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Photo Forum / Film Photography / 35 mm / July 2005Why does 2x converter reduce brightness by 4x?
Hi, I'm confused by something and would appreciate any assistance. Does doubling the focal length, while leaving the aperture unchanged, change my f# by a factor of 2? Does this imply a doubling/halving of irradiance, E? How then to explain this formula [E (image)/4]* f# squared = E (object) where the irradiance changes as the square of the f#? Thanks, JB > Hi, > > I'm confused by something and would appreciate any assistance. > > Does doubling the focal length, while leaving the aperture unchanged, > change my f# by a factor of 2? Yes. f/stop is calculated by dividing focal length by aperture (or is that backwards?). I don't know about the irradiance thing. > Does this imply a doubling/halving of irradiance, E? > [quoted text clipped - 7 lines] > > JB  SignaturePaul Furman http://www.edgehill.net/1 san francisco native plants >> Hi, >> [quoted text clipped - 17 lines] >> >> JB The "brightness" of the image at the focal plane is a function of the square of the aperture size. So a 2X reduction in aperture diameter results in a 2 x 2 or 4 times reduction in the energy density at the focal plane. >>> Hi, >>> [quoted text clipped - 22 lines] > results in a 2 x 2 or 4 times reduction in the energy density at the focal > plane. Yes. In general, the area of any opening is proportional to the square of any linear dimension, regardless of the shape of the "hole". So, if you double any distance, and keep the same shape, you will raise your light level by a factor of four. >Hi, > [quoted text clipped - 14 lines] > >JB What the TC does is effectively take one half of the length and one half of the width of the image and magnifies it to fill the frame. 1/2 times 1/2 = 1/4 Thus there is only 1/4 of the light of the original image, now filling the frame. I'm not sure I understand the formula you quote. BTW, the usual term for the luminous power per unit area is illuminance, usually measured in lux. 1 lux = 1 lumen/metre^2. Irradiance is the term for radiant power per unit area, usually in watts/metre^2. The difference is that this can include non-visible energy. David SignatureDavid Littlewood : Does doubling the focal length, while leaving the aperture unchanged, : change my f# by a factor of 2? : Does this imply a doubling/halving of irradiance, E? : How then to explain this formula : [E (image)/4]* f# squared = E (object) : where the irradiance changes as the square of the f#? Don't think too hard. The f-stop is *defined* to be the ratio of the focal length to the (effective) diameter of the aperture. The amount of light transmitted through the aperture is proportional to the *area* of the aperture, which goes as the *square* of the diameter. Thus, going from f/8->f/16 makes the aperture 2x smaller diameter, and 1/4 as much light. That's why full "stops" go by 1.4, or an approximation to sqrt(2). Putting a doubler on does exactly that... doubles the focal length. It does not change the aperture diameter, so thus the f-stop ratio doubles (i.e. from f/2.8->f/5.6). -Cory Signature************************************************************************* * Cory Papenfuss * * Electrical Engineering candidate Ph.D. graduate student * * Virginia Polytechnic Institute and State University * ************************************************************************* Now I see--one f stop doubles/halves the area, not the diameter. Thanks. > Now I see--one f stop doubles/halves the area, not the diameter. > Thanks. Right, if you look at the sequence of standard f/#s on most lenses, they go in approximately a 1.4 times increment. 1.4 x 1.4 is approximately 2. Forget the formula if you double the focal length without changing the aperture you lose two stops. It is quite simple. A 50mm f1 lens has an aperture of 50 mm. Add a 2X TC to that and you have a 100mm lens with an aperture of 50mm -- which is f2. Start with a 25mm aperture (f2) on the 50 and you get a 100mm lens with a 25mm aperture or f4. Signaturehttp://www.chapelhillnoir.com home of The Camera-ist's Manifesto The Improved Links Pages are at http://www.chapelhillnoir.com/links/mlinks00.html A sample chapter from "Haight-Ashbury" is at http://www.chapelhillnoir.com/writ/hait/hatitl.html > Hi, > [quoted text clipped - 14 lines] > > JB > Forget the formula if you double the focal length without changing the > aperture you lose two stops. It is quite simple. > A 50mm f1 lens has an aperture of 50 mm. Add a 2X TC to that and you have > a 100mm lens with an aperture of 50mm -- which is f2. Start with a 25mm > aperture (f2) on the 50 and you get a 100mm lens with a 25mm aperture or f4. Or start from high school physics: light falls off proportionately to the square of the distance. Double the distance (the focal length) and you cut the light by 4. SignatureMichael | "He's dead, Jim." >> Forget the formula if you double the focal length without changing the >> aperture you lose two stops. It is quite simple. [quoted text clipped - 7 lines] > square of the distance. Double the distance (the focal length) and you cut > the light by 4. That's what I was thinking--inverse square law of light. SignatureRegards, Matt Clara www.mattclara.com >>> Forget the formula if you double the focal length without changing the >>> aperture you lose two stops. It is quite simple. [quoted text clipped - 9 lines] > > That's what I was thinking--inverse square law of light. I would suppose this is true regardless of the f stop scale on the lens....IOW, if you put a 2x teleconverter on your lens, then the f stop scale engraved on the side of the lens is off by a factor of 2 f stops while the teleconverter is mounted. And, I would guess that if you mount a 1.4 x converter on your lens, this throws the scale off by one f stop.......... >I would suppose this is true regardless of the f stop scale on the >lens....IOW, if you put a 2x teleconverter on your lens, then the f stop >scale engraved on the side of the lens is off by a factor of 2 f stops >while the teleconverter is mounted. And, I would guess that if you mount a >1.4 x converter on your lens, this throws the scale off by one f >stop.......... The EF lenses have a nifty solution to this problem... I feel like being a Devil's Advocate... If you add a TC to the FRONT of a lens, eg the type of teleconverters used on point&shoots and prosumers, there is effectively no light loss. Well, there is, but it is just a very minor, almost immeasurable, transmission loss. It is *not* a loss of one or two fstops as you might expect from the multiplier number. I think I understand why, but I'm sure someone here can better explain it than me...! > I feel like being a Devil's Advocate... > [quoted text clipped - 6 lines] > I think I understand why, but I'm sure someone here can better explain > it than me...! I think it's because they are not changing the focal length of the lens, but just acting as magnifying lenses in front. SignatureMichael | "He's dead, Jim." > > I feel like being a Devil's Advocate... > > [quoted text clipped - 9 lines] > I think it's because they are not changing the focal length of the > lens, but just acting as magnifying lenses in front. Afocal front teleconverters do change the focal length, but they also increase the entrance pupil diameter by the same amount. Hence the f/# of the system is not altered. |
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