I've never seen a period (.) used for multiply before. A middle dot (·) yes. So 1/2·NA then? And shouldn't that be 1/(2·NA)? Otherwise, what you have would simply be NA/2.
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Article about light loss on various sensors
- Thread starter Horshack
- Start date
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David J. Littleboy
Senior Member
But the whole "loss caused by the sensor" bit is nuts. The article claims that the effect is due to the angle of incidence (which, by the way, is in itself bogus (hint 44 mm is a long way when the sensor diagonal is also 44mm)), and then shows the effect being larger for small sensors.
It's completely, 100% batguano crazy.
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David J. Littleboy
Tokyo, Japan
zerozeronine
Leading Member
Yes, it does seem weird because the smaller sensors would generally have a less acute angle of incidence. On the other hand, the smaller sensors tend to have smaller pixels, and that's where the article's claim lies (I believe). They're trying to say that the smaller pixels are not able to get the the rays at more acute angles of incidence.
This is why they really need to better control their experiment.
Kaz
This is why they really need to better control their experiment.
Kaz
bronxbombers
Forum Pro
although the crop sensor with low pixel pitch also used older tech with smaller microlenses, etc.
bronxbombers
Forum Pro
hmm that might be a good point although why then don't they get the same for all camera models? that correction should be fixed, no?
Steen Bay
Veteran Member
The LL/DxO graphs show the light loss in the center of the image/sensor.
Bogdan_M
Senior Member
Hmm, do you mean that camera manufacturers aren't aware of this approximation/mislabelling and they fool themselves into calculating the wrong exposure, based on their own mislabelling?
This is really disappointing...
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http://www.flickr.com/photos/bogdanmoisuc/
Bogdan_M
Senior Member
Not at all.
Read this :
http://www.imageval.com/public/Papers/EI%205678-01%20Peter%20Catrysse.pdf
the relevant part is in the chapter 3. Microlens design.
Look at this post if you need a very brief summary
http://forums.dpreview.com/forums/readflat.asp?forum=1018&message=36025858
Smaller sensors have deeper wells (relative to their surface), therefore obtaining fast enough microlenses is not always possible.
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http://www.flickr.com/photos/bogdanmoisuc/
Steen Bay
Veteran Member
Lenses have a 'natural' light loss at wide apertures (like you explained above), that'll be the same on all cameras, but the LL/DxO graph shows that the (center) loss at f/1.2 is app. 0.42ev with 5D and 0.94ev with 7D. That shows that there is an additional light loss caused by the sensor/camera, and the cameras use different amounts of digital gain to compensate for that additional loss.
clk_walker636320
Senior Member
Can you reference anywhere in the LL article what exactly is the 0 EV baseline? And after that please explain exactly how DXO measured the EV loss and concluded that it was due to angle dependence of the light impinging on the sensor and not something due to the design of the sensor itself like less efficency.
To indicate that the 1DsIII has a .55EV loss due to using a f1.2 lens is not established by the DXO data at all. Why does the 30D have .8EV loss compared to the .95EV loss of the 20D?
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A bird in the viewfinder is worth...
Steen Bay
Veteran Member
Good questions, but sorry, don't know what the 0 EV baseline is. Guess it's something like the EV at f/5.6, and if so, then it follows naturally that the (additional, see above) loss at wide apertures must be caused by the sensor/microlenses.
Again, I can only guess. Could be that the 30D has better (faster) microlenses than 20D.
David J. Littleboy
Senior Member
This batguano craziness implies that exposure should change with focus. Does it?
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David J. Littleboy
Tokyo, Japan
Bogdan_M
Senior Member
Steen Bay
Veteran Member
The craziness was confirmed by John Sheehy in the link below (see his next post too). And focus? Well, since focus affects the local contrast (sharpness) in the image, then I guess you could say that the local exposure changes too, but that has nothing to do with this.
http://forums.dpreview.com/forums/read.asp?forum=1018&message=36777132
What keyboard are you using that has a middle dot?
Also, you can't always be certain that non-standard characters will reproduce as originally intended (although in this case it is readable here). I was once accused of being 6 orders of magnitude out simply because the "µ" character I typed had been propagated to some recipients as a space. Others managed to read it OK and the whole discussion descended into chaos.
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Its RKM
OK, although it was a bit dull outside today I ran off a few exposures and had a look at them in Rawnalyse - sure enough, there are missing codes at fast apertures, indicative of digital gain being applied to the RAW files.
Delving into this a bit more, from the missing codes over the entire RAW range I could calculate, to 4 decimal places, the gain applied (total ADU/used ADU) for each aperture in 1/3rd stop steps. This seems to be considerably less than the values presented in the LL article - even for the 5D.
f/# : Gain
1.4 : 1.0426
1.6 : 1.0283
1.8 : 1.0199
2 : 1.0147
2.2 : 1.0098
2.5 : 1.0078
2.8 : 1.0049
3.2 : 1.0020
3.5 : 1.0010
4 : 1
By the way, the "bumps" seem to be real - I checked those twice.
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Its RKM
Bogdan_M
Senior Member
Nice test. So we know have "independent" numbers for the 7D, 5D and 5D mk II
.But why do you say this is less than the values presented by LL for the 5D. They only present the gain for 3 cameras (picked, of course, among those who behave pretty badly):the Canon 550D, the Sony A350 and the Nikon D200.
Or have you seen something that I've missed?
And could you please elaborate on this
http://forums.dpreview.com/forums/read.asp?forum=1032&message=36785148
I find it pretty hard to believe camera manufacturers would fall into their own trap like this...
--f/# : Gain
1.4 : 1.0426
1.6 : 1.0283
1.8 : 1.0199
2 : 1.0147
2.2 : 1.0098
2.5 : 1.0078
2.8 : 1.0049
3.2 : 1.0020
3.5 : 1.0010
4 : 1
By the way, the "bumps" seem to be real - I checked those twice.
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Its RKM
http://www.flickr.com/photos/bogdanmoisuc/
True. I'm using a Windows machine, so I used charmap (although I happen to know it can be entered using ALT-250, since I program in a language that uses those high valued ASCII characters as delimiters). Using ASCII only, I prefer to use the " " character for multiplication, although no character at all also works (parentheses added if necessary).
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They don't have "deeper wells", as you put it, and nothing in that paper suggests they do.
They have smaller active areas, since they still have to fit in the same connections and circuitry as larger pixels. Therefore they require more optical concentration from the microlens.
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Its RKM
Bogdan_M
Senior Member
I said "deeper wells relative to surface", which means either that they are effectively deeper, or just that theyir surfaces are smaller, and therefre their relative depth is bigger. If we really want to be pedantic, I should have said have smaller surface to stack height ratio, but I was aiming for a more intuitive explanation.
Anyway in the article they suggest that usually what happens when technology advances is that stack height increases (the wells get deeper
).Quote from page 10:
"The dilemma in pixel scaling for CMOS imagers is the following. Small pixel sizes require imaging lenses with small f/#s to obtain a sufficient photon supply at the sensor surface (see section Photon noise). Concentration of light onto the photosensitive area in small pixels requires a microlens with even smaller f/#. Given the small pixel aperture, the
smaller f/# means shorter focal length. The focal length is determined by the sensor thickness (stack height). When CMOS technology scales, the lateral dimensions (e.g., transistor size and pixel size) shrink but the stack height usually increases."
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http://www.flickr.com/photos/bogdanmoisuc/
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