what determines the minimum aperture on any lens

<p>I have read, that aperture opening size bears a mathematical relationship to the lens focal length. That would kind of suggest that longer focal lengths should more easily accommodate apertures in the 1/45 to 1/64th range. However, many lenses only stop down to 1/22. Also, many Nikon 50mm lenses only stop down to 1/16th.<br>

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So, is there a limitation on the smallest aperture?<br>

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Some of the older M42 designs eg. 50 f/2.8 do seem to stop down to f/22, and the lens diameters are not larger than more modern designs.</p>

<p>Carey,</p>

<p>I have no experience in the specific physics of lenses, but you may be on the right track. The longer the focal length, the smaller the aperture tends to be able to get. Perhaps it has something to do with the fact that the blades are further from the sensor, so the light has more elements/more space to pass through to create a large enough image for the sensor to gather? Might explain why flatter lenses only go to f/16 sometimes. If I remember correctly, my kit lens will only stop down to f/22 at 18mm, but will go to f/32 or f/36 at 55mm. I'll have to check that.</p>

<p>--Ryan</p>

<p>2 cents worth; It is easier technically to have smaller aperture on a long telephoto lens but I think here on earth, the sun is not bright enough for f45 with useable shutter speed. There are also gravity that bounce the lens arround and things just move :-) Plus folks buy expensive super tele and use them wide open or 1 stop to wide open. The other end of the aperture never wears out.</p>

<p>As for the 50mm's f16. I think the lens maker sold them too cheap and they want you to buy a macro lens or wide angle too.</p>

<p>Too small an aperture introduces increasing diffraction into the results.</p>

<p>I bet it`s a construction and optical limitation. Wider aperture lenses ask for higher diameter diaphragms, that need more complexity to be closed to small diameters. That could be the case of fast primes. I try to remember a "compact" LF 210mm f5.6 lens I have who needs near 20 blades to be closed up to f90.<br /> In the other hand diffraction could make smaller apertures useless. Think that on current digital cameras, diffraction limits are on the f16-22 range for most lenses. Image quality at f45 could not be worth to have that aperture, I think.</p>

<p>

<p>The only mathematical relationship that I can think of regarding aperture is the following...<br>

f1, f1.4, f2, f2.8, f4 etc<br>

which is the square root of 2 to the power of n. Where n is a positive integer..<br>

ie, <br>

(√2)^3= f2.8<br>

An increase of √2, of 1.4 in aperture will reduce the aperture by half. As in the ‘hole’ will reduce in area by half, and thus let in half the light. <br>

This may well not be related to what you are talking about, but, well I am bored at work. <br>

<br>

R</p>

<p > </p>

<p> </p>

 

<p> </p>

</p>

 

<p>Yes, the limitation is how small the aperture can be due to mechanical design, or most likely how small the lens designer allows it to be due to optical limitations or convention.</p>

 

<p>The limiting factor is related to the diffraction limited spot size at the film plane. Large format lenses often allow f64/f90/f128 apertures simply because the spot size, although the same as in 35mm photography, is a smaller area of the image and not as much of an issue.</p>

<p>See: <a href="http://www.bobatkins.com/photography/technical/diffraction.html">http://www.bobatkins.com/photography/technical/diffraction.html</a> for more info...</p>

<p>as far as i know, the fstop number is a ratio of the focal length to the radius of the aperture opening. this suggests to me that constant aperture lenses make no sense; because f/2.8 at 17mm is a physically smaller opening than f/2.8 at 55mm. Thus why not let the aperture open up to the absolute size that it opens to at 55mm and thus make it a variable aperture lens instead of limiting it to only f/2.8 at the wide end. but i imagine that practical concerns arise in this such as design, cost and optical quality.<br>

so as to your question, yes, it should seem the case that a longer length should accomodate the aperture to be stopped down to the same physical size as at a shorter focal length, which would be the same physical size but a smaller f/number on the longer focal length. i imagine that if we were ever to design something as complicated as a zoom lens, we would immediately understand why this is so.</p>

<p>Good point Dan,<br />I have been wondering this myself. Keen to see if anyone knows the answer.</p>

