Digital Depth of Field

[Al]

It is approximately 1/5 the dimensions of a 35 mm frame. Thus when the
lens is set to a "35 equivalent" focal length of 50 mm, the true FL is
about 10 mm. At a given subject distance and f/number, you'll have
about 5 times as much DOF from the digicam. To get the same DOF from
the 35 camera, you need to multiply the f-number by about 5.

For example, f/2 on the digicam gives the same DOF as f/10 (round to
f/11) on the 35 mm camera.

Dave

This is a very good way of calculating DOF.

However, the actual DOF is greater than the formula suggests because
of the limits of resolution of the image sensor.

One practical way of determining DOF is to find the two points at
which the resolution falls to 1/2 of that of the focal point. This can
be done at any distance - just use a sliding log scale to extrapolate
it across the entire focusing range once you have determined any two
points that are 1/2 resolution.

 

[Gisle Hannemyr]

The listed size of the CCD is misleading. I don't know what that
are specifying,

Useless trivia: They actually specify the tube diameter of a Vidicon
video tube (used in the 1950ies) with an equivalent sensor size.

but it doesn't tell you the actual dimensions of the sensor array.
True.

You can find an approximation to the actual size by looking at their
equivalent 35 mm focal lengths. For my Olympus 3040, they tell me
that an actual focal length of 7.1 is equivalent to a 35 mm focal
length of 35 mm and an actual focal length of 21.3 is equivalent to
105 mm. I think the 5050 is similar. 35/7.1 = 105/21.3 ~ 4.93.

I've got a table converting those weird Vidicon designations to actual
dimensions. For 1/1.8" works out as 7.2x5.3 mm.

Since the aspect ratios are different for the Oly and a 35 mm full
frame, it is not clear which lengths they are comparing, but the
answers would still come out in the range 4-5.

The best way to compute the multiplier is to compute the ratio
between the diagonals. 1/1.8" sensors has a diagonal of 8.9 mm,
and 35 mm has a diagonal of 43.3 mm. 43.3/8.9 ~ 4.87. Not
that it matters much, but I think 4.87 it as tad more correct than
your 4.93 figure.

So - returning to the original question - the formula for computing
the f-stop shift on a 35 mm camera that will give identical DOF to a
camera with a different sensor size is log(2)R*2 (where log(2) is the
base 2 logarithm and R is ratio between the sensor diaognals).

Log(2) of 4.87 times 2 yields 4.568, so an Oly C5050z will yield a
DOF that is equivalent to a 35 mm camera stopped down aproximately
an additional four and half stops.

If we do the same computation with a Canon 1D Mk II (1.3 ratio), we
find that the DOF adjustment is 0.757, or ~3/4 of a stop,

 

[Gisle Hannemyr]

Just read an article (Digital Photography Made Easy, UK publication,
pg 31) that basically says the depth of field is different for images
taken with a digital camera than for a film camera. It said that if I
had a f2.8 lens with a digital camera that the final image would have
more of a depth of field near f14.

Is that correct?

It depends on the sensor size of the digital, but assuming they are
talking about typical consumer digicams with 7.2x5.3 mm sensors, you
get a DOF adjustment with respect to 35 mm of about 4.5 stops (see my
other post in this thread how you compute this), so, yes: f/14 looks
about right.

If so, is there a link to the technical reason that the sensors behave
differently than film in this regard?

No link, but it follows from the fact that the a small sensor uses
lower image magnification to capture the same scene than a large
sensor and thus yields a greater DOF.

But it has nothing to do with digital vs. film. You can observe the
same effect on film: Given the same f-stop, subject distance and angle
of view, 110 has greater DOF than 135, which has greater DOF than 120.

Would the same be true for a digital SLR with a f2.8 marco lens,
basically meaning you cannot do shallow depth of field with digital?

Digital SLRs have larger sensors and shallower DOF than consumer
digicams. If you use a Canon 1Ds, it has a sensor which is the same
size as a 135 film frame, and will have exactly the same DOF as an
EOS-1V.

Btw: DOF is also a function of subject distance. I know from
experience that it is not a problem to get a shallow DOF with
7.2x5.3 mm sensor at macro distances.

 

[Rudy Garcia]

It is approximately 1/5 the dimensions of a 35 mm frame. Thus when the
lens is set to a "35 equivalent" focal length of 50 mm, the true FL is
about 10 mm. At a given subject distance and f/number, you'll have
about 5 times as much DOF from the digicam. To get the same DOF from
the 35 camera, you need to multiply the f-number by about 5.

For example, f/2 on the digicam gives the same DOF as f/10 (round to
f/11) on the 35 mm camera.

Dave

Hmmm,

How small is the CoC used to determine DOF on a digital camera?

How small is the effective pixel size in a sensor?

