Technical question - diffraction
R. Saylor
have my digital camera set at f/4 with the zoom at 50mm (equivalent
focal length for a 35mm camera).
Now, with these settings, the actual focal length is about 10mm, so
the diameter of the lens opening is about 10mm/4 = 2.5mm. On a 35mm
camera with 50mm focal length, a 2.5mm lens opening would occur at
f/20.
Since diffraction depends only on the absolute size of the lens
opening, is it then correct to assume that my digital camera at f/4
would exhibit about the same image deterioration due to diffraction as
a 35mm camera at f/20?
If this is the case, then diffraction kicks in significantly before
f/5.6, which, according to the usual rules of thumb, should be the
"sweet spot" for my lens. Kinda disturbing.
Richard
Marvin Margoshes
"R. Saylor" <rlsa...@ix.netcom.com> wrote in message
news:cdnhbtc4b9l1a4596...@4ax.com...
Is diffraction really a problem with cameras? Remember, the aperture
diameter needs to be comparable to the wavelenght of light for diffraction
effects to become significant, and the wavelength of yellow light (approx.
the middle of the visible spectrum) is 0.5 micrometers.
Michael McGuire
>
> Is diffraction really a problem with cameras? Remember, the aperture
> diameter needs to be comparable to the wavelenght of light for diffraction
> effects to become significant, and the wavelength of yellow light (approx.
> the middle of the visible spectrum) is 0.5 micrometers.
his is not true. Diffraction can be a problem for cameras at high f
numbers. The diameter of the diffraction spot is given by
d = 2.44fL
where f is the F# and L is the wavelength of light. If we take L as 0.5
micrometers as suggested, and f 22, we get a spot diamter of about 27
microns. For 35mm cameras the diameter of the circle of confusion that
is used for calculating depth of field is 20 micrometers, a rather
comparable number. To get onto the first question, at f4 the diameter of
the spot is about 5.3 micrometers at f4. But to get the same size final
image, you are going to have to enlarge by a factor of 5 more, so indeed
you are back where you started.
Mike
--
Michael McGuire Hewlett Packard Laboratories
email:mmcg...@exch.hpl.hp.com P.0. Box 10490 (1501 Page Mill Rd.)
Phone: (650)-857-5491 Palo Alto, CA 94303-0971
************BE SURE TO DOUBLE CLUTCH WHEN YOU PARADIGM SHIFT.**********
Tony Spadaro
free with small sensor digitals is pretty much bushwaa. It is also why 35mm
pictures don't look like large format pictures.
The human eye can be easily fooled though. so it's not as bad as it
sounds - as you can tell by looking at the pictures.
--
Tony Spadaro
Chapel Hill artist Tony Spadaro
The Camera-ist's Manifesto
http://home.nc.rr.com/tspadaro/
Dark Alley Photography
http://www.homeusers.prestel.co.uk/magor/tony
a Radical approach to photography
"R. Saylor" <rlsa...@ix.netcom.com> wrote in message
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Molly Bender
effect. To calculate what the actual spread is on film, you have to
multiply it by the distance. (i.e focal length) thus it is proportional to
focal length/aperture or the f stop number. For a given F stop, the effect
at the film/CCD plane is the same regardless of the focal length. However:
since CCDs are smaller than 35 mm film, they are magnified by a factor 5
greater than 35 mm film to get the equivalent picture size. (Note this is
not true for those of you fortunate enough to own a professional camera). I
suppose that corresponds to about 3.5 f stops and I believe that the 0.035
mm COC occurs at about F-22 so I guess you would start noticing it at about
f-8 on a digital camera.
"Tony Spadaro" <tspa...@ncmaps.rr.com> wrote in message
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R. Saylor
It seems to me that you are saying that one can notice the effect of
diffraction on sharpness at f/22 with the 35mm format. I.e., points
in the plane of focus produce a COC on the film which is noticeable
in, say, an 8x10 at normal viewing distance. It seems to me, then,
that with the average consumer digicam, the COC on the sensor would be
1/5 the diameter of the COC for 35mm, and this would occur between
f/4 and f/5.6 for the corresponding focal length (i.e., at the same
absolute aperture diameter). When multiplied by 5, this would be the
same size COC as that for 35mm, and thus the effect on an 8x10 would
be the same. I.e., at some aperture slightly smaller than f/4,
diffraction effects would affect image quality.
Richard
On Thu, 22 Mar 2001 06:48:33 GMT, "Molly Bender" <sewm...@home.com>
wrote:
R. Saylor
Leica 50mm/2 (and I'm sure there are many others), are essentially
diffraction limited. I.e., image quality is about the same from wide
open until an aperture is reached at which diffraction effects begin
to deteriorate image quality. This occurs at about f/5.6 with my
50mm/2, not that this would necessarily be noticeable on an 8x10, but
it shows up on the test bench. Many consumer lenses are not like
this, including those on consumer digicams. Such lenses are somewhat
softer wide open and get better as the aperture size is decreased.
