# Light fall off on dSLRs - an experiment

Discussion in 'Digital SLR' started by Kennedy McEwen, Mar 16, 2006.

1. ### Paul FurmanGuest

here's another:

Paul Furman, Mar 17, 2006

2. ### Kyle JonesGuest

But by removing the photons that would be arriving from many diffeent
angles might he also have removed some side-effect of photons
interacting that would change the result? I'm not saying, I'm asking.
I know I don't really understand quantum mechanics, but what I've been
able to understand tells me that deconstructing a particle system in an
experiment like this _might_ produce a counterintuitive result.

Kyle Jones, Mar 17, 2006

3. ### StaceyGuest

Why not just shoot the same scene using a wide angle lens on both a digital
and a film camera body, scan the film and compare them like that? If they
both end up with the same fall off, you've answered the question.

This other test sounds like a BS way of "proving" this, especially given you
wanted a specific outcome to your test before you did it.

Stacey, Mar 17, 2006
4. ### bjwGuest

When the sensor is at an angle to the source, foreshortening
means fewer photons are incident on the sensor per area.
If the surface is matte, the reflected photons are redistributed
over all angles, with an intensity given by the cosine law, which
is a statement about Lambertianness (Lambertiality?) But
this can't change the fact that there are fewer photons incident;
the incident intensity depends on the angle of illumination,
although the fraction reflected may not.

<http://en.wikipedia.org/wiki/Lambert's_cosine_law>
"This means that although the radiance of the surface
depends on the angle from the normal to the illuminating
source, it will not depend on the angle from the normal
to the observer." (in the section on Lambertian reflectors)
Independent of Lambertianness, think of how it behaves before
getting all the way to the extreme angle. Imagine that the angle
of incidence is 70 degrees off normal - there are cos(70) = 0.34
times as many photons incident per area. This is hard to see
by eye, but it's real. It's also why incident light meters use those
little white domes rather than a flat white disc.

It's like the higher latitudes in winter. The sun's rays are
incident on the earth at a greater angle, so there's less
energy/area, so it gets colder.
All I'm questioning is the mapping between the pixel numbers
in the jpeg and the number of photons detected per pixel, which
is what we really want to know. One could take an underexposed
noisy picture, multiply all the pixel values by 100, and it would still
be an underexposed noisy picture.

The other possibility is that due to fluctuations in whatever
(shutter speed? LED output?) the test didn't detect the ~15%
difference that should be there due to a statistical fluke.
I kind of doubt this. But if it were true, it would mean you'd
have to take a number of exposures to figure out what the
noise in each measurement was. (Maybe you did this and
didn't say.)

bjw, Mar 17, 2006
5. ### Kennedy McEwenGuest

No, I am not. There don't seem to be many sensor designers who do say
that there should be light fall-off due to microlens arrays - only those
in the 4/3 consortium, and none of the articles saying that come from
the sensor manufacturers, only the camera manufacturers. Also, as
already mentioned, the problem might even be more of an issue with their
smaller pixels than it is with larger, less noisy pixels.
I would be very surprised if some sort of similar test wasn't actually
performed at least on a batch basis by the sensor manufacturer. I know
that we do something similar to this with our sensor production, which
is what prompted the idea in the first place. Then again, our sensors
routinely operate with f/1 lenses and faster, so extreme angular
response is important.

Kennedy McEwen, Mar 17, 2006
6. ### Philip HomburgGuest

Except that a sensor element with a microlens is a bit more complex than
a little square hole.

Does the anti-alias filter have any effects that depends on the angle of
light?

Philip Homburg, Mar 17, 2006
7. ### Kennedy McEwenGuest

No - the light from the lens is incoherent so constructive and
destructive interactions cancel each other out. If the lens was
photographing a laser illuminated scene then perhaps there could be some
issues, but that is rather unusual and people are claiming this effect
is significant in normal light.

Kennedy McEwen, Mar 17, 2006
8. ### Philip HomburgGuest

But if his test is repeatable, he does have point. How can a DSLR show
additional light fall-off compared to film when the sensor itself show
not exhibit that property?

