lens will produce twice the irradiance and hence twice the signal of a 35mm f/2 lens on a given pixel of the sensor with the sampling the same in object space. The focal length is the same - the only difference between the two systems is aperture. |
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John Hayes: Bjorn, That equation (at 42:32) flows from the camera equation taking into account possible losses in the imaging system. It can be also recast as a statement of Etendue but that's not important for this discussion. As for your thinking, first, you have to start with the radiance of the object; not the irradiance, which doesn't capture angular dependence. Rather than trying to pick apart your argument, I think that it's better to first look at how the camera equation is derived from simple geometric optics using the correct radiometric quantities. John Greivenkamp does a superb job of deriving the equation using two different methods in his class notes available here: https://wp.optics.arizona.edu/jgreivenkamp/wp-content/uploads/sites/11/2019/08/502-11-Radiative-Transfer.pdf John's approach is very simple and clear--just be sure to pay close attention to the definition of the various radiometric quantities (and their units) shown on slide 10-2. Slides 10-7 through 10-9 are the most relevant to your idea of how to do the derivation and you should study them carefully. The thing that conservation of energy requires is the conservation of radiance (not irradiance) through an optical system. That means that for a lossless optical system, the radiance of the image will be the same as the radiance of the object. Grievenkamp also shows how the camera equation falls out of the conservation of radiance in slide 10-18. Once you fully grasp how the camera equation is derived, the rest of what I've said easily follows. Just remember that the camera equation deals strictly with geometric imaging. When you consider the process of imaging stars (or any object smaller than what can be described using geometric imaging), you have to involve diffraction, which ultimately does introduce a factor proportional to the square of the diameter of the entrance pupil. That's why it's important to remember that when I'm talking about the camera equation, I'm also talking about extended objects--not stars. John PS. As an aside, I want to mention that Greivenkamp and I were students together and he was a very smart guy! Unfortunately, he recently passed away after a long battle with cancer. He won many awards, published a lot of papers, started a great book series, he was a past president of SPIE, and at the time of his passing, he was a professor emeritus at the Wyant College of Optical Sciences. I think that it is fitting to say that he really knew his stuff and that he had a great impact on optics education! His slides typify the simple way that he could present complex subjects. |
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If you use sensors with equal QE that sample the same patch of sky in object space, the signals will be identical regardless of the focal ratio or aperture size. Yes, if you use the same sensor at different focal ratios, the signal will indeed change...exactly as shown on the slides. You are missing the important part here-- namely that the sensors have to "sample the same size patch of sky in object space." That requires different sensors. John |
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John Hayes: Thanks John glad you like them. Missed you at AIC but saw the excellent presentation. The Mel 15 example was linear (STF stretched)a and the SNR calculations were based off raw stacks. |
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You are missing the important part here-- namely that the sensors have to "sample the same size patch of sky in object space." That requires different sensors. John - in the example I gave, both lenses have the same focal length - 35mm. Hence for a given sensor, they will sample exactly the same patch of sky. In the write up you linked, the magnification “m” is purely dependent on object distance and focal length which is the same in both cases. Hence, the image produced on the sensor will be exactly the same by both lenses in terms of field covered. As shown by the camera equation, the 35mm focal length f/1.4 will have higher irradiance than the 35mm focal length f/2 because of the lower f/# but this is purely by virtue of greater aperture since the focal length is the same in both cases. This is consistent with the camera equation in the technical paper you linked. I have changed the f/# which changes the image irradiance. But I am sampling exactly the same way in object space because focal length is the same - sampling in object space is governed purely by the focal length if the sensor is fixed. Therefore the signal on the pixel is doubled, in this case by doubling the area of the lens while keeping the focal length fixed. |
