is DOF affected by cropping factor?

i can see how it is, but someone is arguing with me..

it seems to me, DOF is a product of focal lenght and aperature

opening..it shouldn't make a difference the size of the capture

medium..

am i wrong?

i mean i cant see how that is...

if you crop a picture, does the depth of field change?

http://www.photo.net/learn/optics/dofdigital/index.html

You're both right (or, for the half-empty crowd, both wrong), because the problem was under-specified.

 

Cropping does not change DoF, all things being equal.

 

However, to maintain a constant angle of view from a given perspective in a decreasing-size frame, you need a longer focal length lens and a lower reproduction ratio from subject to sensor. This gives more depth of field than the image taken on a larger sensor from the same position with the shorter lens.

 

This is something you either love or hate about smaller sensors. (I found it disconcerting to switch between MF/LF and 35mm, and now find it much more so when switching between MF/LF and APS sensors).

 

-matt

And of course, switch "longer" and "shorter" in my response above.

No. Depth of field depends on focal length, aperture and magnification ratio (or distance to the subject). In practice, a smaller area of light capture medium means that to frame the same subject you would use a lower magnification ratio or a shorter focal length lens, and this is what makes the depth of field increase.

Oops Matt squeaked in ahead of me!

Assuming the same lens, aperture and focus distance, Depth of Field is governed by circle of confusion size and circle of confusion size is determined by the degree of enlargement needed to get to a given print size.

 

When you crop you need a greater degree of enlargement, hence the acceptable value for the COF decreases hence the DOF is affected (reduced).

 

So yes, format size DOES affect DOF.

On this topic, many people enter the circle of confusion and never leave.

Jeff: LOL!

The primary dependency in DoF calculation is not the circle of confusion, it's the physical

size of the lens opening. If you are comparing 35mm film cameras with the 16x24 sensor

format of the DSLRs and normalizing field of view, you'll find you get an effective gain in

Depth of Field equivalent to about one stop.

 

That is, using a 35mm lens to obtain the same Field of View that you would get with a

50mm lens on 24x36 format, you'll get the same DoF at f/8 with the DSLR that you would

get with f/11 on the film SLR.

 

Godfrey

Check this..(ensure frame size reflects the largest dimension of your sensor)...

 

<a href="http://www.bep.co.uk/photo/Docs/pcalc99cxls.XLS">click here</a>

Read Bob Atkins article on DOF at www.photo.net (this website). He explains it all correctly and completely.

For "normal" distances (neither macro nor infinity), DoF can also be thought of as related to the angle of view and the diameter of the aperture (not the f-stop).

 

For a given film format and a given f-stop, DoF at a given distance diminishes as the square of the focal length (once again, for "normal" distances. For a given focal length and a given f-stop, DoF increses as the size of the film (proportionally smaller enlargement). In the end, it means that for "normal" distances and a given f-stop, DoF increases proportionally to the cropping factor.

 

Worse, with many small-sensor digital cameras, you're almost never shooting "normal" distances (often defined as magnifications between 1:8 and 1:20), being rather "close to infinity" (a "normal" angle of view on a 1/3.2" sensor 6 ft away is a magnification of about 1:150).

 

Put it another way, the diameter of the aperture is what "sees" the world, and the angle of view (combined with the print size) defines how the image is magnified. If you see the world through a 80mm f/2.8 lens, you're "capturing" cones of light that have a base diameter of 28mm. With a 1/3.2" sensor (4.25x cropping), you're "capturing" cones of light 4.25 times narrower, i.e. 6.7mm in diameter. Because those cones of light are narrower, along one of those cones you need to move much further away from the tip to get to a given diameter, which explains why those small-sensor cameras have a large depth-of-field. (yes, I know, the reasoning isn't quite valid in object space, it applies better to image space, but for small depths around normal magnifications it works well enough).

 

The reasoning works the other way: large format photographers take pictures at apertures that often range between f/11 and f/22, sometimes even smaller, and still manage to get a shallow depth-of-field at those apertures.

Good discussion, but sometimes I wonder if the translations between 35mm and various digital formats is that useful. For me anyway, having sold any film gear I had, now I find I'm less interested in what would be what if it were that etc.

