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Nov 18, 2007, 6:50 PM  #1 
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I bought this lens a few years back attached to a KM1000 camera....
Can anyone tell me anything about it..... I understand it's 50mm with a minimum aperture of 1.4 but what I don't understand is the 1 prior to the colon "1:1.4" Thanks Mike 
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Nov 18, 2007, 6:52 PM  #2 
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This is the other half of the front.......

Nov 18, 2007, 6:56 PM  #3 
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The "f" number is the ratio of the diameter to focal length, so in spoken words your lens is "one to one point four". The colon is used to denote a ratio.
When talking about lenses just about everyone assumes the '1'. 
Nov 18, 2007, 6:59 PM  #4 
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this is a very highly rated lens. it'sthe firstK series made.. i think the first '1' is to show a relationship as what a f1 lens would be. very good question as i don't think i've seen it addressed before.. i can't answer 100% sure about this. i'll also wait with you for a proper answer.
roy 
Nov 18, 2007, 8:18 PM  #5 
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My old optics engineering textbook had a crummy explaination (unless your deep into math). However Alan Marcus posted an excellent description at....
http://mybetterphoto.com/forms/QnAde...threadID=26655 The f/numbers are not fractions they are a ratio. Sorry to be long winded but I can't help it the teacher in me takes over.Sorry this is so long winded. You guys might not like it. That's OK, I take criticism well. During exposure the camera lens projects an image of the outside world on the film/chip. This position is known as the image plane. The ability to accurately and consistently control the brightness of the image is a major ingredient for successful photography. The photographer can regulate image plane brightness by somehow changing scene brightness or by varying the camera lens to somehow increase or decrease brightness. It seems the lens operates like a funnel gathering light. The larger the lens diameter, the more light gathered and the brighter the image will be. Thus exercising control over lens diameter is a nobrainer. This adjustment is complicated however by the fact that the projection distance inside the camera (focal length) has a surprisingly big influence. Consider what happens when you move a slide projector further back from the screen. Doubling the projection distance causes the image on the screen to become twice as big (2x larger). Twice as big means the light from the unchanged projector lamp must now cover four (4) times more surface area (height x width). You guessed it, the image brightness at the screen is reduced by a factor of four (4). The same thing happens when double the focal length with your zoom camera lens. The zoom lens mechanics changes lens diameter at the same time as you focal length to compensate. If curious about the effects of distance changes, you can look up the rule in physics know as the law of the inverse square. Now consider that the two most important inthecamera factors controlling brightness is the working diameter and the focal length of the lens. With thousands or perhaps millions of camera designs that bring into play a hodgepodge of lens dimensional combinations. Each will presents a different brightness at the image plane. You guessed it â€“ disorder results. Without a solution, most pictures you take will would be under or over exposed. Ratio to the rescue: The ratio of two numbers is pure because it is devoid of dimension. In the case of the camera lens we divide the focal length by the working diameter to get a ratio which we call the focal ratio or in other words the f/number. The beauty of the f/number system is, when any lens is set to a specific f/number it delivers the same image brightness at the image plane as any other lens set to the same f/number regardless of different dimensions. Simply stated, the focal ratio or f/number gets rid of the chaos revolving around the setting of the lens aperture. Other systems have been proposed and tried but we still elect to keep the one that has works best. (You're free to invent a better system). The problem is: The number set used seems illogical. Who would have thunk up this sequence. 1 â€“ 1.4 â€“ 2 â€“ 2.8 â€“ 4 â€“ 5.6 â€“ 8 â€“ 11 â€“ 16 â€“ 22 â€“ 32 â€“ 45 â€“ 64 â€“ 90 Not all can be used on a single lens so you camera utilizes a span of some of these values. Well the span has logic. Any number in the set multiplied by 1.4142 yields the neighbor going right, conversely, dividing by 1.4142 yields the neighbor on the left. 1.4142 is the square root of 2 and its OK to round it to 1.4. So what's the significance? Say you have a circle (the lens is a circle) and you desire to change the lens diameter in order to alter image brightness. To increase lens surface area by the most logical increment of adjustment which is a doubling (2x), you multiply its diameter by 1.4. The product is a revised diameter that produces twice the light energy at the image plane. Conversely, divide the diameter of the lens by 1.4 yields a revised aperture with half the surface area thereby reducing brightness by half. As to the two engraved f/numbers: It has become customary to label a lens with its maximum f/number. When calculating the f/number of a zoom lens you must take into account that the maximum diameter (aperture) is fixed but the focal length is a variable. Thus the zoom lens will be labeled with two different maximum focal ratios, one for max zoom and a different one when operating at minimum zoom Sorry to report that us poor photographers must forever cherish the lowly focal ratio (f/number) system until you think of something better. The fnumber is an expression of relative aperture derived by dividing the focal length by the effective aperture. We need a universal system to express settings that will produce a predictable brilliance at the film/chip plane. Universal means the brilliance will be the same for lens x or y, when set to a common value. The snag is; lens x and lens y have different dimensions (focal length and aperture). Ratio to the rescue. The ratio is a dimensionless quantity. You are correct; a fraction is a specific application of the ratio as it relates the part (the numerator) to the whole (the denominator). It seems to me that using the ratio expresses an fnumber best as apposed to the fraction. Indeed the camera manufacturing segment agrees with me as nearly all lenses are engraved 1:1.4 or 1:1.8 etc. Two values separated by a colon is the universal notation for the ratio. One can win the Pulitzer Prize with a simple nonadjustable camera. The odds of success improve when adjustments are available, these expand the picture taking environment. Over the last umpteen years cameras have evolved with chip logic that greatly reduce the photographers need to know all this technical stuff and I applaud. The perfect camera would have no controls, knobs, or buttons. One would simple think (previsualize) the finished picture and the camera would respond telepathically. The real question is; how do we teach people who want to learn these concepts? I applaud you for teaching and helping others to understand. You have my blessing to use whatever works for you. 
Nov 18, 2007, 9:00 PM  #6  
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robar wrote:
Quote:
I second that . According to Stan lens summary ******************************** '50 mm f/1.4 Mark Roberts  . . . it's a beauty. Great optically of course, but the heft and "feel" of the thing is just unsurpassed. . . [and in a later thread . . .] It is a *great* lens optically, and what's more, it's mechanical construction is simply magnificent. Manual focus at its finest. ********************************* I would say in pristine condition it is worth around at least 50 times (could be a 100) what you paid. Good deal Daniel 

Nov 19, 2007, 2:41 AM  #7 
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Very highly regarded lens indeed!
I had the later Mseries for a while (sold it to make some profit), and it was very sharp, with great colors and built like a tank. I'm sure you'll have fun with this one. Tom 
Nov 19, 2007, 5:40 AM  #8 
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Thanks folks.......A tank indeed....it's one heavy lens.....I'll lie and say I understood what you have written Interest Observer.....
Maybe after a thousand reads it might sink in..... :?:lol::idea:...... Mike 
Nov 19, 2007, 7:50 AM  #9 
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now i'm really confused.. naw...
basically , IO, that's pretty much what i said in my assumption. roy 
Nov 20, 2007, 12:22 AM  #10 
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Hi Mike,
Well I just copied and pasted (with attribution) because it was a much better (and easier) explaination than the math in my old optics engineering textbook (going back and deriving the entire theory) had to offer. Its been 30 years and I have even forgotten a lot of the math  so its getting to be all squiggles now..... I am to the point of just wanting to go out and take pictures and let the magic happen! 

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