Depth of field - how best to explain it?

[alanr0]

As Gary2306 points out, something amiss here

fPerusing many posts on depth-of-field, I observe facts, data and lore, I now add my 2¢.

A wearisome dialogue on Depth-of-Field:

<snip>

How does all this square with the camera and the displayed image? Suppose we mount a 50mm lens on a 35mm full frame. The format size is 24mm height by 36mm length. Our desire is to make an 8X10 inch print for display. Now the 35mm format is tiny and thus we must enlarge the camera image to make an 8x10. The degree of enlargement is 8.5X.

If the print is to be viewed from standard reading distance the maximum circle size is 0.5mm. To tolerate the 8.5X enlargement, the circle size at the focal plane of the camera must be 0.05 ÷ 8.5 = 0.0059mm. We have thus discovered the required circle size for this set-up at the focal plane of the camera.

All this rather complicated. Typically, for depth-of-field calculations I adopt a circle size of 1/1000 of the focal length. Thus for a 50mm lens, the circle size allotted is 50 ÷ 1000 = 0.05mm.

This is a highly unconventional approach.

You suggest a circle of confusion of 1 milliradian - 3.4 arc minutes. This is larger than commonly used, but not entirely unreasonable.

At a viewing distance of 500 mm, the CoC is 0.5 mm.

If you print from 135 film format (or a 24 x 36 mm FF sensor) to 8 x 12 inches, the magnification is 8.47.

Image-plane CoC is 0.5 ÷ 8.5 = 0.059mm. (you dropped a factor 10 in your example).

By making the circle of confusion diameter a fixed proportion of the lens focal length, you are fixing the angular extent in object-space.

With a 500 mm focal length lens, 500/1000 = 0.5 mm CoC in the image plane requires that your enlargement factor is 1:1 - so you are viewing an image only 1 inch high at 20 inches distance.

This method is convenient plus it allows for typical enlargement based on a “normal” focal length which is the corner to corner measure of the format. To see how this works, consider a compact digital APS-C format size 16mm height 24mm length. Diagonal measure =30mm. Thus the industry assigns 1/1000 of 30mm = 0.03mm as the circle size in camera. To make an 8x10 from this format the degree of enlargement is 12.7. The circle size on the 8x10 will be 12.7 X 0.03 = 0.38mm (within tolerance of 0.5mm.

Let me add that the 1/1000 of the focal length rule-of-thumb is not engraved in stone. Kodak often set this value at 1/1750 and Leica uses 1/1500 for critical work.

The conventional rule-of-thumb is 1/1000 to 1/2000 of the sensor diagonal, NOT the lens focal length.

Nasse (Zeiss) "Depth of Field and Bokeh" (page 13) recommends Diagonal / 1500

This is typically rounded to 0.03 mm for 24 x 36 mm full frame, and 0.02 mm for APS-C

A different rule-of-thumb is that a "normal" field of view is presented by a focal length roughly equal to the sensor diagonal. Is this where your suggestion comes from? Diagonal of 24 x 36 mm is 43.3 mm, so 50 mm is not too far off.

Take c = 0.03 mm as a popular CoC for a full frame sensor.

For a 200 mm lens at f/11, the entrance pupil diameter, A = 18.2 mm

Hyperfocal distance Uh = f (A/c + 1) = 121 m.

Or with Gary2306's 0.02 mm CoC, Uh_apc = 182 m.

See here for a derivation of my hyperfocal formula (equation 21), and a comparison with approaches described by Tom Axford, Bill Claff, and Jeff Conrad.

Some on-line DoF calculators, such as https://www.dofmaster.com/dofjs.html make more approximations, but typically give broadly similar (though not identical) results

How do we calculate depth-of-field?

Hyperlocal distance: Let’s use 1/1000 of the focal length to obtain the hyperlocal distance. We mount a 200mm, 1/1000 of this focal length conveys a circle size of 200 ÷ 1000 = 0.2mm. This day we set the aperture at f/11. We find the working diameter to the 200mm lens set to f/11 = 200 ÷11 = 18.2mm. Now we multiply this value by 1000 = 18.2mm X 1000 = 18,200mm. This value is the hyperlocal distance in millimeters. This value is 18.2 meters or 57.7 feet.

 

[Garry2306]

@alanr0

Re your

"Or with Gary2306's 0.02 mm CoC, Uh_apc = 182 m."

I personally do not worry about FF or APS-C or MF or MFT etc.

I choose an infinity blur that I wish to see in the final image.

For me, as I can do it in my head, The Rule of Ten is a very useful estimating tool, ie hyperfocal_in_m = focal_length_in_mm/10, assuming the CoC_in_um is the focal_length_in_mm

Thus for the case in hand the RoT hyperfocal estimate for 200mm lens 200/10 = 20m, but at a CoC of 200um. It's easy to adjust this to a 20um infinity blur, by simply multiplying by 10.

Of course, I could have chosen any infinity blur I wished, say, 10um, giving me a hyperfocal for my 200mm lens, at a CoC of 10um, of (20/10)*10*2 = 400m.

All in my head

Personally, I wouldn't worry about hyperfocal details for long telephoto lenses, eg 200mm, as gauging such distances is not easy/accurate.

Hyperfocal distances work well for retrofocus/WA lenses and focus bracketing in your head.

For example, take a 16mm lens.

The ROT hyperfocal, at a solid CoC (infinity blur) of 16um, is 16/10 = 1.6m.

Once I know this, I also know where to focus if I need an additional image for the foreground, ie at H/3 = 1.6/3 ie just over half a meter, which, in turn, tells me, using the odds/even rule, I have a near DoF, of the nearest focus bracket, of H/4 = 1.6/4 = 0.4m.

All in my head