 

Having said that I'll contradict myself and ask, to sum things up for my own frame of reference:

 

If I have a 50mm lens at f/1.8 on a 35mm camera or "full frame", what focal length and aperture will give me the identical angle of view and DOF on my Canon 10D with 1.6 "crop" factor?

I guess what I'm asking, is does one apply the crop factor, in 10D's case 1.6, to the aperture. I know 31mm is roughtly equivalent to 50mm angle of view on a 35mm, so would the f/1.8 similarly translate to roughly f/1.0 and create about the same DOF? Or is this too simplistic?

The "crop factor" changes nothing about the lens. A 31mm f/1.8 lens remains a 31mm f/

1.8 lens. All the "crop factor" allows you to do is to understand what field of view a given

lens for your DSLR equivocates to in terms of a 35mm film SLR. Period.

 

So ... all I said was that if you take a DSLR and a 35mm film SLR, fit them with lenses that

provide the same field of view, the DSLR will have approximately one f/stop more DoF

than the 35mm film SLR at the same distance and other settings.

 

Godfrey

Sadly you are wrong, although it is not obvious until you go back to first principles and learn how depth of field was originally visualised/conceived.

My reference was the Ilford Manual of Photography a 1948 edition which I bought as a student photog about then.

One of the principle parameters of depth of field is the magnification introduced to produce an image/print of a given size. The smaller the 'negative' area the greater the magnification, the less the DoF.

 

With film to make a 10x8 print from a 10x8 camera was a contact print but from a Leica it involves enlarging ... so a much smaller circle of confusion is required in the Leica so that when it is enlarged it still looks like a dot to the eye viewing the print at the appropriate distance.

check .... http://www.geocities.com/nikon5700itee/dof.html if you do not truely understand circle of confusion

Dean: Yes, for "normal" magnifications, a 31/1.1 would get you similar results ("normal" being "not macro and not close to infinity". 1:8 to 1:20 magnifications are "normal" in my opinion).

 

Notice that the 50/1.8 (on 24x36) and the 31/1.1 (on 15x22.5) will actually have the same aperture diameter (about 28mm) and project the exact same overall amount of energy on the frame - it all fits together nicely.

 

(At infinity the behavior is too non-linear to be able to use a linear approximation, and at macro sizes the extra relative extension needed to fill the larger frame will introduce differences as well).

Thanks for the clarification, and I see that earlier Godfrey also had addressed this, and had effectively answered it, so sorry if I was being redundant.

 

The reason I asked is I want an effect on my 10D that is equivalent to the 50mm f/1.8 on a 35mm SLR. I like the shallow depth of field I can achieve with my 50mm f/1.8, so I want to widen the field of view without increasing the DOF.

 

It'll be interesting to see any reviews of the Sigma DC 30mm f/1.4 when it debuts.

On the 10D, I use a Canon 28/1.8 for this purpose. It is as close to a 50mm f/1.4 on film

as you can get in the Canon lens line and is quite a nice lens on the 10D. DoF at f/1.8 is

so insignificantly different from what you get with a 50mm lens at f/1.4-1.8 that it would

be hard to tell it apart .. <br><br>

<center>

<img src="http://homepage.mac.com/ramarren/photo/PAW4/large/18.jpg"><br>

<i>© 2004 by Godfrey

DiGiorgi<BR>

Canon 10D + 28/1.8</i><BR>

</center><br>

enjoy, <br>

Godfrey

Thanks for that suggestion Godfrey, I'm quite used to the 28mm f/2.8 on the 10D. I'll look into the 28 f/1.8.

-

 

Bob Atkins has got it exactly right. DOF is ultimately based upon how much out-of-focus you can stand, or in other words, how large a circle-of-confusion (COF) you can stand. To get, for example, an 8x10 from a 90mm lens using a full frame sensor requires less enlargement than from an APS sized enlarger --- so the COFs are made larger with the APS sized sensor --- in other words the DOF decreases with an APS sized sensor.

 

Note my assumption that you will be producing the same sized print. Hence, the print from the APS sized sensor will be more telephoto --- if you made the 2 prints so that the subject size stayed the same then the DOF would stay the same, but the print from the smaller sensor would be smaller.

 

As an aside --- DOF is a subjective judgement based upon the physics of light.

 

 

BTW to determine what size COF is acceptable it is required to specify a viewing distance --- this is usually taken to be the same as the diagonal of the print.