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## The Technical Side of Depth of Field

This afternoon I decided to take advantage of the rare June rain to break out my macro lens to practice photographing some water kissed flowers from our garden.

I like to think I have a pretty good handle on my most used lenses – the 30mm and 50mm primes. I know what shutter speeds about I need to freeze fast moving toddlers (1/320s-ish), and what aperture I need to bring their whole faces into focus (f2.8 – f3.2 usually does the trick). I use my 60mm f/2.8G Micro for macro photography. With a focal length so close to my 50mm prime, I expected it to handle about the same. One thing that immediately struck me was how large an f-stop I needed to get the whole flower in focus.

Even at f/8 little of the flower was in focus, and flowers are not that big!

Here’s another example, at f/18

So why the incredibly shallow depth of field? Distance to subject seems an obvious culprate. A quick search lead me to the mathematical formula for calculating the nearest in focus point, and farthest in focus point using depth of field from distance, focal length, aperture and something called circle of confusion.

Approximations:

NearestInFocusPoint = s x f

^{2}/ (f^{2}+ N x c x (s – f))

FarthestInFocusPoint = s x f^{2}/ (f^{2}– N x c x (s – f))Where

s is distance to the subject being focused on

f is Focal Length

N is the f-Number and

c is the Circle of Confusion

Then the total depth of field can be calculated as the nearest in focus point subtracted from furthest in focus point and simplifying. For simplicity, let D_{n} be the denominator of NearestInFocusPoint, and D_{f} be the denominator of FarthestInFocusPoint:

DoF = FarthestInFocusPoint – NearestInFocusPoint

DoF = [s x f^{2}/ D_{f}] – [s x f^{2}/ D_{n}]

DoF = [s x f^{2}x D_{n}/ (D_{f}x D_{n})] – [s x f^{2}x D_{f}/ (D_{f}x D_{n})]

DoF = [s x f^{2}x (D_{n}– D_{f})] / [D_{f}x D_{n}]

DoF = s x [f^{2}x (D_{n}– D_{f})] / [D_{f}x D_{n}]Where

D_{f}x D_{n}= (f^{2}– N x c x (s – f)) x (f^{2}+ N x c x (s – f))

From the above equation my hypothesis appears to be correct; as distance approaches zero, the numerator approaches zero and the denominator approaches (f^{2} + N x c x f) x (f^{2} – N x c x f) or (f^{4} – (N x c x f)^^{2}). For my camera, c = 0.02mm with a maximum aperture of Æ’/32, far less than the 60mm focal length. Thus (N x c x f)^^{2} is far, far, smaller than f^{4}. Thus the denonimator, D_{f} x D_{n}, is positive and non zero. Thus as distance approaches zero, DoF approaches zero.

Of course there are practical limitations and focusing distance to subject cannot be zero. My 60mm lens has a minimal focusing distance of about 7.5 inches, or 190mm. Using my camera’s specs, if the flowers were only 10 inches from my lens, at f/3.2 the dept of field is just .1 inches. At f/8 the depth of field is just 0.17 inches. I’d have to back up another foot to have a depth of field at least 1 inch wide.

Values of 0.02 – 0.03mm seemed to be pretty typical circle of confusion values, at least according to the sites I visited while researching this blog post. Even with a very wide angled lens, capable of a very narrow aperture, I suspect my hypothesis would still generally be true. Nikon has a 6mm fisheye lens, and even for that lens the math holds.

Posted in Photography | Tags: Doing the Math, Photography Equipment

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