 |
 | Home | Contact Us | FAQ | Search & Site Map | Link to Us |
 | Sign In | Join | Other 45 Sites in Network |
| PhotoKB HomeDiscussion Groups Digital Photography Digital PhotoDSLR CamerasZLR CamerasPoint & Shoot CamerasFilm Photography 35 mmLarge FormatMedium formatDarkroomFilm and LabsOther EquipmentPhoto Technique Nature PhotographyPeople PhotographyTechnique GeneralGeneral Photo Topics General TopicsAustralian PhotographyUK PhotographyDirectoryPhoto Clubs |
Photo Forum / Digital Photography / DSLR Cameras / August 2006Which Polarizer?
I have a Canon XT with a Sigma DC 18-200mm 1:3.5-6.3 lens, so obviously I'm not a Pro! However I'm trying to decide on a circular polarizer filter for enhancing sky colors and minimizing reflections. I hear that circular filters can knock 2 f-stops off a lens. Since I already have a slow lens, would a more expensive filter take less of a toll than a cheap one? What are the considerations? Any recommendations? Roger > I have a Canon XT with a Sigma DC 18-200mm 1:3.5-6.3 lens, so obviously > I'm not a Pro! [quoted text clipped - 5 lines] > > Roger Polarizers do take about 2 stops compensation. Light from the subject is randomly polarized, and the filter passes only the light that happens to be in the same polarization plane as the filter, so about 75% of the light from the subject is rejected. There is also a small further transmission loss from the polarizing medium iself, so there's your 2 stops. There can be variations due to the 'tightness' of the polarization, and the transmission characteristics, and with cheaper filters there can be some color shift as well. Get the best you can afford. Colin D.  SignaturePosted via a free Usenet account from http://www.teranews.com >> I have a Canon XT with a Sigma DC 18-200mm 1:3.5-6.3 lens, so obviously >> I'm not a Pro! [quoted text clipped - 15 lines] >filters there can be some color shift as well. Get the best you can >afford. Not quite - a theoretically perfect polariser would only take 50% of the light (1 stop). Deviations from perfection in the polariser material do indeed show up, and absorb a little more light, so the typical metered attenuation is about 1.5 stops. As you say, the more expensive ones tend to use better polarising material - or perhaps I should say that the ones using better materials tend to be more expensive (the old adage "you get what you pay for" should IMO be more accurately re-written "you don't get what you don't pay for"). They may thus have slightly better transmission and less colour shift. However, this 1.5 stops (+/-) is for light that has no polarisation whatever - i.e. is totally randomly polarised. In reality, light from the sky or reflected from non-metallic surfaces (i.e. most surfaces, including water and glass) has some component of polarisation. Thus the observed attenuation may be greater or less than 1.5 stops, depending on the orientation of the filter with respect to the predominant polarisation of the light. David SignatureDavid Littlewood > Polarizers do take about 2 stops compensation. Light from the subject > is randomly polarized, and the filter passes only the light that happens > to be in the same polarization plane as the filter, so about 75% of the > light from the subject is rejected. There is also a small further > transmission loss from the polarizing medium iself, so there's your 2 > stops. OK, this makes sense, but I still have a stupid question: doesn't the camera meter notice this reduction in the amount of light entering the lens and compensate accordingly? Or is the meter not operating through the lens? Signature-Take out Ron to reply- My random photostream: http://www.flickr.com/photos/boggissimo/ >> Polarizers do take about 2 stops compensation. Light from the subject >> is randomly polarized, and the filter passes only the light that happens [quoted text clipped - 7 lines] >lens and compensate accordingly? Or is the meter not operating through >the lens? Yes, the meter is operating through the lens in your Rebel XT. The meter will notice the reduction in light levels and approximately compensate. However, it may not get it quite right. For example, if you want to increase the saturation of a blue sky, and use a polariser, you actually want the photo to be a little darker, but the meter might increase the exposure too much in order to raise the overall illumination level back to what it was. Clearly a little trial and error may be required to get exactly the effect you want - and clearly there is a great advantage in using a DSLR that you can review the results (though only rather crudely) on the screen straight away. For most occasions the reduction given by the meter will be reasonably OK. The purists in film days advised: meter without filter, fit filter, and apply compensation of +1.5 stops (or slight variant of this according to taste, determined by trial and error). This approach will still work; it is more time-consuming than the simpler approach of sticking on the filter and letting the meter do its thing, but perhaps slightly more reliable. David SignatureDavid Littlewood I always shoot at ISO200 when using a polarizer (and long telephotos) and have never had a problem. Most of the time they are used in scenics where a fast shutter speed is not really necessary. I find them great when I travel and want to photograph through a car windshield or window and have only had one serious problem -- over polarizing at high altitudes. > I have a Canon XT with a Sigma DC 18-200mm 1:3.5-6.3 lens, so obviously > I'm not a Pro! [quoted text clipped - 5 lines] > > Roger >> I have a Canon XT with a Sigma DC 18-200mm 1:3.5-6.3 lens, so obviously >> I'm not a Pro! [quoted text clipped - 7 lines] >and have never had a problem. Most of the time they are used in scenics >where a fast shutter speed is not really necessary. That's been my experience too. A scenic can usually be done with a tripod, which means you can shoot in almost any light conditions. As for recommendations, I like high quality optics and prefer brands like Hoya, B/W, etc. They cost more, but the coatings seem to be better. >I have a Canon XT with a Sigma DC 18-200mm 1:3.5-6.3 lens, so obviously >I'm not a Pro! [quoted text clipped - 5 lines] > >Roger The cam will adjust, so will you. Get a cheep hoya for $40 and see if you like it. If ya move up - sell it with the lens. cheers Ken > I have a Canon XT with a Sigma DC 18-200mm 1:3.5-6.3 lens, so obviously > I'm not a Pro! > However I'm trying to decide on a circular polarizer filter A linear polarizer is better. > for enhancing sky colors and minimizing reflections. What I like most about polarizer is to enhance the color of grass, flowers, and trees, ... > I hear that circular filters can knock 2 f-stops off a lens. Since I > already have a slow lens, This can be a problem. > would a more expensive filter take less of a > toll than a cheap one? No. > What are the considerations? Any recommendations? Just get a cheap polarizer, to see whether you like the effect. If you want a better one later, maybe with a better lens, get a multicoated linear polarizer. http://digitcamera.tripod.com/#slr >> I have a Canon XT with a Sigma DC 18-200mm 1:3.5-6.3 lens, so obviously >> I'm not a Pro! >> However I'm trying to decide on a circular polarizer filter > A linear polarizer is better. Unfortunately, it'll mess up metering and AF with a lot of cameras, so 'better' is ... a rather relative term. http://www.luminous-landscape.com/tutorials/polarizers.shtml http://www.photofilter.com/cir_lin.html http://www.mat.uc.pt/~rps/photos/filters_uv_pol/#polq2 -Wolfgang > > I have a Canon XT with a Sigma DC 18-200mm 1:3.5-6.3 lens, so obviously > > I'm not a Pro! [quoted text clipped - 24 lines] > > http://digitcamera.tripod.com/#slr Anyone else posted to say that a linear polariser is a bad idea with an autofocus / TTL metering body? AFAIK, such bodies use beam splitters and so have to be used with circular polarisers. To cope with the loss of light, load the camera with a faster film than usual (or set a higher ISO than usual if you're using a difital body). Note also that the effect of the filter will depend on the relative position of sun, camera and subject. I found, taking the "classic" photo of boats in a small harbour on a sunny day, that in one position