<p>EDIT: a 200-400mm f4...<br>

@ 400mm apperture = 100mm [diameter or radius, im not sure]<br>

@ 200mm apperture = 50mm [diameter or radius, im not sure]</p>

<p>So it should be a...</p>

<p>200-400mm f2-4</p>

<p>Hmmmmm....</p>

<p>Dan is right. The f/stop is focal length/effective aperture diameter.</p>

<p>So an effective minimum diaphragm opening of 2mm would be f/15 on a 30mm lens or f/150 on a 300mm lens. A 200mm f/2 lens needs an opening of 100mm, which is the reason such a lens is as large as it is. </p>

<p>At f/32, a 35mm lens has an aperture opening of barely more than 1mm. Now think about diffraction. The smaller an orifice is compared to the wavelength of light, the more diffraction occurs. This means two things: with pixels being <em>very tiny</em> things, it doesn't take that much diffraction to blur an image, and at that size you are definitely seeing the effects. Second, notice that it is wavelength-dependent. Because of that, the diffraction is different for each color of light, leading to chromatic aberrations. If I'm not mistaken, this effect can be corrected for in the optical formula, but not easily or cheaply.</p>

<p>As for "fixed" aperture lenses, I believe the actual diameter of the aperture diaphragm remains constant throughout the range, but the optics of lens alter it's "effective" size as it is zoomed. Look into the long end of a constant-aperture lens and see how the diaphragm appears to be magnified as you zoom out. This is why "effective aperture" and actual diaphragm opening are not the same thing.</p>

 

<p>Thank you Glenn. I had a suspicion that "effective" was the killer in the logic!!</p>

<p> </p>

<p>For anyone who feels like reading it, there is a nice review of some of this information here...</p>

<p><a href="http://en.wikipedia.org/wiki/F-number">http://en.wikipedia.org/wiki/F-number</a></p>

<p>like richard pointed out, the smaller the focal length to the diameter of the lens the smaller the f# is.<br /> successive apertures halve the amount of incoming light. To achieve this, the diaphragm reduces the aperture <strong>diameter</strong> by a factor <strong>1.4 </strong> (square root of 2) so that the aperture <strong>surface</strong> is halved each successive step as shown on this diagram<br>

<img src="http://www.dpreview.com/learn/articles/glossary/exposure/images/123di_aperture.gif" alt="" width="544" height="200" /></p>

<p>large appertures are affected by spherical aberrations, small by diffraction</p>

<p>Carey, the answer to your question is really...the lens designer determines the minimum aperture. This would be within the practical physical limits of the lens, and the optimization of optical performance based on the physics of light. So, there is no reason to push the envelope to very small apertures on small sensors(DX or FX) if diffraction will degrade the image so much that the aperture is essentially unusable.</p>

<p>Here's the math: The aperture number is the ratio of the focal length to the aperture's diameter. For example, f/2 on a 50mm lens has an aperture diameter of 25mm.</p>

<p>Now here's where your old geometry class comes in handy. To open up 1 "stop" means you're doubling the amount of light coming through the lens. So the aperture diameter doubles, right? Wrong. The area of a circle grows in relation to Pi, not just doubling the diameter. Remember the formula for the area of a circle: Pi times the radius, squared. (Pi*r)^2. Knowing this, you can figure out the diameter of an aperture that lets in twice as much light for a given lens. In this example, an aperture of f/2 on a 50mm lens has a diameter of 25mm. That means the area of the aperture opening is (Pi * 12.5mm)^2. Note that a circle's radius is equal to half its diameter. Thus, an aperture of f/2 has an opening area of 1542 mm^2.</p>

<p>Opening up 1 full stop will double the area of the aperture opening. So the new area will be 1524 * 2 = 3084 mm^2. What diameter produces a circle with an area of 3084 mm^2? Using the same formula, but solving for radius:</p>

<p>(Pi * r)^2 = 3084<br /> (Pi * r) = sqrt(3084) = 55.5<br /> r = 55.5 / Pi = 17.68 mm</p>

<p>So the next-widest aperture (by 1 full stop) will have a radius of 17.68 mm, or a diameter of 35.35 mm.<br /> Finally, go back to the focal length / aperture diameter ratio to find the name of the corresponding f-stop:<br /> 50mm / 35.35mm = 1.4</p>

<p>So the next-widest aperture from f/2, in full-stop increments, is f/1.4. This works for any focal length, and any aperture.</p>

<p>i'm gonna disagree with your formula. you have (pi*r)^2, when the area of a circles is the classic pi*®^2. pi is not squared.</p>

<p>Dan: You're right, of course, my mistake. I find it clumsy to type math formulas in plain-text. :(</p>