However, a point is reached where diffraction cuts in, and further
improvement in sharpness (due to smaller aperture) is offset by
increased diffraction.
My feeling (based on mathematics and physics, not the test bench) is
that diffraction effects will be more severe at a given f-stop with
digicams because the diameter of the aperture opening is much smaller
(on the average about 1/5) than with the 35mm format. Thus, maybe the
"sweet spot" rules (about 2 stops down from full open), which are
applicable to 35mm, need to be revised for typical consumer digital
camera lenses.
For example, without doing a really careful test, it seems that my
digital camera at f/2.8 (wide open) is sharper than at f/8, which
would seem to indicate that there are diffraction problems at f/8 (and
I suspect before that). By the way, I'm using a tripod, so shutter
speed is not a factor.
Richard
Kennedy McEwen
Margoshes <physche...@telocity.com> writes
>
>
>Is diffraction really a problem with cameras?
Yes
> Remember, the aperture
>diameter needs to be comparable to the wavelenght of light for diffraction
>effects to become significant,
Wrong - see below.
>and the wavelength of yellow light (approx.
>the middle of the visible spectrum) is 0.5 micrometers.
>
Also wrong. Yellow light is closer to 0.6microns
(cf. the bright yellow sodium D-line doublet at 0.589um & 0.5896um)
The simple explanation for diffraction is that the radius of the Airy
disc (defined as 1/e of the peak) is 1.22*L*F, where L= wavelength of
light and F is the f/# of the lens. Using your figure of 0.5um, an
f/5.6 lens will have an Airy disc radius of 3.5um - roughly the same
size as some of the pixels in digicams. (Note that for red light,
slightly common in flesh tones, this disc radius is around 5um!).
More specifically, the MTF of a diffraction limited circular lens
aperture is given by the equations:
MTF = (2H - sin(2H))/pi
where cos(H) = L*S*F radians
L= wavelength of light in mm
S= spatial frequency at the focal plane in cy/mm
F= f/# of lens.
This is a near triangular function which starts at MTF=1 for 0cy/mm and
decreases almost linearly with increasing spatial frequency until
approximately 10% MTF, where the rate of reduction flattens off.
Now, it is fairly clear from the second equation that when S=1/(L*F),
cos(H)=1, so H=0 and the MTF of the diffraction limited lens, given in
the first equation, is also zero.
So the maximum spatial frequency a perfect lens can resolve is given by
1/LF. Using your figure for light wavelength, this results in an f/5.6
optic resolving up to a limit of 357cy/mm on the focal plane.
For the focal plane width of approximately 7mm, this equates to a
maximum lens resolution of 2500cycles/picture width, or roughly speaking
5000pixels. Since this is many more pixels than you will find along one
side of a digital camera image, you might think that the lens
diffraction would be less than significant - wrong! This is the
diffraction limited resolution - where the lens ceases to resolve
anything at all.
You really need an MTF of approximately 20% in the image for something
to be visibly resolved - and that is the product of the MTF of the lens
and the sensor pixel.
The pixel MTF is also well defined by the equation:
MTF= sin(pi * A * S)/(pi * A * S)
where A is the width of the pixel in mm.
It shouldn't take you too long to figure out that above f/4.5 the lens
diffraction MTF begins to dominate for the 3um pixels found in the
cheaper 3Mpix cameras. Under this condition, the MTF of the pixel and
the lens are approximately 40%, giving a total MTF for the camera of
around 20% at 185cy/mm. So beyond f/4.5 the lens diffraction becomes
more significant than the geometric MTF of the sensor pixel!
To calculate the limiting f/#, look at the number of pixels present in
the highest resolution image from the camera. Say 2048x1536 in a 3Mpix
camera. So the highest resolution from the sensor is 2048 pixels in
approximately 7mm - or about 150cy/mm, assuming 2 pixels per cycle.
Using the equations above, this results in a sensor MTF of around 70%,
so the MTF of the lens required to get a total camera MTF of 20% is at
least 30%. An f/8 lens has an MTF of 25% at 150cy/mm.
All this is for a diffraction limited lens, which most optics on amateur
digital cameras don't even approach.
So, in conclusion, Richard's concerns are more than justified,
diffraction MTF begins to dominate at approximately f/4.5 and limits the
camera resolution completely at f/8 - which is why quite a few cameras
limit f/8 as the maximum f/#.
Bear in mind that most lens MTFs get nowhere near diffraction limits,
particularly at low f/#s and away from the centre of the field of view
and you rapidly reach the conclusion that most 3Mpix amateur digicams
have no more resolution than you could get from interpolating images
from a cheaper 1Mpix camera!
Here endeth the optics lesson. ;-)
--
Kennedy
Yes, Socrates himself is particularly missed;
A lovely little thinker, but a bugger when he's pissed.
Python Philosophers (replace 'nospam' with 'kennedym' when replying)
R. Saylor
Richard
Lennart Nilsson
"Kennedy McEwen" <r...@nospam.demon.co.uk> skrev i meddelandet
news:wtRWyXCa...@kennedym.demon.co.uk...