the filter had very little effect whereas in another position the effect was mich more noticeable. Regards, Ian. > > > I have a Canon XT with a Sigma DC 18-200mm 1:3.5-6.3 lens, so obviously > > > I'm not a Pro! [quoted text clipped - 28 lines] > AFAIK, such bodies use beam splitters and so have to be used with circular > polarisers. The beam splitter is partially polarizing. That's why a linear polarizer might interfere. But the AE metering is affected at most about 1 stop. Since the meter is not perfect anyway, I might want that 1 stop exposure bracketing. And under certain circumstances (color, angle, ...), circular polarizer will have exact the same AE problem as linear polarizer. For AF, the differences between circular and linear polarizer is very little. And because of the partially polarizing beam splitter, with a circular polarizer, sometimes the color effect I see in the viewfinder is different from that the sensor records when the mirror gets out of the way. I noticed this problem with circular polarizer and switched to linear polarizer. http://digitcamera.tripod.com/#slr >The beam splitter is partially polarizing. That's why a linear >polarizer might interfere. But the AE metering is affected at most [quoted text clipped - 11 lines] >way. I noticed this problem with circular polarizer and switched to >linear polarizer. Hard to go along with what you are saying here. First, a circular polariser can not have the same problems with a beam splitter as a linear one; the polarisation direction is effectively re-randomised after it has done its work. There may be some slight residual effect (the effect of the quarter wave plate is somewhat frequency sensitive) but nothing like that from a linear polariser. And as for the sensor seeing something different from what your eye sees, this may be so, but it could not be because of the type of polariser and the effect of the mirror. The mirror is one of the types of surface ("specula", i.e. reflective metal) which has no surface polarisation effect. I suppose if you had an SLR with a glass mirror (Canon Pellix, EOS 1nRS etc) or an ancient rear-silvered glass mirror (Zenit) you may see some slight effect, but 99.9% of SLRs in the last 30 years use front silvered mirrors. Colour memory is notoriously fickle, and it is far more likely that it is playing tricks on you. And, as for the 1-stop effect of polarisation on the meter beam-splitter, I cannot even begin to understand why the fact that you *may* want to bracket should make this 1 stop factor unimportant. Metering is difficult enough to get right without cavalier dismissal of such an obvious source of unpredictable error. David SignatureDavid Littlewood >And as for the sensor seeing something different from what your eye >sees, this may be so, but it could not be because of the type of >polariser and the effect of the mirror. The mirror is one of the types >of surface ("specula", i.e. reflective metal) which has no surface >polarisation effect. I Sorry, "specular" David SignatureDavid Littlewood > >The beam splitter is partially polarizing. That's why a linear > >polarizer might interfere. But the AE metering is affected at most [quoted text clipped - 15 lines] > splitter as a linear one; the polarisation direction is effectively > re-randomised Not random, but the polarisation is rotated a certain angle. Each color is rotated a different angle. But a certain color is rotated a certain angle. If your scene is mostly a single pure color, e.g., green, the green color is just rotated, the result is the same as a linear polarizer, just a different angle. You can try this yourself by stacking 2 polarizers together, and see this effect. Circular polarizer works best on greyish scenes. > And as for the sensor seeing something different from what your eye > sees, this may be so, but it could not be because of the type of [quoted text clipped - 4 lines] > (Zenit) you may see some slight effect, but 99.9% of SLRs in the last 30 > years use front silvered mirrors. The beam splitter affects the AE/AF sensors, and the viewfinder. It is partially polarizing. > Colour memory is notoriously fickle, and it is far more likely that it > is playing tricks on you. Yes, but again you can stack 2 polarizers together and see the exaggerated effect instantly, not on memory. > And, as for the 1-stop effect of polarisation on the meter > beam-splitter, I cannot even begin to understand why the fact that you > *may* want to bracket should make this 1 stop factor unimportant. > Metering is difficult enough to get right without cavalier dismissal of > such an obvious source of unpredictable error. The 1 stop is about the max. On average it is much less. And even without polarizer, I change EC most of the time. With polarizer, my changing EC is not that different. http://digitcamera.tripod.com/#slr >> First, a circular polariser can not have the same problems with a beam >> splitter as a linear one; the polarisation direction is effectively [quoted text clipped - 7 lines] >stacking 2 polarizers together, and see this effect. Circular polarizer >works best on greyish scenes. If you think a quarter-wave plate simply rotates the plane of polarisation, then you have clearly not understood birefringence; it is rather more complex than you suggest, and certainly does *not* involve the rotation of polarised light (which is an entirely different property). However, for all practical purposes you can regard the light, after passage through the quarter-wave plate, as *effectively* re-randomised (which is why I used exactly those words earlier). More precisely, the polarisation vector is rotating trillions of times a second, far faster than can be seen by any meter, so it sees the average, i.e. "random", vector direction. As I said (in the bit you carefully snipped) there is a residual effect as the birefringence is somewhat colour sensitive, but this is a much smaller effect for the issue here. >> And as for the sensor seeing something different from what your eye >> sees, this may be so, but it could not be because of the type of [quoted text clipped - 6 lines] > >The beam splitter affects the AE/AF sensors, Yes; I think we all agree on this. > and the viewfinder. It is >partially polarizing. By what mechanism do you believe this to occur? Not something I have seen suggested before. The beam splitter is not in the optical path to the viewfinder - you did realise that? >> Colour memory is notoriously fickle, and it is far more likely that it >> is playing tricks on you. > >Yes, but again you can stack 2 polarizers together and see the >exaggerated effect instantly, not on memory. No; I just tried it, and could only see a slight decrease in density (because of the absorption of the polarisers). No colour change was apparent. If you use polarisers with a slight tint, this would obviously double if you use 2. Using 2 polarisers does not even increase the polarising effect (unless they are very inefficient polarisers). Otherwise, I think it's in your imagination. There is also, BTW, no obvious theoretical reason I can see why your claim should be valid, but if you know one I would be interested to hear it. >> And, as for the 1-stop effect of polarisation on the meter >> beam-splitter, I cannot even begin to understand why the fact that you [quoted text clipped - 5 lines] >without polarizer, I change EC most of the time. With polarizer, my >changing EC is not that different. I don't disagree with you assessment of the amount - I just did a test and found the difference was typically 1/3-1/2 stop. However, there are enough variables without ignoring things easily controlled. David SignatureDavid Littlewood > there is > a residual effect as the birefringence is somewhat colour sensitive, but > this is a much smaller effect for the issue here. This color sensitive effect is very obvious when I stack 2 polarizer together, the front being a circular polarizer. If I rotate the rear polarizer, (so the effect is entirely the interaction between the circular polarizer and the rear polarizer, not the effect of the polarizer on the scene), the color changes dramatically between warmer and colder. > The beam splitter is not in the optical path to the viewfinder Is it behind a half mirror? http://digitcamera.tripod.com/#slr >> there is >> a residual