MGrayson
Subscriber and Workshop Member
Playing around with old lenses got me thinking about diffraction. Why? Because we often need to use f/11. When I was shooting a tech cam with a large sensor, f/11 ws kind of standard. And then there's the whole f/64 thing with large format.
But calculations of diffraction blur show that the size of the Airy disk (blurred image of a point) depends only on wavelength and f-number. For 5µ pixels, pixel level blurring becomes dominant at the same f-number independent of sensor size.
And THIS is why viewing at 100% on a monitor is dangerous. Because what matters is the final image size or, given a sensor size, magnification. Large sensors don't have to be magnified as much to make a final image, so they can afford larger Airy disks. The formula for disk size, BTW, is 1.22*f*w, where f is the f-number and w is the wavelength.
Let's make a 16"x20" print from 8"x10" film using f/64. Pick a wavelength of 6,000 Å, AKA 600nm, AKA 0.6µ, a nice Yellow. This gives us an Airy disk of size 1.22*64*0.6µ = 47µ. That looks like a large number, but when we make our 16"x20" print, that becomes only twice as large, so 94µ = 0.1mm = 1/250". In other words, we get 250ppi resolution.
Let's do the same thing with a 645 digital back at f/11. The disk size is 1.22*11*0.6µ = 8.1µ, and that's like one to two pixels depending on the back. But our sensor is a bit over 2 inches and we need a magnification of at least 9 to make the print, so the spot becomes 75µ. Better, but not hugely better than Ansel Adams at f/64. We could even get away with f/13.
And, as most of us have 44mmx33mm sensors, we get a magnification of 12 and so a disk size of 96µ in the final print. Almost exactly the same as the 8"x10" film at f/64.
Of course, modern lenses are sharp at lower f-numbers, but I wanted to see what corresponded to the historical gold standard. And if you've ever seen a 16"x20" print in person of "Clearing Winter Storm", or "Moonrise, Hernandez", the last thing on your mind is "meh, could have been sharper."
Matt
This, BTW, is George Biddell Airy. He's also the guy who chose Greenwich as the location of the Prime Meridian.
But calculations of diffraction blur show that the size of the Airy disk (blurred image of a point) depends only on wavelength and f-number. For 5µ pixels, pixel level blurring becomes dominant at the same f-number independent of sensor size.
And THIS is why viewing at 100% on a monitor is dangerous. Because what matters is the final image size or, given a sensor size, magnification. Large sensors don't have to be magnified as much to make a final image, so they can afford larger Airy disks. The formula for disk size, BTW, is 1.22*f*w, where f is the f-number and w is the wavelength.
Let's make a 16"x20" print from 8"x10" film using f/64. Pick a wavelength of 6,000 Å, AKA 600nm, AKA 0.6µ, a nice Yellow. This gives us an Airy disk of size 1.22*64*0.6µ = 47µ. That looks like a large number, but when we make our 16"x20" print, that becomes only twice as large, so 94µ = 0.1mm = 1/250". In other words, we get 250ppi resolution.
Let's do the same thing with a 645 digital back at f/11. The disk size is 1.22*11*0.6µ = 8.1µ, and that's like one to two pixels depending on the back. But our sensor is a bit over 2 inches and we need a magnification of at least 9 to make the print, so the spot becomes 75µ. Better, but not hugely better than Ansel Adams at f/64. We could even get away with f/13.
And, as most of us have 44mmx33mm sensors, we get a magnification of 12 and so a disk size of 96µ in the final print. Almost exactly the same as the 8"x10" film at f/64.
Of course, modern lenses are sharp at lower f-numbers, but I wanted to see what corresponded to the historical gold standard. And if you've ever seen a 16"x20" print in person of "Clearing Winter Storm", or "Moonrise, Hernandez", the last thing on your mind is "meh, could have been sharper."
Matt
This, BTW, is George Biddell Airy. He's also the guy who chose Greenwich as the location of the Prime Meridian.
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good catch