Philip Homburg, Mar 17, 2006
9. ### Kennedy McEwenGuest

Not sure I understand what you are describing here, but I measured about
half the loss you are estimating, so it might not be very far from an
emipirical explanation.

Kennedy McEwen, Mar 17, 2006
10. ### Kennedy McEwenGuest

Why not just shoot the same scene using a wide angle lens on both a digital
and a film camera body, scan the film and compare them like that? If they
both end up with the same fall off, you've answered the question.
[/QUOTE]
Because, as others have pointed out, that is much more complicated to
get quantitative results than your nave question suggests.
Actually, I expected to find about half a stop difference - I was
extremely surprised by an entirely negligible difference.

Kennedy McEwen, Mar 17, 2006
11. ### mark.thomas.7Guest

(sorry - long post) (O:
To do that I would need to know what angle you attained - you haven't
said yet..?
And then I would have to examine a typical wide angle lens for that
camera to determine where the *opposite* outer edge of the rear element
would be - when it was focused at whatever distance puts the rear
element closest to the sensor (infinity I presume).. Not owning a 5d
and suitable wide angle, I can't do that.

And to show I'm being fair, I would also need to know how much light
from various areas of the lens will actually *hit* the sensor from
those angles - does the opposite edge actually contribute much? I
suspect not, but without knowing the makeup of the light spread across
the sensor, it remains an unknown variable that would, or at least
could, affect the result..

And no matter how you look at it - there is a bit of a problem with
your results - (compounded by the fact that you didn't actually state
*what* incident angle you achieved)... That problem is that all of the
actual, published technical data that I can find (which isn't much
admittedly) on ccd and cmos sensors, indicates that they suffer
noticable and significant light fall off at angles over 10-15 degrees
away from incident. Losses of 15% and more, not 2.6%...

Your description of the experiment raised some other issues, too, eg:
But hang on - the lens *does* in fact restrict the light very
effectively to the image circle of that lens - how was your light
source collimated (which is not the same as a 'point source')? Could
there have been reflections from the chamber, focusing screen, etc? To
do this properly you would surely want a thin 'tube' of light.
This concerns me, because that description shows you have a significant
amount of light hitting the sides of the mirror chamber, the focussing
screen, etc and bouncing around onto the sensor- this would not happen
(as much) with a lens, because there is effectively nothing much
outside the image circle.
OK, but what angle *is* that?
I'm not sure I am convinced of that. I don't have a DSLR in front of
me, but when a wide angle lens is fitted and focused to infinity, and
the *opposite* side of the rear element is used as a potential light
source for an opposite-edge-located sensor, I would have thought it
would be quite a bit steeper angle than you could achieve without the
lens. But without some much more precise detail, i can't really argue
this.
Like I said, it sounds like there is a lot of light bouncing around!!!
This is a rather worrying statement, and as I stated above, it suggests
that the sensors may be getting illumination from areas other than your
point source.
Could we see the images and exif data? I don't usually work in 'PS
levels'! I would imagine by their nature they are not ridiculously
large files. It would also be interesting to see what the 'unexposed
areas' actually look like.. As was stated above though, RAW data would
be more interesting.
That's where I disagree. I don't think this is a simple test at all!!
Categorically? Hmm. I used to work in the sciences, and I have heard
people get very excited before.... and then realise that somewhere in
their data there is a significant 'whoops' (or several..)...

And it flies in the face of the figures I have seen, eg Figure 5 on
that kodak spec sheet I linked to. I'll try to find another set of
data for Canon or Sony to see if that is just an unusual/poor sensor
design.

Please don't get me wrong, Kennedy - I appreciate the effort you have
gone to and it's a fascinating topic, but be open to suggestions to
improve the experiment, and don't dissmiss the naysayers. Sometimes,
naysayers are very useful people to have around! I'm happy to have my
arguments shredded, and if they are wrong I'll give up gracefully and
congratulate the 'winners', *but I want links and references and
verifiable data*! Too many people here with 'opinions'... Me
included!