<p>The mathematical relation ship I was thinking of was focal length/diameter of aperture or f-number, so f/8 becomes a "constant" across lens' focal lengths. The symbol is a constant, although the physical outcome in terms of the size (diameter) of the aperture opening is not.</p>

<p>So, for 50mm focal length lens f/2, as Kevin said is 25mm. f/8 is about 6.25mm diameter. f/32 about 1.5625mm, f/45 about 1.11mm diameter and f/64 about 0.78125mm diameter.</p>

<p>The average wavelength of visible light being 420nm, that is 0.00042mm, which is 1000 times smaller than the f/64 aperture opening for a 50mm lens, would suggest that that opening size would be transparent to light. But the problem with resolution is the size of the Airy disc, not the wavelength of light. Therefore the size of the Airy disc must then approach the size of the aperture opening in order for this interference to occur.</p>

<p>http://en.wikipedia.org/wiki/Airy_disc</p>

<p>This link also suggests the minimum spacing for digital sensors is 0.004mm, based on the size of the Airy disc and the wavelength of light.</p>

<p>I noticed since I posted the question that the Nikon 55mm and 60mm Micro lenses do stop down to f/32. So, if not purely a marketing reason why normals are limited to f/16, then maybe there is some design correction for diffraction up to f/32 in 35mm format, in any case. Diffraction and marketing must be the answer to the question, since f numbers up to 32 are at least possible for 35mm format lenses in the 50mm focal range.</p>

<p>Here's an example with simplified numbers.</p>

<p>Let's say your focal length is 200 mm. 200 mm is almost exactly 8 inches. I'm American, so I'll inches for familiarity.</p>

<p>If you set your lens to 200mm @ f/4, how big does the aperture have to be?</p>

<p>200 mm/ 4 = 8 inches / 4 = 2 inches</p>

<p>Your aperture is 2 inches in diameter.</p>

<p>Now, let's say that your 200 mm lens is VERY fast, and it lets you open up to 200mm @ f/2. How big is the aperture now? Remember, 200 mm = 8 inches, more or less.</p>

<p>8 inches / 2 = 4 inches. That's a BIG aperture, but let's use this example to make the math as easy as possible.</p>

<p>So, 200 mm @ f/4 = 50 mm = 2 inches.<br>

And, 200mm @ f/2 = 100 mm = 4 inches.</p>

<p>So far, so good!</p>

<p>Now, what's the difference in the amount of light that passes through f/4 versus f/2. Keep in mind that the focal length doesn't matter at all, but for illustration we'll stick with 200 mm = 8 inches = 20 centimeters.</p>

<p>f/4 = 8 inches/4 = 2 inches</p>

<p>f/2 = 8 inches/2 = 4 inches</p>

<p>What's the difference in area between a circle with a 2-inch diameter and a circle with a 4-inch diameter? Here's the one formula we have to deal with, but it's very simple.</p>

<p>Area or a circle = PI times the radius squared = PI * radius * radius.</p>

<p>The radius is half of the diameter. If the aperture is 2 inches, the radius is 1 inch. If the aperture is 4 inches, the radius is 2 inches.</p>

<p>Also, PI is approximately equal to 3, and in our example the extra decimal places don't matter, because we're comparing two numbers multiplied by a constant, and the constants cancel out. (Trust me on this.)</p>

<p>Aperture A (f/4) = 200 mm/4 = 8 inches/4 = 2 inches<br>

Radius A = 2 inches/2 = 1 inch<br>

Area A = PI * radius * radius = 3 * 1 * 1 = 3 square inches.</p>

<p>Aperture B (f/2) = 200 mm/2 = 8 inches/2 = 4 inches<br>

Radius B = 4 inches/2 = 2 inches<br>

Area B = PI * radius * radius = 3 * 2 * 2 = 12 square inches.</p>

<p>So, f/4 at 200mm has an aperture that covers an area of 3 square inches.</p>

<p>f/2 at 200mm has an aperture that covers an area of 12 square inches.</p>

<p>The ratio is 12 / 3 = 4. f/2 allows 4 times as much light to hit the film as f/4. Hence, we call the difference between f/4 and f/2 a "two-stop difference." Each stop represents doubling (or halving) the amount of light that reaches the film, so it takes two stops (double and double again) to multiply the amount of light 4 times.</p>

<p>How to we find what's in between f/2 and f/4? From experience, we know that there's something called f/2.8, but why is it 2.8 and not 2.6 or 3.1 or some other number? Let's look at the area formula again.</p>