effect as the birefringence is somewhat colour sensitive, but [quoted text clipped - 6 lines] >polarizer on the scene), the color changes dramatically between warmer >and colder. Sorry, I misunderstood your previous post - I thought you were referring to 2 linear polarisers. The colour does indeed change with a CP in front (though I would not call it dramatic). This is the second-order effect I mentioned before - the fact that the effect of the quarter-wave plate is somewhat dependent on the colour of the light. It does not seem to have a material effect on the light intensity. The slight colour effect only shows up if the light subsequently passes through another polariser - as in your experiment. If you use a single CP (as one would in real life) the partial polarisation at the beam splitter would I think not show any noticeable effect on the meter. And as for the effect on the image, this would be zero; the slight colour effect of the QWP will not show up, as the light does not pass through another polariser. You will find the best explanations of birefringence and the effect of wave plates (more commonly referred to as retardation plates) in texts covering chemical microscopy through crossed polarisers - a fascinating exercise in its own right. >> The beam splitter is not in the optical path to the viewfinder > >Is it behind a half mirror? Only in the (very rare) pellicle-type cameras I mentioned is there a semi-reflective mirror in the optical path of the viewfinder. Otherwise (i.e. 99.9%+ of cases) definitely not; it is a fully-silvered mirror, which should not show any polarising effect. And of course the mirror is also lifted when taking the photo, so the image is totally unaffected. David SignatureDavid Littlewood > >> there is > >> a residual effect as the birefringence is somewhat colour sensitive, but [quoted text clipped - 36 lines] > which should not show any polarising effect. And of course the mirror is > also lifted when taking the photo, so the image is totally unaffected. Can you explain why the beam splitter does not affect the viewfinder? Where does it split the beam to the AE/AF sensors? http://digitcamera.tripod.com/#slr >> >> The beam splitter is not in the optical path to the viewfinder >> > [quoted text clipped - 8 lines] >Can you explain why the beam splitter does not affect the viewfinder? >Where does it split the beam to the AE/AF sensors? Good question. Actually, I was guilty of a slight over-simplification. About 20% of the main mirror is semi-reflective (you can often see this square in the centre). This passes light to the various beam splitters and optics for AF and metering, usually in the space below the mirror. About 80% of the light to the viewfinder hits a fully reflective mirror and will show no polarisation artefacts; the 20% in the centre would be too small to be seen, especially since the polarisation efficiency of a 45 degree semi-reflective mirror is less than that of a polarising filter. And, as I said, none of this affects the actual image taken on exposure. David SignatureDavid Littlewood >> >The beam splitter is partially polarizing. That's why a linear >> >polarizer might interfere. But the AE metering is affected at most [quoted text clipped - 23 lines] >stacking 2 polarizers together, and see this effect. Circular polarizer >works best on greyish scenes. Not rotated, but rotating thus its angle is continuously changing; in other words "circularly polarized". SignatureIan G8ILZ >>> >The beam splitter is partially polarizing. That's why a linear >>> >polarizer might interfere. But the AE metering is affected at most [quoted text clipped - 26 lines] > Not rotated, but rotating thus its angle is continuously changing; in > other words "circularly polarized". It probably isn't clear what you are saying. (not that this will help much, but here goes) When light goes through a linear polarizer, the light (e.g. the electric vector of the electromagnetic wave) vibrates in a single plane. Light vibrating 90 degrees from that are blocked from passing through the polarizer. Think of carrying your tripod with the legs extended, and you walk through some vertical bars. If you have the tripod vertical, you can go through, if horizontal, you get blocked. Now think if the tripod legs as the direction of vibration of light waves. In a circular polarizer, you have a linear polarizer in front and a "quarter wave plate" in back. The light that passes through the linear polarizer then passes through the plate, which has indices of refraction different along two orthogonal axes. The thickness of the plate is such that along one axis, light of a specific wavelength is slowed 1/4 the wavelength. Then when the polarized light passes through the plate, oriented so the axes are 45 degrees to the linear polarizer, the electromagnetic wave vectors end up rotating. Back to the tripod analogy: after passing through the vertical bars, with the tripod vertical, you then pass through the quarter wave plate and come out the other side twirling the tripod as you continue to walk. (you are twirling the tripod orthogonal to the direction you are walking). A subsequent encounter with a polarizer (linear or circular) acts as if the incident light is unpolarized. Regarding other statements made, Mirrors, even 100% aluminum, still polarize light, as do beam splitters. I have the data in my office,and if I remember I'll look it up. It is significant enough that I have had to compensate for it in spectrometers I have designed. That is the reason why a circular polarizer is needed: the multiple reflections, and the AF and meter system can be influenced (i.e. get the wrong data) with polarized light. The circular polarizer reduces that problem. Regarding the two circular polarizer experiment: if you take two circular polarizers and stack them then rotate one, you should see NO change in intensity, and NO color change. If you do, they are not quality circular polarizers. See: http://www.clarkvision.com/photoinfo/evaluating_polarizing_filters Roger Photos at: http://www.clarkvision.com SNIP > In a circular polarizer, you have a linear polarizer in front > and a "quarter wave plate" in back. The light that passes [quoted text clipped - 5 lines] > plate, oriented so the axes are 45 degrees to the linear > polarizer, the electromagnetic wave vectors end up rotating. Could it be helpful to also remind that natural light is usually in-coherent, so we're talking about multiple electromagnetic waves that arrive out-of sync/phase. Bart > Could it be helpful to also remind that natural light is usually > in-coherent, so we're talking about multiple electromagnetic waves > that arrive out-of sync/phase. > > Bart I know that this isn't what you meant, but natural light isn't going to be phase coherent. What you meant is that the incoming light is usually not polarised in one direction (at least I assume this is what you meant). >> Could it be helpful to also remind that natural light is usually >> in-coherent, so we're talking about multiple electromagnetic waves [quoted text clipped - 5 lines] > to > be phase coherent. That's indeed what I said. Natural light is mostly not (phase) coherent (=in-coherent). > What you meant is that the incoming light is usually not polarised > in one direction (at least I assume this is what you meant). I meant that incoming light is not only un-polarized (but it largely *should be* after the stretched polarization foil passage), but it is *also* (unlike e.g. laser light) not in a single phase. Isn't it the phase difference that allows the quarterwave plate to produce more than a simple 90 degree rotation? That is a question, rather than a statement (I'm a photographer, not an opto-electronic scientist). SignatureBart > <achilleaslazarides@yahoo.co.uk> wrote in message > > What you meant is that the incoming light is usually not polarised > > in one direction (at least I assume this is what you meant). [quoted text clipped - 4 lines] > phase difference that allows the quarterwave plate to produce more > than a simple 90 degree rotation? I misunderstood. My mistake. It seems to me that it would work also for coherent radiation. However, I only learnt how a circular polariser works 10min ago by reading RN Clark's post here, so may be wrong (it is far too late here