And I have to say that the more research I do into this (yes, I have no
life!), the more conviced I am that there *is* an issue with light
incidence. How much, I don't know, and I am frustrated by the fact
that it should be fairly easy to test *properly*, using the same wide
lens on a full frame Canon-DSLR and film SLR. But has anyone done it?
- if they have, I can't find it.

PS There's another interesting read here...

http://www.swissarmyfork.com/digital_lens_faq.htm

Scroll down to "3) Do we really need digital lenses?" It's an old
document, and I'm not suggesting that Mr Wisniewski is a foremost
authority, but it sums up my argument pretty well, and his numbers are
backed up by what I read from the manufacturers and designers of
sensors. In simple terms, they *are* sensitive to angle of incidence
where film is not, and from what I have seen and read, this *can* be a
problem that is worse on digital than on film.

Just depends on the lens and the sensor. (o: Like I said, I'd *love*
to see the specs of the Canon CMOS, but I haven't found it yet - and
I'm not sure, if I was Canon, that I would want it to be 'out
there'....

In the same way that Olympus want to push their telecentric approach to
help boost the 4/3 system, Canon and Nikon and maybe Sony might be
biased in exactly the reverse way...

(O:

http://www.dalsa.com/dc/documents/I..._Whitepaper_Digital_Cinema_00218-00_03-70.pdf
Check pages 7 thru 9. Before dismissing it, find out who 'Dalsa' are..

http://oemagazine.com/FromTheMagazine/feb02/detectors.html
"Microlensed imagers ... show a strong sensitivity dependence on
incident photon wavelength and angle."
OK, it's an old article, but the basic CCD and CMOS design hasn't
changed much. And again, find out who James Janesick is...

mark.thomas.7, Mar 17, 2006
12. ### Kennedy McEwenGuest

You don't understand this, do you. I didn't *want* to "know the
difference between using film and digital"!
Because that "something else entirely" *is* what I wanted to know!

That "something else entirely" was precisely how much the sensitivity to
incident light angle the digital sensor (which you and many others
continually cite without any reference quantifying it) actually was. So
I set out to measure that - not the difference between film and digital,
but to quantify how big this effect that *you* were claiming dominated
corner light fall-off on the 5D actually was. Quantify the cause, not
the symptom.

The *implication* of this being zero is that there is no difference
between film and digital sensors *because* that is the only difference
people like you ever do cite. Over to you Batman, come up with some new
excuses why digital full frame sensors should be different to film,
since your original justification is now in flames!

Kennedy McEwen, Mar 17, 2006
13. ### mark.thomas.7Guest

But if his test is repeatable, he does have point.

ahem. repeatability is *not* the only criteria for proof of a
postulation..!! I reckon I would probably get similar results if I
repeated the test, but I am not at all convinced that this test is
valid. See above posts for details.

mark.thomas.7, Mar 17, 2006
14. ### w.beckleyGuest

One problem that I see from such a test is that film itself is an
unreliable medium for this kind of test. Which film stock would you use
for the test? How would you expose the test? How would you process the
test? How would you print the test? Each stage of this could change the
outcome of your test, to exagerate or hide exposure variations in the
corners, and it would be hard to find a neutral happy medium.

Shooting negative / print film, for example, would mask vignette and
other optical defects more than, say, Velvia would. Which one is our
benchmark for what "film" is?

Kennedy's test, which is very sound in its approach, not only removes
the variables of photochemical finishing, it also removes the variable
of the lens, whose inherent vignette is going to make it difficult to
imperically measure any exposure variation introduced by either medium
independent of the lens.

Which comes to the final reason a film-to-digital comparison would be a
waste: how do you imperically measure either of these in a way that is
relevant to the other? Print both of them? How? Scan the
slides/negatives? How? Printing and scanning both change the outcome as
much as everything else.

There are simply too many variables to make a worthwhile test by
shooting the two media against one another. Empirical testing is only
useful if there is one variable at play. And in this test, there is:
the angle of incidence of the light striking the sensor.

