p.41 #1 · Pre-order: Sony FE 200-600 f5.6-6.3 G OSS ($1,998)
Steve Spencer wrote:
What you said above is only true if you keep the amount of light constant, but if the brightness of the light is controlled by the aperture, which it is, then that formula does not apply.
For any given light intensity (i.e., a fixed aperture size), a larger sensor area means less light reaches the sensor. So, for the larger sensor, you would need to open the aperture much larger to get the same amount of light as you would with a smaller sensor, according to your logic...
p.41 #2 · Pre-order: Sony FE 200-600 f5.6-6.3 G OSS ($1,998)
molson wrote:
When the light is spread over a larger sensor area, the intensity of the light reaching the sensor is reduced according to the inverse square law. So, for a sensor twice as big, the light intensity is reduced by four times, and you are in fact losing two stops of light with the larger sensor. Simple physics...
? In your world of "Simple physics" Smart phones would rule in low light
p.41 #3 · Pre-order: Sony FE 200-600 f5.6-6.3 G OSS ($1,998)
Pre-ordered as well. I have no idea what I'm going to use this lens for (don't do a lot of wildlife), but I've got a bad case of GAS and I'm not afraid to admit it - this is the 5th lens I've bought this year. And I already have the 100-400 and 1.4x.
Jun 13, 2019 at 11:13 AM
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p.41 #4 · Pre-order: Sony FE 200-600 f5.6-6.3 G OSS ($1,998)
codyconway wrote:
kind of - logic says that the area of light that is reflected through an optic is not collected light, but projected. Thus the projection of the light is the area that is applied. In a smaller sensor, you get that "crop" factor because the light is only being captured from the central zone. The excess light is then difussed by the housing / padding, etc. You do not gain or lose light based on sensor size. I wish people would understand that better. Then again, perhaps I'm the one with the twisted view on it?
Just apply the formula I provided above. The mistake many people make is equating brightness of the light with amount of the light. They are in fact two very different concepts. The amount of light (i.e., the number of photons hitting the sensor) is a function of the brightness of the light, but also the time that light is collected, and the size of the sensor. So, the amount of light is a function of Aperture X Shutter Speed X Area of the sensor that is being used. When people substitute brightness for amount of light they get all mixed up. If you keep them separate and realize that brightness is a function of the aperture all the relations are simple and straightforward.
Jun 13, 2019 at 11:18 AM
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p.41 #5 · Pre-order: Sony FE 200-600 f5.6-6.3 G OSS ($1,998)
molson wrote:
For any given light intensity (i.e., a fixed aperture size), a larger sensor area means less light reaches the sensor. So, for the larger sensor, you would need to open the aperture much larger to get the same amount of light as you would with a smaller sensor, according to your logic...
No Cliff, it is a simple mathematical formula. Brightness X Time X Area = amount of light hitting the sensor as a whole. At the same brightness (i.e., aperture) a larger sensor has more light (i.e., a greater number of photons) hitting the sensor as a whole. So, it is not true that you need to open the aperture to get the same amount of light. It is actually the opposite you could close the aperture by the amount that the area is bigger to get the same amount of light. Note this is for the amount of light hitting the sensor as a whole not about exposure of the individual pixels. When we are talking about exposure of the individual pixels we are modifying the basic formula to basically be holding area constant. Then we can simply talk about brightness (i.e., aperture) and time (i.e., shutter speed).
p.41 #6 · Pre-order: Sony FE 200-600 f5.6-6.3 G OSS ($1,998)
molson wrote:
When the light is spread over a larger sensor area, the intensity of the light reaching the sensor is reduced according to the inverse square law. So, for a sensor twice as big, the light intensity is reduced by four times, and you are in fact losing two stops of light with the larger sensor. Simple physics...
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Steve Spencer wrote:
No Cliff, you are wrong about that. The brightness is the same at the same aperture. The formula for the amount of light is Brightness X Area X Time. Brightness is controlled by the aperture. Area is controlled by the size of the part of the sensor you are using, and time is controlled by shutter speed. That give you the amount of light the sensor has to work with.
What you said above is only true if you keep the amount of light constant, but if the brightness of the light is controlled by the aperture, which it is, then that formula does not apply....Show more →
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codyconway wrote:
kind of - logic says that the area of light that is reflected through an optic is not collected light, but projected. Thus the projection of the light is the area that is applied. In a smaller sensor, you get that "crop" factor because the light is only being captured from the central zone. The excess light is then difussed by the housing / padding, etc. You do not gain or lose light based on sensor size. I wish people would understand that better. Then again, perhaps I'm the one with the twisted view on it?