for me to be thinking). Please do correct me if I am mistaken. Cheers. >> What you meant is that the incoming light is usually not polarised >> in one direction (at least I assume this is what you meant). [quoted text clipped - 7 lines] >That is a question, rather than a statement (I'm a photographer, not an >opto-electronic scientist). You can have a simple answer to that question, but not a simple explanation. The circular polariser works exactly the same for coherent and incoherent light. That's the easy bit. The problem with explanations is that they usually ignore the wave/particle dual nature of light. Explanations of polarisation always focus on the wave part and ignore the photons. The simplest way to think of it is that each photon behaves like a little burst of waves, with a wavelength, a polarisation direction etc. In random light, each photon will have its polarisation direction and the phase of its light different. It will pass through polarisers, or not, according to the relative orientations. Coherent light consists of photons whose polarisation direction and phase are all the same. However, each one follows the same rules when it passes through a polariser; it's just that the results will all be the same for each photon because of their identical polarisation/phase. However, the "phase" in the description of retardation refers, not to the phase of one photon relative to its neighbours, but to the phase relationship between its O and E components as it passes through a birefringent material. It is, if you like, its internal phase, not its phase relative to its, er, relatives. Hope this has not terminally confused you (it certainly makes my head spin). David SignatureDavid Littlewood SNIP > Hope this has not terminally confused you (it certainly makes my > head spin). No doubt from using a circular polarizer ;-) Your explanations are appreciated. SignatureBart > You can have a simple answer to that question, but not a simple > explanation. Hello, Could you please explain what is wrong with or complicated about the following explanation: Consider a completely polarised wave, the axis of polarisation of which is vertical (along the z-axis). That is, the electric field is always in the direction of the z axis. It now passes through a plate with the property that, along a direction at 45 degrees clockwise from the z axis and perpendicular to the direction of propagation, it passes light with no change; along a direction at 45 degrees anticlockwise from the z-axis and in the same plane, it retards the phase of this wave by 1/4 wavelength. So, assume that the moment our wave arrives at this plane, the electric field is at its maximum (top) position (this amounts to choosing the origin of time, in this case). Its component along the +45 degree axis is transmitted unchanged, but the one along the -45 degree axis is retarded by 1/4 wavelength, ie, since it was at the top of its cycle, it ends up at exactly zero and going up (of the previous cycle). So there is a phase difference of pi/2. After the plane, the +45 component starts decreasing from the max, while the -45 increasing from zero, and, in general, they just continue evolving as before. Some thought will show that their resultant will rotate (clockwise). Hence circularly polarised light. I know this is wordy, but if it's drawn, it's pretty simple really. Of course, I may misunderstand what a "quarter wave plate is". > The circular polariser works exactly the same for coherent and > incoherent light. That's the easy bit. [quoted text clipped - 7 lines] > different. It will pass through polarisers, or not, according to the > relative orientations. But why is it a problem? That light is made of particles is irrelevant to this, as far as I can see. Could you explain why you think it is? > Coherent light consists of photons whose polarisation direction and > phase are all the same. However, each one follows the same rules when it [quoted text clipped - 6 lines] > birefringent material. It is, if you like, its internal phase, not its > phase relative to its, er, relatives. Well, I don't see how it is either useful or meaningful to talk of photons passing through birefringent materials. Since, as far as I (vaguely) remember, such materials are formally ones for which the refractive index is tensorial, it seems the whole thing can be understood using nothing but Maxwell's eqns. Furthermore, you talk of the phase of a photon relative to others. But, if you are going to think of this at such a microscopic level, photons don't pass through matter, they get absorbed and reemitted and so on, and in general, do very complicated things in there (if we insist of thinking of photons in more or less classical terms; we don't insist, the picture is completely different). So, which photon, and which relative? Or maybe you meant something completely different and I misunderstood? Thanks. >>You can have a simple answer to that question, but not a simple >>explanation. [quoted text clipped - 72 lines] > > Thanks. See: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polclas.html If you like math: http://farside.ph.utexas.edu/teaching/qm/lectures/node7.html Now it gets really "heavy:" http://en.wikipedia.org/wiki/Photon "However, in the framework of special relativity, this is not the case for massless spin-1 particles, such as the photons. They have only two spin projections, or helicities, which correspond to the right- and left-handed circular polarizations of classical electromagnetic waves. Linear polarizations are produced by the superposition of the two spin projections of a photon." In some places you'll find the description of a photon as always being circularly polarized, which is what the above statement alludes to. Roger > See: > http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polclas.html [quoted text clipped - 14 lines] > being circularly polarized, which is what the above statement > alludes to. Roger, Thanks for the links. I am in fact a field theorist (many-body field theory though) so I think I understand (more or less) what is explained in the links. I must admit I fail to see what this has to do with the question in my post, though. What I did not understand is the reference to the need to speak of the particulate nature of photons to explain polarisation (ie I do not see what was mentioned in this thread that needs qm to be understood). David Littlewood wrote "The problem with explanations is that they usually ignore the wave/particle dual nature of light", and I do not see why this is a problem (I mean that the mere existence of a more microscopic point of view does not invalidate the higher-level point of view, so why call it a problem that the lower level is not invoked?). Basically, I very strongly object to this paragraph, written by David Littlewood: > The problem with explanations is that they usually ignore the > wave/particle dual nature of light. Explanations of polarisation always [quoted text clipped - 4 lines] > different. It will pass through polarisers, or not, according to the > relative orientations. Eg it is not a good idea to think of a photon as a little burst of waves with a wavelength, polarisation direction etc (it's either a very large spatially burst of waves with an almost definite wavelength or a little burst of waves with lots of wavelengths as you well know, and I suppose David too). You can conclude many incorrect things this way. Also, saying of a photon "it will pass through polarisers..." is hardly a good way to explain things. Basically, I was asking why he feels that explanations of polarisation that do not involve qm are unsatisfactory. > I must admit I fail to see what this has to do with the > question in my post, though. You are correct: this tangent has nothing to do with your original post. But sometimes theses threads result in a lot of information transfer, whether about a particular subject, or in understanding how others see and react to other posts, in which case we get to know the personalities behind the names. Concerning your original post. For modern cameras with AF and and TTL metering, only get circular polarizers (for all the macro to quantum mechanical explanations previously described by several posters). Roger > For modern cameras with > AF and and TTL metering, only