> You really need an MTF of approximately 20% in the image for something
> to be visibly resolved - and that is the product of the MTF of the lens
> and the sensor pixel.
>
> The pixel MTF is also well defined by the equation:
> MTF= sin(pi * A * S)/(pi * A * S)
>
> where A is the width of the pixel in mm.
>
> It shouldn't take you too long to figure out that above f/4.5 the lens
> diffraction MTF begins to dominate for the 3um pixels found in the
> cheaper 3Mpix cameras. Under this condition, the MTF of the pixel and
> the lens are approximately 40%, giving a total MTF for the camera of
> around 20% at 185cy/mm. So beyond f/4.5 the lens diffraction becomes
> more significant than the geometric MTF of the sensor pixel!
>
> To calculate the limiting f/#, look at the number of pixels present in
> the highest resolution image from the camera. Say 2048x1536 in a 3Mpix
> camera. So the highest resolution from the sensor is 2048 pixels in
> approximately 7mm - or about 150cy/mm, assuming 2 pixels per cycle.
> Using the equations above, this results in a sensor MTF of around 70%,
> so the MTF of the lens required to get a total camera MTF of 20% is at
> least 30%. An f/8 lens has an MTF of 25% at 150cy/mm.
Good explanation, just a minor completion:
You have used the same MTF for the whole sensor as for a single pixel and
neglect the "phase-factor" (sensor-MTF = pixel-MTF * phase-factor). In this
case (2 pixels per cycle) the phase-factor varies between 0 (!!!) and 1, so
a sensor-MTF of 70% is a bit optimistic, an average of roughly 50% seams
more reasonable.
Lennart
Kennedy McEwen
<lenna...@telia.com> writes
>
>Good explanation, just a minor completion:
>You have used the same MTF for the whole sensor as for a single pixel and
>neglect the "phase-factor" (sensor-MTF = pixel-MTF * phase-factor). In this
>case (2 pixels per cycle) the phase-factor varies between 0 (!!!) and 1, so
>a sensor-MTF of 70% is a bit optimistic, an average of roughly 50% seams
>more reasonable.
>
expect to achieve. It also neglects readout MTF etc, which is almost
the same as the geometric MTF for large detectors or the typical MTF of
a wide open lens. I also used Marvin's 500nm light wavelength
throughout, where the peak wavelength in the visible spectrum is 10%
longer.
Practical results are considerably worse than the optimal rosy picture I
painted. ;-)
Steve West
the nikon 950 and 990 on resolution targets. I expected that a low f#s,
the resolution would be worse, since I believed that lens aberration
would overwhelm diffraction effects when the aperture was larger. I was
wrong, and the resolution clearly improved as the effective aperture was
increased.
swest
Don Stauffer
One of the common design techniques in optical systems is to set up so
that blur circles from each source of blur is about equal. This is a
useful rule-of-thumb.
BUT!- the shape of the spot profile from diffraction is quite different
than from most geometrical aberrations. The effect of diffraction is
more of a softening than a blurring. Some people LIKE the effect, and
use it artistically.
There used to be (may still be) 'soft focus' filters available that make
use of diffraction. Some photographers made theirs by placing a piece
of very sheer nylon stocking in front of lens. Each hole in the piece
acts as a diffraction source.
Diffraction-based softening is very useful for portrait work. In fact,
some portrait lenses are DESIGNED to have enhanced diffraction blur.
Also, some portrait photographers, for same reason, do portraits at
higher f/#, which does require faster film and slower shutter speed.
--
Don Stauffer in Minnesota
stau...@usfamily.net
webpage- http://www.usfamily.net/web/stauffer
Charlie Ih
Even if I do, I cannot not explan as good as you have done and less complete.
This is fundamental reason why most DC's having maximum F# around 5.6.
There have been very few DC's having a F# around 1.4 because that would make
the camera too bulky and more costly. Thanks again for the excellent job.
In article <wtRWyXCa...@kennedym.demon.co.uk>,
--
Charles S. Ih
302-831-8173, FAX 302-831-4316
e-mail, i...@ee.udel.edu
Dave Martindale
>My feeling (based on mathematics and physics, not the test bench) is
>that diffraction effects will be more severe at a given f-stop with
>digicams because the diameter of the aperture opening is much smaller
>(on the average about 1/5) than with the 35mm format.
It's not the diameter of the opening that matters, it's the f/number.
So the 50 mm lens with 25 mm entrance pupil (f/2) on your 35 mm camera
and the 10 mm lens with 5 mm entrance pupil (also f/2) on your digicam
have the same diffraction limit - despite the one aperature being
much smaller than the other.
On the other hand, you enlarge the digicam image 5 times as much for
a given print size, so the same amount of diffraction blur looks
worse for the digicam image. And, thus, you'll want to operate the
digicam at generally a larger aperture (smaller f/number) to reduce
the diffraction. This also means that the digicam needs to have less
aberration-caused blur than the 35 mm camera lens, too.
Dave