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Speaking of the physical aperture.

When looking at the 18-200 mm, f/3.5-5.6, I wonder how this works. With diameter D = f/N I get:

  • 18 mm / 3.5 = 5 mm
  • 200 mm / 5.6 = 35 mm

So, does the aperture diameter change when zoomed in? I assume not, so how does this work?

Edit: I now took a picture of the visible size of the iris, or entrance pupil (which was what I meant when speaking of the physical aperture). I focused at the point where I took the picture from. The lens is a Sigma 70-300 f/4-5.6. The tape I used is gaffer tape. Ok, all important details mentioned ;) The results are closer to the lense's f-number (now calculating the other way round than above):

  • 23 mm / 70 mm = 1/3.0
  • 51 mm / 300 mm = 1/5.9

Is this method of measurement more or less valid?

70 mm 300 mm

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3 Answers 3

The physical apterture opening is the same size. It's fully open in both cases.

The reason that the f numbers are different, is that they don't measure the physical size of the aperture, they measure how much light the aperture lets through. If you look into the lens from the front, you see that the aperture appears to become larger when you zoom in. The light that falls into the lens is also affected the same way, so that more light can pass through the same physical opening.

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Thanks for the hint about the visible size of the aperture! This makes completely sense. –  Simon A. Eugster Jul 8 '11 at 13:20

It depends where the iris (the aperture diaphragm) sits in the light path. At the 200mm end of the zoom range, the iris is closer to the pointy end of the light cone (relatively speaking) than it is at the 18mm end of the range. The actual physical iris size doesn't matter so much in terms of the depth of field, since the effective optical aperture is the same. The same goes (more or less) for the bokeh artifacts, except that the physical iris size can affect the character of the bokeh (if not the overall size of the out-of-focus highlights). Where it does make a difference is in diffraction, since that's a property of the physical iris size and the wavelength of light -- expect more diffraction at a given aperture setting at the long end of the zoom range than you would see with a prime lens of the equivalent focal length or with a zoom that starts at a longer focal length (like a 70-200mm).

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f/N is not the diameter of the physical aperture, it's the diameter of the entrance pupil, i.e. the image of the aperture as seen from the front of the lens. And yes, it changes as you zoom: look at the front of your lens while you zoom and you will see.

Edit 1: A related question is “How come a 5 mm aperture is ‘faster’ than a 35 mm aperture?” I don't know whether this was somehow implied by the original question, but anyway, here it goes:

At the tele end, the lens only collects those photons that come from a very narrow field of view, while at the wide end it collects light from a much wider FoV. Then, although the tele setting gets 48 times more light from any given element in the scene, the wide setting collects light from a solid angle 123 times larger. It thus ends up having 2.5 times more light gathering power.

Edit 2: Here is where the numbers come from: I compared the two ends of a 18-200 mm f/3.5-5.6.

The amount of light collected from a given small element of the scene (small enough to always fit within the FoV) scales like the surface area of the entrance pupil, i.e. like D², where D is the diameter. If we compare the 200/5.6 setting to the 18/3.5 setting, the ratio is

(35.7 mm / 5.14 mm)² = 48 (the tele end collects more light)

The solid angle FoV scales like 1/f², then, for this same zoom, the ratio is

(200 mm / 18 mm)² = 123 (the wide end has more FoV)

The amount of light collected from an extended scene (as oposed to a small scene element) of known average luminance scales like (D / f)² = 1/N², where N is the aperture number. For this zoom you get

(5.6 / 3.5)² = 2.5 (the wide end collects more light)

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Thanks for the entrance pupil link! I took pictures of it and added them above. Thanks also for the related question. Can you explain how you get those numbers, or are they just for illustration? –  Simon A. Eugster Jul 8 '11 at 13:16
    
OK, I edited the post adding the details on the calculations. –  Edgar Bonet Jul 8 '11 at 19:14

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