get circular polarizers My understanding so far: Circular polarizer's quarter wave plate makes one wavelength perfectly circular polarized. Other colors are partially linear polarized. The amount and orientation of the linear polarization depends on the color that differ from the quarter wave plate wavelength. The optical path to the viewfinder is partially polarizing. So it may interact with the circular polarizer. Depending on the polarizer orientation, some color may be reduced more than other colors, because of the above color dependant partial linear polarization of circular polarizer. So I may see a color change. I want to avoid this color change caused by the interaction of the viewfinder and the circular polarizer. I want to be able to depend on the viewfinder to rotate the polarizer to see the color effect I want that will be recorded on the sensor. So I choose linear polarizer. I choose to live with the AE inaccuracy caused by linear polarizer, because the AE error amount is small. And since I do not use spot metering, average or matrix metering is not perfect anyway. >>For modern cameras with >>AF and and TTL metering, only get circular polarizers [quoted text clipped - 19 lines] > because the AE error amount is small. And since I do not use spot > metering, average or matrix metering is not perfect anyway. Your logic is flawed. The main effect of the circular polarizer is FIRST as a linear polarizer, so the effect on the scene is the same whether or not you use a linear or circular polarizer. The effect of the quarter wave plate in the circular polarizer does not change the effects of the linear polarizer on the scene. It does not change the color of the light you perceive, nor what the camera records (unless your eyes are polarized). The quarter wave plate is reducing the linearly polarized light entering the camera (and via the viewfinder, your eye) by changing it to circular (or elliptically polarized light. The fact that some colors are not perfectly at the 1/4-wave plate wavelength is irrelevant regarding color change on the scene. It doesn't change the scene's colors; it only changes the camera's electronic sensors sensitivity to be more accurate. Use of a linear polarizer simply increases the risk of AF and or exposure errors. Roger > >>For modern cameras with > >>AF and and TTL metering, only get circular polarizers [quoted text clipped - 23 lines] > is FIRST as a linear polarizer, so the effect on the scene is the > same whether or not you use a linear or circular polarizer. Linear polarizer makes every color linear polarized. The ratio of polarization and the orientation of polarization is the same for every color. > The effect of the quarter wave plate in the circular polarizer > does not change the effects of the linear polarizer on the scene. Circular polarizer does not make every color perfectly circular polarized. The ratio and orientation of partial linear polarization changes based on the color wavelength. > It does not change the color of the light you perceive, nor what > the camera records (unless your eyes are polarized). The viewfinder is partially polarized. > The quarter wave plate is reducing the linearly polarized light > entering the camera (and via the viewfinder, your eye) by [quoted text clipped - 3 lines] > color change on the scene. > It doesn't change the scene's colors But it changes the color of the viewfinder display, because of the systematic partial linear polarization based on color wavelength caused by the quarter wave plate, and the partially polarizing viewfinder. >>>I choose to live with the AE inaccuracy caused by linear polarizer, >>>because the AE error amount is small. And since I do not use spot [quoted text clipped - 7 lines] > polarization and the orientation of polarization is the same for every > color. No. It depends on the polarization of light coming from the scene. >>The effect of the quarter wave plate in the circular polarizer >>does not change the effects of the linear polarizer on the scene. > > Circular polarizer does not make every color perfectly circular > polarized. That does not matter. > The ratio and orientation of partial linear polarization > changes based on the color wavelength. It has no effect on the scene and how the linear polarizer in front of the quarter wave plate selectively blocks light from the scene. >>It does not change the color of the light you perceive, nor what >>the camera records (unless your eyes are polarized). > > The viewfinder is partially polarized. So the use of a straight linear polarizer maximizes the problem whereas the quarter wave plate in the circular polarizer minimizes the problem. >>The quarter wave plate is reducing the linearly polarized light >>entering the camera (and via the viewfinder, your eye) by [quoted text clipped - 7 lines] > systematic partial linear polarization based on color wavelength caused > by the quarter wave plate, and the partially polarizing viewfinder. See the above statement. The polarization of optical components in the viewfinder, typically a small fraction of an f-stop, will have small effects. Those effects would be maximized by a linear polarizer and minimized by a circular polarizer. The greater issue is not the polarization induced by the pentaprism, which you would need an instrument to measure, it is the effect on AF and exposure metering. Roger > >>>I choose to live with the AE inaccuracy caused by linear polarizer, > >>>because the AE error amount is small. And since I do not use spot [quoted text clipped - 9 lines] > > No. It depends on the polarization of light coming from the scene. I am not discussing the effect of the polarizer on the outside scene. I am discussing the effect of the partially polarized viewfinder on the light behind the polarizer. > >>The effect of the quarter wave plate in the circular polarizer > >>does not change the effects of the linear polarizer on the scene. [quoted text clipped - 3 lines] > > That does not matter. That is the most important basis of the discussion. > > The ratio and orientation of partial linear polarization > > changes based on the color wavelength. > > It has no effect on the scene and how the linear polarizer > in front of the quarter wave plate selectively blocks light > from the scene. Again, I am not discussing the effect of the polarizer on the outside scene. I am discussing the effect of the partially polarized viewfinder on the light behind the polarizer. > >>It does not change the color of the light you perceive, nor what > >>the camera records (unless your eyes are polarized). [quoted text clipped - 4 lines] > problem whereas the quarter wave plate in the circular > polarizer minimizes the problem. With a linear polarizer, every color will be darker (or not) by the same amount. You think that is the problem. For me, this is not the problem, since there is no color change, only brightness change. My eyes are not that sensitive to slight brightness change. But they are sensitive to slight color change. With a circular polarizer, some color will be darker than other colors. Which color is darker depends on the relative orientation of the circular polarizer and the partially polarizing viewfinder. So as you rotate the circular polarizer, the viewfinder will show different color. To control the variables better, keep the circular polarizer stationary, so its effect on the outside scene is fixed. Then you rotate the camera viewfinder, the color shown in the viewfinder will change. This is the problem that linear polarizer does not have. > >>The quarter wave plate is reducing the linearly polarized light > >>entering the camera (and via the viewfinder, your eye) by [quoted text clipped - 7 lines] > > systematic partial linear polarization based on color wavelength caused > > by the quarter wave plate, and the partially polarizing viewfinder. >> >>>I choose to live with the AE inaccuracy caused by linear polarizer, >> >>>because the AE error amount is small. And since I do not use spot [quoted text clipped - 59 lines] > rotate the camera viewfinder, the color shown in the viewfinder will > change. This is the problem that linear polarizer does not have. Uh, why do you care about color shifts showing up in the viewfinder? >> >>The quarter wave plate is reducing the linearly polarized light >> >>entering the camera (and via the viewfinder, your eye) by [quoted text clipped - 7 lines] >> > systematic partial linear polarization based on color wavelength caused >> > by the quarter wave plate, and the partially polarizing viewfinder. Signature--John to email, dial "usenet" and validate (was jclarke at eye bee em dot net) > >> >>>I choose to live with the AE inaccuracy caused by linear polarizer, > >> >>>because the AE error amount is small. And since I do not use spot [quoted text clipped - 61 lines] > > Uh, why do you care about color shifts showing up in the viewfinder? Because I rotate the polarizer, while looking into the viewfinder, in search for the color effect I like. If the color shown in the viewfinder is not the same as the color that will be captured by the sensor, then it is a problem. This is the problem with circular polarizer that I can avoid by using linear polarizer instead. >>Uh, why do you care about color shifts showing up in the viewfinder? > [quoted text clipped - 3 lines] > sensor, then it is a problem. This is the problem with circular > polarizer that I can avoid by using linear polarizer instead. Q: Which produces a greater change in transmitted color: a) crossed linear polarizers? b) two circular polarizers? c) circular polarizer followed by a linear polarizer? A: a. If you try this experiment and find a larger shift with b or c, then your circular polarizers are crap. Roger >>>Uh, why do you care about color shifts showing up in the viewfinder? >> Because I rotate the polarizer, while looking into the viewfinder, [quoted text clipped - 13 lines] > >Roger Roger, I'm afraid you are quite wrong here. The CPs in (b) and (c) *must* create more colour variation than the two LPs in (a), simply from the physics of how CPs work. A pair of ideal LPs will produce no visible colour variation (and do not do so in the ones I have tested) though the strong intensity variation makes it hard to observe perfectly. A CP followed by any polariser (CP or LP) will show slight colour variations, since only light of the wavelength for which it has a retardation of one quarter of a wavelength will be perfectly circularly polarised. Light of other wavelengths will remain (slightly) elliptically polarised and will show some variation in intensity as the two filters are rotated. Try it and you will see it is so. Where I differ from Aaron is whether this is an effect of any significance in use, and whether it outweighs the effect on metering and AF systems of using a linear polariser. Aaron thinks the polarisation of the reflex mirror is enough to give him an unacceptable level of colour change in the viewfinder, but does not mind the risk of errors in metering etc. I take the opposite view on both points (the polarisation effect of the mirror is smaller than the use of a second polarising filter). What cannot be denied, however, is that the effect does exist when using two filters. David SignatureDavid Littlewood >>>> Uh, why do you care about color shifts showing up in the viewfinder? >>> [quoted text clipped - 19 lines] > create more colour variation than the two LPs in (a), simply from the > physics of how CPs work. I disagree. I agree about the points that CP + LP produces color. On a white background, my CPs produce a slight warming or orange cast. Linear polarizers when crossed typically produce a very strong color, and in the ones I have with me at the moment, a very very very strong purple. > A pair of ideal LPs will produce no visible colour variation (and do not > do so in the ones I have tested) though the strong intensity variation > makes it hard to observe perfectly. See above. Ideal does not exist. > A CP followed by any polariser (CP or LP) will show slight colour > variations, since only light of the wavelength for which it has a > retardation of one quarter of a wavelength will be perfectly circularly > polarised. Light of other wavelengths will remain (slightly) > elliptically polarised and will show some variation in intensity as the > two filters are rotated. Try it and you will see it is so. I agree with this. The quality of the CPs will vary how much this happens. In one of my original posts I stated how two quantary CPs were horrible, but two Hoyas were very good (almost no variation, depending on what you look at it is obvious or not). > Where I differ from Aaron is whether this is an effect of any > significance in use, and whether it outweighs the effect on metering and [quoted text clipped - 5 lines] > filter). What cannot be denied, however, is that the effect does exist > when using two filters. On this point I agree. I didn't deny effects didn't exist, I argued which was the greater effect. Again read the question again: "Which produces a greater change" Roger >>> Q: Which produces a greater change in transmitted color: >>> a) crossed linear polarizers? [quoted text clipped - 15 lines] >a very strong color, and in the ones I have with me at the moment, >a very very very strong purple. I agree that when real-life polarisers are totally crossed there is a residual colour - mine (a pair of Zeiss LP filters mounted in a unit for microscopy, which I have used to test DSLR noise on 30-second exposures) are typically blue - but this is usually only observable on photos, and is not visible when I can see through them. Lesser LPs are of course likely to show this earlier. >> A pair of ideal LPs will produce no visible colour variation (and do >>not do so in the ones I have tested) though the strong intensity >>variation makes it hard to observe perfectly. > >See above. Ideal does not exist. Indeed; however, we were talking about what was visible through the camera viewfinder, which pre-supposes they are not completely crossed. >> A CP followed by any polariser (CP or LP) will show slight colour >>variations, since only light of the wavelength for which it has a [quoted text clipped - 8 lines] >no variation, depending on what you look at it is obvious or >not). I agree that if the CPs are crap they are likely to show *more* variation in colour than using top class CPs. This does not prove the contrary: that the existence of a greater colour change using a CP/LP pair than with a pair of similar-quality LPs means the CP is "crap" (see below). >> Where I differ from Aaron is whether this is an effect of any >>significance in use, and whether it outweighs the effect on metering [quoted text clipped - 9 lines] >argued which was the greater effect. Again read the >question again: "Which produces a greater change" Well, what I was actually disagreeing with was your statement (left in the quotes above for reference) that if the CP/LP combination produced more colour variation, "your CPs are crap". Assuming the polarising layers are identical, the 1/4 wave plate will probably introduce more colour variation on rotation than the LPs alone. (I was about to say "must inevitably, but there exists the possibility that the effects may happen to cancel out.) That is the thesis which I think is wrong.. Never mind, I don't think we are far apart, and it was a useful bit of mental exercise! David SignatureDavid Littlewood > Well, what I was actually disagreeing with was your statement (left in > the quotes above for reference) that if the CP/LP combination produced [quoted text clipped - 3 lines] > "must inevitably, but there exists the possibility that the effects may > happen to cancel out.) That is the thesis which I think is wrong.. I have updated my evaluating polarizing filters page with data using both linear and circular polarizers. See: http://www.clarkvision.com/photoinfo/evaluating_polarizing_filters I show extreme color shifts with crossed linear polarizers and metering errors of more than 5 stops with linear polarizers. I also show the color shifts you get with circular polarizers. > Never mind, I don't think we are far apart, and it was a useful bit of > mental exercise! Yep. Roger >> Well, what I was actually disagreeing with was your statement (left >>in the quotes above for reference) that if the CP/LP combination [quoted text clipped - 7 lines] >data using both linear and circular polarizers. See: >http://www.clarkvision.com/photoinfo/evaluating_polarizing_filters Very good page, Roger, and I don't disagree with any of it. >I show extreme color shifts with