 

[Dave Martindale]

How small is the CoC used to determine DOF on a digital camera?

Depends on who's doing the calculations, but twice the pixel pitch
is a good starting point.

How small is the effective pixel size in a sensor?

Depends on the sensor. In the 4-5 MP consumer cameras that are being
discussed in this thread, the pixel pitch is about 2.5-3 microns.

Dave

 

[Roger Riordan]

The Olympus web site list the size as (in inch) 1/1.8” (.55”) CCD which
is about .99 sq inches a 35 mm images size is about 1.75" x .75" or about
1.31". That computes to about 3/4 not 1/5.

It's the camerar makers that use the Irish math. As a typical example, the
Coolpix 995 data sheet lists the sensor as 1.125" high-density CCD, but
specifies the lens as focal length: 8-32mm (equivalent to 38-152 on 35mm
format). So the lens gives the same coverage as a 35 mill lens with almost five
times the focal length. This implies that the sensitive area is approximately
5mm by 7.5mm. Apparently there is a lot of dead space around the sensitive part
of the sensor.


Roger Riordan AM

 

[Leonard Evens]

The best way to compute the multiplier is to compute the ratio
between the diagonals. 1/1.8" sensors has a diagonal of 8.9 mm,
and 35 mm has a diagonal of 43.3 mm. 43.3/8.9 ~ 4.87. Not
that it matters much, but I think 4.87 it as tad more correct than
your 4.93 figure.

I don't understand your arithmetic. How do you get 8.9 mm from 1/1.8
inches? 1/1.8 inches is about 14.1 mm. Just what does the 1/1.8
inches refer to?

So - returning to the original question - the formula for computing
the f-stop shift on a 35 mm camera that will give identical DOF to a
camera with a different sensor size is log(2)R*2 (where log(2) is the
base 2 logarithm and R is ratio between the sensor diaognals).

An equivalent formula is 2*log(R)/log(2) and you can use the logarithm
to any base you choose, including the common log to base 10 you will
find on most scientific calculators.

Log(2) of 4.87 times 2 yields 4.568, so an Oly C5050z will yield a
DOF that is equivalent to a 35 mm camera stopped down aproximately
an additional four and half stops.

If we do the same computation with a Canon 1D Mk II (1.3 ratio), we
find that the DOF adjustment is 0.757, or ~3/4 of a stop,

Don't some of the newer DSLRs actually use a full 35 mm frame of 24 x 36?

 

[Gisle Hannemyr]

Gisle Hannemyr wrote:

I don't understand your arithmetic. How do you get 8.9 mm from 1/1.8
inches? 1/1.8 inches is about 14.1 mm. Just what does the 1/1.8
inches refer to?

The 1/1.8" refer to the diameter of an ancient Videcon video tube
that in the 1950ies happened to contain a sensor area with a 8.9 mm
diagonal. (I guess the glass tube enclosing the sensor had to be
somewhat larger than the sensor.)

There is no artihmetic. To get to 8.9 mm for 1/1.8" - you have to
look it up in a table of sensor specs. The table tells you that a
so-called «1/1.8" sensor» has a 8.9 mm diagonal.

(Why the camera manufacturers use this confusing and outdated measure
for modern CCD sensors - I can only guess. And my guess is that the
marketing droids like it, 'coz it gives a bigger measure making the
sensors appear to be larger than they actually are.)

Don't some of the newer DSLRs actually use a full 35 mm frame of
24 x 36?

Yes, and they don't have to be that new. The Canon 1Ds (from 2002) and
the Kodak DCS-14n (also announced in 2002) both use full frame. The
Canon will set you back US$ 8000 for body alone.

 

[Kennedy McEwen]

I don't understand your arithmetic. How do you get 8.9 mm from 1/1.8
inches? 1/1.8 inches is about 14.1 mm. Just what does the 1/1.8
inches refer to?

The 1/1.8" refers to the outside diameter of a glass vidicon tube, if
one existed, that would have an active focal plane of the same size as
that on the CCD. It's an archaic scheme, as is the convention of using
1/x sizes instead of actual size, but it has been kept going by the
video world where lens sizes are defined by the vidicon tube diameter
that they were compatible with rather than the faceplate size - when it
started, nobody considered that a solid state TV sensor would ever
exist.

The glass can be thinner on smaller tubes but needs to be thicker for
larger tubes to prevent implosion, so the standard "inch" used in these
measurements is about 16mm of active diagonal on the focal plane.

Hence a 1/3" CCD has a diagonal of around 5.3mm, a 1/1.8" CCD has a
diagonal of 8.9mm. At a diagonal of 43.3mm, a 35mm frame is equivalent
to a 2.7" CCD. Hence the ratio between that an a 1/1.8" CCD is around
4.9.

 

[Mike]

Found some good information on this at

http://www.wrotniak.net/photo/dof/index.html