<p>Area = PI * radius * radius<br>

AreaA = 3 * 1 inch * 1 inch = 3 square inches<br>

AreaB = 3 * 2 inches * 2 inches = 12 square inches</p>

<p>How can we find an area that's exactly twice as much as 3 square inches?<br>

AreaC = 6 square inches = 3 * radiusC * radiusC</p>

<p>Obviously, if we DOUBLE the radius (or the diameter) we QUADRUPLE the area. Doubling is too much. We have to find some other number that's larger than 1 but smaller than 2 that we can multiply our smaller radius of 1 by in order to exactly double the area.</p>

<p>Let's call this number M, for Magic. What is the magic number that DOUBLES when you SQUARE it?</p>

<p>AreaC = 3 * (1*M) * (1*M) = 6</p>

<p>M must satisfy the following condition.</p>

<p>1 squared = 1 * 1 = 1</p>

<p>(1*M) squared = (1*M) * (1*M) = 2<br>

1*1*M*M = 2<br>

M*M = 2/1 = 2</p>

<p>What number multiplied by itself yields a products of 2? M, our "magic" number, is the square root of 2.</p>

<p>The square root of too is approximately equal to 1.4.</p>

<p>Let's see if a value of 1.4 works.</p>

<p>AreaC = 3 * (1*M) * (1*M) = 3 * (1*1.4) * (1*1.4) = 3 * 1.4 * 1.4 = 5.88</p>

<p>It's not exactly 6, because we rounded the square root of two by dropping some decimal places, but 1.4 is almost the correct number, and it's close enough to be within the mechanical tolerances of a lens aperture (especially given that lens apertures aren't true circles anyway).</p>

<p>So, 1.4 is our magic number that allows us to increase or decrease an aperture by one stop. For example, if we have an aperture of f/2 and we want to decrease the light by one stop, we can recalculate our aperture as</p>

<p>f/( 2 * 1.4) = f/2.8</p>

<p>Continuing:<br>

f/( 2.8 * 1.4 ) = f/4 (approximately)<br>

f/( 4 * 1.4 ) = f/5.6<br>

and so on up the list of familiar f-stop numbers, each one separated from the last by a factor of the square root of two (our "magic" number).</p>

<p>Finally, in our example, the aperture between f/2 = 4 inches and f/4 = 2 inches works out to be f/2.8.</p>

<p>200 mm/2.8 = 8 inches/2.8 = 2.86 inches (approximately).</p>

<p> </p>

<p>Any given f/ number's physical size will be relative to focal length, as f/ refers to the ratio of the lens opening to its focal length. For example, f/4 means "focal length divided by 4". To make it easy, think of the f/ number as meaning "fraction". So, f/4=1/4. On a 100mm lens, f/4 will be 25mm; f/2 on a 50 mm will also be 25mm; f/4 will be 12.5mm.<br /> The smaller the physical size of the opening, the more diffraction affects the image. It's always there, as it's an edge effect, but at small apertures fewer undiffracted rays remain. If an opening gets small enough, any two points opposite each other will be close enough that no undiffracted rays remain.<br /> At f/32 on a 100mm lens the aperture diameter is 3.12mm; on a 50mm it's 1.56mm; on a 25mm, it's .78mm- really tiny.<br /> The smaller that the physical size of the aperture gets, the more it becomes like the one on a pinhole camera, and it begins to produce similar results; i.e., lots of depth of field, with overall softness.<br /> Macro lenses have smaller apertures to provide desirable depth of field at high magnification levels, though at the expense of overall optical quality.</p>

<p>f/32 on a 28mm lens results in an aperture diameter of 0.875mm. that is small. Pin hole equivalent is about 0.19mm diameter... it's about 5 times larger than pin hole though.</p>

<p>f/22 on a 28mm lens results in a diameter of 1.27mm. Pin hole equivalent is still 0.19, but f/22 aperture diameter is about 7 times larger than pin hole in this case.</p>

<p>What follows are 3 images shot at f/11, f/16 and f/22 on a 28mm lens. Based on the images I would not guess that f/32 on a 28mm lens would be that much worse than f/22, much less, f/32 on a 50mm lens. I'm still thinking marketing is the answer as to why 50mm lenses do not stop down to f/32...</p><div></div>

<p> f/16</p><div></div>

<p>f/22</p><div></div>