The number of photons per unit area per unit time at the sensor depends on the f/number. That's the simple physics. Consequently the total number of photons (and photo-electrons) in the image is proportional to the area of the sensor. Larger sensor means more photo-electrons, lower shot noise, and a higher signal to noise ratio.
p.41 #8 · Pre-order: Sony FE 200-600 f5.6-6.3 G OSS ($1,998)
Steve Spencer wrote:
Just apply the formula I provided above. The mistake many people make is equating brightness of the light with amount of the light.
And this is where the old adage "a little knowledge is a dangerous thing" might apply...
You're confusing terms and formulae that might be appropriate for designing solar panels, but have little or nothing to do with digital camera sensor performance. I'm just using the same (misapplied) physics, in a tongue-in-cheek way of demonstrating how silly some of these logic constructs are... but I suppose I've failed because it seems it's still way over many people's heads.
Now on the other hand, if in your arguments you replace the term "sensor" with "photosite" you will put yourself on the right track...
Jun 13, 2019 at 11:54 AM
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p.41 #9 · Pre-order: Sony FE 200-600 f5.6-6.3 G OSS ($1,998)
molson wrote:
And this is where the old adage "a little knowledge is a dangerous thing" might apply...
You're confusing terms and formulae that might be appropriate for designing solar panels, but have little or nothing to do with digital camera sensor performance. I'm just using the same (misapplied) physics, in a tongue-in-cheek way of demonstrating how silly some of these logic constructs are... but I suppose I've failed because it seems it's still way over many people's heads.
Now on the other hand, if in your arguments you replace the term "sensor" with "photosite" you will put yourself on the right track......Show more →
Nope, sorry Cliff the formula holds true regardless of the size of whatever we are talking about sensors, football fields, planets, photosites or viruses, and it is a very general formula for the way light works. Some might even call it a physical law of the way the universe works.
p.41 #10 · Pre-order: Sony FE 200-600 f5.6-6.3 G OSS ($1,998)
I preordered. I’ll be evaluating against a D500/200-500 combo. In the end the reduced weight, internal zoom and possibility of using animal eye AF for static wildlife put me over the edge. I was in a meeting and didn’t my order in until almost noon, so I’m hoping I’m not too low on the list...
p.41 #11 · Pre-order: Sony FE 200-600 f5.6-6.3 G OSS ($1,998)
For those who pre-ordered the 200-600 and are concerned about the lack of a rear drop-in slot for a CP filter, I just pre-ordered the Gobe 95m CP filter. Recently, I got my first Gobe CP (72mm) and confirmed that it has no color cast. Its also not as dark as my B+W or Hoya CP filters. The Gobe is only $83 vs the $179 B+W filter or $181 Hoya.
p.41 #18 · Pre-order: Sony FE 200-600 f5.6-6.3 G OSS ($1,998)
Steve Spencer wrote:
Nope, sorry Cliff the formula holds true regardless of the size of whatever we are talking about sensors, football fields, planets, photosites or viruses, and it is a very general formula for the way light works. Some might even call it a physical law of the way the universe works.
You seem to be forgetting that the units of measurement for light gathering are all based on intensity per unit area - so in any equation regarding sensor area, those area units cancel out. You can't ignore that simple rule of mathematics... but I guess since you're already ignoring (actually, just completely misunderstanding) so many laws of physics, one more doesn't matter.
p.41 #19 · Pre-order: Sony FE 200-600 f5.6-6.3 G OSS ($1,998)
kdrk888 wrote:
Guy, you need to get one for the sake of doing the big bronco test, you will need stand pretty far away to test the corner sharpness at 600mm.
Agreed, Guy the challenge is on, Bronco from 1/2 mile!
The number of photons per unit area per unit time at the sensor depends on the f/number. That's the simple physics. Consequently the total number of photons (and photo-electrons) in the image is proportional to the area of the sensor. Larger sensor means more photo-electrons, lower shot noise, and a higher signal to noise ratio.
Dave
P.S. To be clear, I am agreeing with Spencer.
This is only correct if your sensor consists of just a single pixel, unless you are talking about pixel binning to increase light gathering ability.