Will

w.beckley, Mar 17, 2006
15. ### David J TaylorGuest

Kennedy,

Thanks for that interesting test - it has provoked a lot of discussion!

One factor I have not seen mentioned is the light spectrum. I am guessing
that people are seeing light fall-off in blue skies with wide-angle
lenses. It would be interesting to see if the fall-off differed between
the red and blue ends of the spectrum.

I would also suggest measuring with RAW rather than JPEG data, as the
differences you are seeing are rather small.

Cheers,
David

David J Taylor, Mar 17, 2006
16. ### Andrew HaleyGuest

It is showing around 5%.

(73.80 / 71.82) ^ 2.2 = 1.06.

Andrew.

Andrew Haley, Mar 17, 2006
17. ### mark.thomas.7Guest

..good, trivial, and robust..

Oh, really? You seem to be carefully avoiding some posts above that
show how non-trivial and non-robust this experiment is.

And please post just ONE SINGLE link to one of the 'people' who know
sensor design and state that incident angle is irrelevant. I've posted
several that state the exact reverse, including a graph from a sensor
manufacturer... It appears their method must have used a slightly
different methodology, or perhaps Canon CMOS sensors are just
completely different to other sensors..?

And to go back to basics (feel free to argue these point by point):

1. The problem is 'light fall off on DSLR's' - yes or no?

2. Hands up those who shoot their DSLR without a lens?

3. If the lens is kept the same, and you use it on a film camera (eg a
Canon SLR) and then a digital camera (eg a Canon FF DSLR) at the same
exposure, same focus setting, same scene, same lighting... You could
simply look at the images and compare. Too tricky? (And yes, of
course you would use transparency film to avoid processing issues)

Can you *please* explain how a 'better' test is to remove the lens
altogether, use a non-collimated light source, don't tell us what
angles were achieved, and not even know what angles *might* be expected
from a typical problematic wide angle lens....? Sheeesh.

If you can't get that, there is no hope for you, and you need *not*
attend any more of my science lectures.......

other problems' are incomprehensible..

The simplest way to identify the effect is by looking at, and
measuring, images that *show the problem*. And doesn't the problem
(allegedly) occur when you use a wide angle lens???? By pulling off
that lens and running this experiment you have *added* a whole new set
of variables and substituted something that is NOT the same. As I and
others have stated, a lens has an image circle - the light is
effectively constrained to that area - a point light source is not
collimated and is quite different - so even silly things like the small
amount of reflected light off the walls of the mirror box *could* be
relevant... But you can't know because you have removed the lens......

sigh.

I give up.

mark.thomas.7, Mar 17, 2006
18. ### Philip HomburgGuest

Can you give a message ID of a post that summarizes your position?

I did not find anything in the 'above posts' that clearly explains why
his experiment is the wrong way of measuring the sensor's response to
light with a maximum angle of incidence.

Philip Homburg, Mar 17, 2006
19. ### Bronek KozickiGuest

[...]

you did not use coherent source of light, thus your test proves nothing.
And, BTW. how is it that sensor vendors document much bigger light
falloff than the one you "detected"? Do you even have idea what's the
actual mechanism of this light fallof, so that you know what to measure?

B.

Bronek Kozicki, Mar 17, 2006
20. ### Roger N. Clark (change username to rnclark)Guest

Yes, you can do this qualitatively. Print film has similar
response curves to digital cameras. I've shown that on my web site.
Your f/# and shutter arguments are irrelevant. What you want to
test is the relative fall of from center to corner. It
matters not what the exposure time is. It does matter what the
f/stop is, but most decent systems should reproduce to a
fraction of an f/stop with the same lens. And if you do the
test with the lens wide open, the stop is not moving.

In my QE testing, I specifically changed aperture, and of course
one must change exposure. Normalizing aperture, the plot
is quite linear, showing that both shutter and aperture
are predictable and reproducible to a fraction of a stop.
See Figure 3 at:
http://www.clarkvision.com/imagedetail/evaluation-1d2

Roger

Roger N. Clark (change username to rnclark), Mar 17, 2006