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You are missing the important part here-- namely that the sensors have to "sample the same size patch of sky in object space." That requires different sensors. Thanks for the clarification Arun. I'm sorry... I misunderstood your comment. You are talking about this particular case and yes, in this case, the angular sampling is held constant. This is what happens in a standard DSLR when you adjust the F/# and it is exactly what the camera equation is used to describe for normal daylight photography. A faster system produces more signal...due to the fact that the sensor (i.e. the pixel) size is held constant. Since you are holding the focal length constant, it forces the aperture to be the determining factor, but that's only because you are constraining the focal length. In truth, the focal ratio is what determines the irradiance in the focal plane and that's the better way to think about it.  |
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Thanks John, I am glad we agree. The fundamental physics is the same and that’s the most critical part. In fact, I can take the camera equation and multiply it by the square of pixel size to give the signal incident on each pixel (irradiance times area). But sampling in object space is proportional to f/p. So if we know how we are sampling in object space, I can write p in terms of f and sampling (arc sec/pixel). Using this, it is possible to show that for constant sampling in object space, the only thing that controls the strength of signal is the clear aperture of the scope - since f/# in the camera equation can be written in terms of focal length and aperture, the equation for signal can be rewritten in terms of aperture and sampling. Focal length drops out, but of course this and pixel size is what photographers will use to determine sampling! Just different ways to think about the same physics 😀 |
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John Hayes: Hi John, We don't need to add more slides or equations here to the discussion. I also didn't claim that irradiance is conserved but energy. My example (point 1 to 4) makes the very same point than @Arun H. does. I keep FL and pixel constant and increase aperture. Same experiment, same conclusion: more signal on the pixel. Regarding my statement: If you keep the image scale constant in object space, a larger aperture collects more signal. My statement is indeed ambigous. What I meant to say is: "If you keep the image scale constant in object space, *increasing* aperture collects more signal." Sorry for giving confusion. Björn |
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Hi, as someone who used the tiny pixel IMX183 sensor, I agree that it’s not usable on slow scopes with a small aperture. I used it once on my 80mm Apo at native focal lenght (560mm), and the results were not good. However, I achieved decent results with a 0.8x reducer. Personally, I think it’s best used either on fast camera lenses, or fast scopes with f/5 and under. A 8” f/4 Newton would be a good match with that sensor for galaxy imaging - given that your seeing plays along. It makes for a decent sampling rate, it’s fast, it has a big aperture. Used with a refractor, I’d say stay away from smallish apertures and slow focal lenghts, but eg paired with a 4” or 5” Apo native f/6, reduced 0.8x, I can imagine it does well, you get a f/4.8 ratio, and have a decent aperture. The best would probably a RASA, but it’s just not a beginner scope at all, and also very expensive. Yeah, the IMX183 sensor is very peculiar, and isn’t exactly an allround sensor, but I’d disagree with saying it’s only usable on a RASA aka big aperture and very fast. Bigger aperture & fast focal ratio definitely helps with that sensor - f/5 and faster is definitely what you should use, 80mm and f/7 did not yield nice results for me. I kinda wish for a 4/3 sensor with 3.x pixel size that is not square like the ASI 533 - kinda not really a thing rn, at least not in ZWO sortiment, which I would use due to my AsiAir Pro. Imo, something in the middle between the tiny but somewhat common IMX183 and the 3.76 Micron sensors would be good, something around 3.0-3.2 would be nice - bigger full well, would make still nice galaxy imaging with 8” and 10” Reflectors. |
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Mina m.b.: Hi Mina, I got acceptable results with the 183 cameras with a few telescopes. For those looking to buy a new camera though, as you said 3.76u pixels are are better. I did review both an APC-s and full frame below, both great in my opinion: APS-C: https://astrogeartoday.com/the-next-generation-of-astro-cameras-testing-the-qhyccd-qhy268m/ Full Frame: https://astrogeartoday.com/the-ultimate-deep-sky-camera-the-qhy600m-reviewed/ Latest image with the camera: https://www.astrobin.com/vxsvh4/ . Rouz |
