DOF = f(TOD)?

[akptc]

This is probably a really dumb question, but I was just browsing for CV 35/1.2 image samples and saw these pics, on a website by Koji Kawakami, which suggest that DOF at maximum aperture is larger than DOF during daytime. I know it sounds silly and physics apply all the time but why do I see great DOF at f1.2 in the night shots but paper-thin DOF at f1.2 in the 3rd, daytime shot?

Image 1 (night)
Image 2 (night)
Image 3 (day)

 

[rogue_designer]

Appears to be a function of subject distance. The closer the subject, the thinner the apparent depth of field will be. I'm sure there is a formula and mathematic reason. But I am too drunk and too tired to effectively track it down.

 

[clarence]

It is just as rogue_designer says. Depth of field is a function of subject distance. I do not know the mathematical formula for this, but if you look at the DOF calculator on the barrels of your lenses, you will notice this. The images you have linked to cannot be used for a depth-of-field comparison because the night shot is focused from a distance while the day shot is focused on a near object. The intensity of light plays no role at all in determining depth of field.

Clarence

 

[akptc]

Thanks, I get it now. Found this site with a DOF calculator. It shows that:

at f/1.2 and object distance of 5 ft, DOF = 0'5.26 (tiny DOF)
at f/1.2 and object distance of 5 ft, DOF = 17'3.43 (large DOF)

Another late-night mystery solved... 🙂