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I'm planning to create a customized aperture effect. Like this one in the Photo.

Courtesy of Lensbaby

I'll be doing a DIY customized aperture that will be attached to the filter. My lens' widest aperture is F/1.8 and so that will be also the widest f-number of my customized aperture or if that's not possible, then the widest possible smaller than 1.8 will be. But I also want it with different f-numbers also, like F/2.8. So my question is What are the diameters of the lens' aperture in physical sizes? (e.g. mm, cm or inch) Do we have any formula for that? My focal length is 35mm if that information is needed.

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I'm not sure what you mean by "pixel size", but is What size hole do I use to make differently shaped bokeh? essentially what you're asking? –  mattdm Aug 8 '13 at 3:00
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Why does the physical size of the aperture matter if you are attaching your mask to the filter? –  Michael Clark Aug 8 '13 at 3:12
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The problem is that pixels are inherently dimensionless. They're just pixels. So that makes no sense. It might be helpful to know what online converter you are using. –  mattdm Aug 8 '13 at 3:23
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actually i just want to get the exact diameter of f-number and using pixel as measurement was just a random selection in photoshop, cause i'm doing it in Photoshop, although i can use mm or inches or cm anyway. such print will be needed in my project. I guess to avoid confusion shall I change it to mm or cm instead? –  Jez'r 570 Aug 8 '13 at 3:32
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We have some questions tagged with "Shaped Bokeh" to explore. –  Esa Paulasto Aug 8 '13 at 6:08

3 Answers 3

up vote 3 down vote accepted

Pretty simple to calculate: the way the aperture is written tells you the diameter. Remember f refers to the focal length:

f/2.0

Means the diameter is the focal length divided by 2.0.

f/4.0

Means the diameter is the focal length divided by 4.0.

Warning: the effective aperture diameter is not necessarily the same as the diameter of the part of the front lens element in use. The aperture in a lens is placed at a point in the lens array where the image is as defocused as possible so as to minimise vignette. Placing a shape over the front lens element will not get you exactly the same effect as placing that same shape in the same plane as the aperture, but it will approximate it. Furthermore if you place the shape in front of the front lens element but you stop down the lens aperture, your effective aperture will still be round.

Thus, for what you are planning to do, the aperture diameter is probably not as relevant as the size of the front lens element.

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To create custom bokeh shapes, the size of the opening in the filter-mounted mask should be based on the apparent or effective aperture, which is more precisely described as the entrance pupil diameter, rather than the actual physical diameter of the diaphragm inside the lens. This is a way of describing the size of the aperture as viewed through the front element of the lens. To calculate the effective aperture in millimeters, divide the focal length of the lens (in mm) by the f-number. D = F/N where D is the diameter of the effective aperture, F is the focal length and N is the f-number.

To get the desired shape, the mask should completely cover the visible parts of the diaphragm when stopped down to the setting you are shooting at as viewed on the optical axis of the lens from the distance the lens is focused. To allow the maximum amount of light through the lens you should make the opening as large as possible without allowing the the diaphragm to be seen. Thus all of the openings in the mask should just barely fit inside a circle with diameter D from the formula D = F/N explained above.

There are a few tricky parts here:

  • Focal lengths and f-numbers for most lenses are rounded to the nearest standardized unit. A 35mm lens may actually have a 33mm focal length when focused on infinity, or it may have a 37mm focal length. It is fairly well known that longer telephoto lenses often fudge on their true focal length. So much so that most lens reviewers that use objective measurements report the actual vs. claimed focal length. It is not uncommon to see a 70-300mm telephoto zoom lens that is actually only a little above 280mm at the long end. Likewise, actual f-numbers are rounded to the nearest standardized unit. Even one-half of the standardized f-numbers are rounded since they are based on the square root of 2 (which is an irrational number). f/5.6 is really closer to 5.7. f/22 is closer to 23 than 22.

  • The focal length of a lens changes as the focus distance changes. When you say you have a 35mm lens, the focal length is only truly 35mm when the lens is focused at infinity. When the lens is focused on a close object, the focal length of the lens changes. As your focus distance moves closer, the opening will need to be smaller - even for the same aperture setting!

  • Most modern lenses are wide open when mounted on a camera. Most cameras focus and meter with the lens wide open, then stop down the aperture immediately prior to the opening of the first shutter curtain. Most cameras have a Depth of Field Preview Button that will allow you to stop the lens down to the selected aperture and hold it there. Some will even let you detach the lens from the camera with the aperture held at a specific f-stop.

  • Most aperture diaphragms, even those that claim to be circular, are not true circles at most f-stops, since the curvature of each of the blades would need to vary as the opening gets smaller and larger. That is why you can often count the number of aperture blades based on the shape of the bokeh a lens produces. If the "short" side of an aperture blade occludes light that is allowed to pass by your mask, it could affect the shape of the bokeh.

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"effective aperture" is a slightly vague term that people use in different ways. The size of the image of the aperture as viewed through the front element (as given by the formula F/N) is the entrance pupil diameter. –  Matt Grum Aug 8 '13 at 9:06

We have a formula for that,

Let: 
F = Focal length
D = Lens Diameter (in mm)
N = F-Number

N = F/D 

So to get your question we changed the formula to this: D = F/N (35mm/1.8 = 19.44mm)

Next is, 1mm = 3.779528px therefore : 19.44mm * 3.779528px = 73.490822px

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Where did "1mm = 3.779528px" come from? That would be a pixel 264µ wide! Typical sensors have pixel pitches between 4µ and 8µ wide. A 4µ pixel pitch would convert to 250px/mm. –  Michael Clark Aug 8 '13 at 3:18
    
got it from here: translatorscafe.com/cafe/units-converter/typography/4-8 –  user19311 Aug 8 '13 at 3:22
    
@MichaelClark That appears to be 1/96th of an inch. 96dpi is a typical screen resolution, so I guess that's where it comes from. –  mattdm Aug 8 '13 at 3:22
    
The whole subject is ridiculously complicated, especially with pixels as a fake unit on new high resolution displays. But basically, the idea that a pixel has an fixed measurement size is not standard. –  mattdm Aug 8 '13 at 3:26
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Okay. I was reading Pixels as in sensor pixels. I still don't see how pixel size has anything to do with it. I also still don't understand exactly what the question is asking. –  Michael Clark Aug 8 '13 at 3:29

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