crossed linear polarizers and >metering errors of more than 5 stops with linear polarizers. >I also show the color shifts you get with circular polarizers. Yes, I can get this with 2 crossed LPs - but only when they are virtually at 90 degrees and the eye can't see through them. Does show on photos though. My best pair (Zeiss) give a rather more pleasing blue colour. At all "viewable" settings I cannot see a colour shift by eye. David SignatureDavid Littlewood > http://www.clarkvision.com/photoinfo/evaluating_polarizing_filters > > metering errors of more than 5 stops with linear polarizers. With 2 linear polarizers crossed, they can almost block out the light. Maybe the relatively very much stronger ambient light got into the viewfinder that threw off the meter by 5 stops. With 1 linear polarizer, as normally used, on your page earlier you showed 1/3 stop difference between 2 extreme positions. So the average metering error may be half of that, 1/6 stop. To me, these small errors are neglectable. >>http://www.clarkvision.com/photoinfo/evaluating_polarizing_filters >> [quoted text clipped - 8 lines] > metering error may be half of that, 1/6 stop. To me, these small errors > are neglectable. The scene was not polarized. With ONE linear polarizer, the polarizer introduced errors of 1/3 stop. There is no average here. On polarized scenes, the errors would be larger. The two linear polarizers simulate that (to an extreme). The two linear polarizer test had no ambient light issues. In fact I was so stunned by the meter reading that I repeated the metering, shielded the camera, put in the viewfinder blocker, all with the same result. The target was in sun but I and the camera were in shade by the roof of a house, further reducing extraneous light on the camera. The test simply illustrates the errors that could be encountered on polarized scenes, and why the manufacturers recommend using circular polarizers. The result: metering errors with linear polarizers can range 0 to 5 stops on a DSLR. Roger > >>http://www.clarkvision.com/photoinfo/evaluating_polarizing_filters > >> [quoted text clipped - 12 lines] > the polarizer introduced errors of 1/3 stop. There is > no average here. You said between the maximum and minimum position the exposure was different by 1/3 stop. So if the polarizer is rotated to any other orientation, the difference between the middle point and any of the 2 extremes would be less than 1/3 stop. > On polarized scenes, the errors would be > larger. The two linear polarizers simulate that (to an > extreme). I don't understand why you think whether the scene is polarized would make a difference on linear polarizer metering error. After the linear polarizer, the light is purely linear polarized to a certain orientation. The relative difference of this and the orientation of the partially polarizing beam splitter would decide how much intensity would be reduced, thus causing the metering error. Whether the scene is polarized would not have any effect on the polarized orientation of the light after the polarizer, thus it would not affect the metering error. The polarizer may reduce the intensity of the polarized scene more than unpolarized scene. But the sensor will see the same amount of this reduction as the meter will see before the beam splitter. So this interaction of the polarizer and the polarized scene will not affect the metering error. I can only think of the extreme darkness caused by 2 crossed polarizer as the cause of the 5 stop metering error, not because of the linear polarizer. You can try to put a linear polarizer in front of a circular polarizer, cross them to block out the light completely, just as with 2 linear polarizers, and see how the meter react. > The two linear polarizer test had no ambient light issues. Maybe the 2 linear polarizers crossed out too much light that the remaining light is too dark and is out of the working range of the meter? Or the camera gets fancy and try to expose this very dark scene as a night scene, as opposed to try to expose it to be medium gray. >>>>http://www.clarkvision.com/photoinfo/evaluating_polarizing_filters >>>> [quoted text clipped - 33 lines] > polarized orientation of the light after the polarizer, thus it would > not affect the metering error. The two polarizer test illustrates the effects of a polarized scene (at the extreme). > The polarizer may reduce the intensity of the polarized scene more than > unpolarized scene. But the sensor will see the same amount of this [quoted text clipped - 16 lines] > meter? Or the camera gets fancy and try to expose this very dark scene > as a night scene, as opposed to try to expose it to be medium gray. The exposure error, as seen in Figure 4C at http://www.clarkvision.com/photoinfo/evaluating_polarizing_filters is underexposed. Basically as the polarizers crossed and the light dimmed, the camera meter did not see much of a reduction in light, even though it was plainly obvious in the viewfinder. Putting on a linear followed by a circular polarizer the meter follows the light dimming just fine. You seem to be arguing that other light is contributing to the image. If so the exposure error would be overexposed. But it is underexposed. The viewfinder looked dark and no extraneous light was visible (the target, as in Figure 4D was dimly visible, and appeared purple, just as the image shows). You can deny the exposure errors and continue to use linear polarizers all you want. But the tests illustrate the real story and are the reason why manufacturers recommend circular polarizers. Roger > >>>>http://www.clarkvision.com/photoinfo/evaluating_polarizing_filters > >>>> [quoted text clipped - 63 lines] > the light dimmed, the camera meter did not see much of a reduction in > light, even though it was plainly obvious in the viewfinder. I just don't understand how this can happen. Can anybody explain? >> > Maybe the 2 linear polarizers crossed out too much light that the >> > remaining light is too dark and is out of the working range of the [quoted text clipped - 6 lines] >> the light dimmed, the camera meter did not see much of a reduction in >> light, even though it was plainly obvious in the viewfinder. > I just don't understand how this can happen. Can anybody explain? Some reasons that pop into my head: - The second polarizer was just right to give the metering system almost all light, causing it to believe the scene brighter (probably 2 stops) - The crossed _linear_ polarizers remove practically all light.[1] My 20D meters for LV-1 (which is the bottom of its metering range) even at LV-6 ... - Light entering through the viewfinder can spoof the metering system into believing in a much brighter scene, especially with the lens being darkened by 2 crossed linear polarizers. -Wolfgang [1] Since circular polarizers de-polarize the light behind them, even crossing them should not cause the second polarizer to see only polarized light --- especially polarized 90 degrees to it. With linear polarizers you should get 100% of the light blocked, if they were perfect and perfectly aligned at 90°. > >> > Maybe the 2 linear polarizers crossed out too much light that the > >> > remaining light is too dark and is out of the working range of the [quoted text clipped - 13 lines] > system almost all light, causing it to believe the scene > brighter (probably 2 stops) I think the theoretical maximum is 1 stop underexposure, if the beam splitter is a complete polarizer. If the beam splitter is a complete polarizer, if the incoming light is completely unpolarized, the completely polarizing beam splitter blocks out half of the light. To compensate this, the metered value is doubled, which is 1 stop. If the linear polarizer in front of the lens is completely aligned with the beam splitter, all light behind the polarizer gets to the meter, but the meter is also compensated to 1 stop brighter, so the exposure will be 1 stop darker. So if this reasoning is correct, the 5 stop underexposure must be caused by something else, not the linear polarizer. In reality, the beam splitter is not a complete polarizer. If it is a 50% polarizer (50% unpolarized light and 50% polarized light), then the maximum underexposure is 1/2 stop, and the maximum overexposure is 1 stop. The beam splitter is probably much less than 50% polarizing. > - The crossed _linear_ polarizers remove practically all light.[1] > My 20D meters for LV-1 (which is the bottom of its metering > range) even at LV-6 ... > - Light entering through the viewfinder can spoof the metering > system into believing in a much brighter scene, especially with > the lens being darkened by 2 crossed linear polarizers. These are what I think, too. http://digitcamera.tripod.com/#slr > >>Uh, why do you care about color shifts showing up in the viewfinder? > > [quoted text clipped - 11 lines] > A: a. If you try this experiment and find a larger > shift with b or c, then your circular polarizers are crap. We went through this before. My polarizers are B+W brand. If I stack two polarizers together, if the front is a circular polarizer, I see color change. If the front is a linear polarizer, I see brightness change, but no color change. Other people have confirmed this observation with high quality polarizers. And the theory discussed in this thread supports this observation. >>>>Uh, why do you care about color shifts showing up in the viewfinder? >>> [quoted text clipped - 20 lines] > > And the theory discussed in this thread supports this observation. There is a color shift along with the intensity shift. Measure it. It is more extreme than any color shift with a circular polarizer, at least in all polarizers I have examined, even ones for scientific instruments. Roger >> You can have a simple answer to that question, but not a simple >> explanation. [quoted text clipped - 28 lines] >I know this is wordy, but if it's drawn, it's pretty simple really. Of >course, I may misunderstand what a "quarter wave plate is". It's fine, provided it is understood that the retardation is proportional to the thickness of the material. It is, BTW, conventional to refer to the first of your two directions as the ordinary (O) ray and the one retarded relative to it as the extraordinary (E) ray. Also, it might be slightly confusing to say the O ray is "transmitted unchanged". Both the O and E rays pass through according to the usual laws, but the refractive index of the material differs in the two directions, so their velocities are different, and hence their wavelengths are different. The E ray is slower than the O ray, and thus the two get out of phase. >> The circular polariser works exactly the same for coherent and >> incoherent light. That's the easy bit. [quoted text clipped - 10 lines] >But why is it a problem? That light is made of particles is irrelevant >to this, as far as I can see. Could you explain why you think it is? Well, the question arose as to "coherent" (in-phase) light, and it seemed that confusion was arising between the phase coherence between different photons, and the phase coherence of the two orthogonal vector directions (the "O and E rays") of any given photon. If you weren't confused, then obviously the clarification was not necessary for you. >> Coherent light consists of photons whose polarisation direction and >> phase are all the same. However, each one follows the same rules when it [quoted text clipped - 12 lines] >refractive index is tensorial, it seems the whole thing can be >understood using nothing but Maxwell's eqns. As I said, I only introduced the photon bit to disentangle a possible confusion between external coherence (i.e. between different photons/quanta) and the phase difference between different field vector angles in a single photon/quantum of light. >Furthermore, you talk of the phase of a photon relative to others. But, >if you are going to think of this at such a microscopic level, photons [quoted text clipped - 5 lines] > >Or maybe you meant something completely different and I misunderstood? Well, I cannot claim to be a particle physicist, only someone who did chemistry many years ago. Most of us need to think of such things in terms of analogues which our brains can grasp, even though we know it is merely a feeble attempt to describe a more complex reality. I find it helpful to think of photons as tiny bundles of energy which have a discrete existence (quantisation) but also behave to an extent like energy waves. As long as one remembers the wave/particle duality, and that one's nice, easy mental pictures are just stories-for-children, then one can mostly get by in real life. However, it is undoubtedly true that different "bits" of light (call them photons, quanta, or what you will) can be either coherent with respect to each other, or not. How you choose to envisage this is up to you. BTW, I like the idea that "the whole thing can be understood using nothing but Maxwell's equations". I think if I tried to get my brain to handle it using nothing but Maxwell's equations (rather than its nice comfortable little mental pictures) I would struggle. David SignatureDavid Littlewood > It's fine, provided it is understood that the retardation is > proportional to the thickness of the material. It is, BTW, conventional [quoted text clipped - 5 lines] > velocities are different, and hence their wavelengths are different. The > E ray is slower than the O ray, and thus the two get out of phase. David, Thanks, I did not know the terminology for O and E rays. And yes, you are right, I should not have said "transmitted unchanged", since the relevant effect of this quarter-wave plate is to introduce a phase difference between the two components. > >But why is it a problem? That light is made of particles is irrelevant > >to this, as far as I can see. Could you explain why you think it is? [quoted text clipped - 4 lines] > directions (the "O and E rays") of any given photon. If you weren't > confused, then obviously the clarification was not necessary for you. However, I do not see the need to discuss photons (in addition to waves). I guess we all think about these things in different ways, > BTW, I like the idea that "the whole thing can be understood using > nothing but Maxwell's equations". I think if I tried to get my brain to > handle it using nothing but Maxwell's equations (rather than its nice > comfortable little mental pictures) I would struggle. I meant that the whole thing may be understood using ideas of waves travelling in space, with no recourse to photons (or any particle aspects), rather than using only equations. Understanding comes by analogy, not calculations (and when calculations lead to understanding, it is through analogy). >> BTW, I like the idea that "the whole thing can be understood using >> nothing but Maxwell's equations". I think if I tried to get my brain to [quoted text clipped - 6 lines] >analogy, not calculations (and when calculations lead to understanding, >it is through analogy). Well, I did rather have my tongue in my cheek there. Of course, the "photon" part of the mental image does come in very useful for dealing with things like the photoelectric effect, and with photochemical reactions - including, especially, silver halide imaging. I agree it is an unnecessary obfuscation when looking at polarisation and birefringence, and only mentioned it to try to distinguish a quite different phenomenon. David SignatureDavid Littlewood > Regarding the two circular polarizer experiment: if you take > two circular polarizers and stack them then rotate one, > you should see NO change in intensity, and NO color change. > If you do, they are not quality circular polarizers. I see very obvious color change when rotating the rear polarizer behind a circular polarizer. It is a B+W multicoated circular polarizer that supposedly won quality award in a German photo magazine. Which circular polarizer does not have this color change when rotating a rear polarizer? BTW, if you screw the 2 polarizers together, there are 3 rings that can be rotated. Rotating the last ring does not rotate any filter. You need to rotate the middle ring to rotate the rear filter. >>Regarding the two circular polarizer experiment: if you take >>two circular polarizers and stack them then rotate one, [quoted text clipped - 6 lines] > polarizer does not have this color change when rotating a rear > polarizer? Do you have the two polarizers in the same direction? For example, one threaded into the other, and both are circular polarizers? My Hoya multicoated filters do not show intensity or color changes as one polarizer is rotated with respect to the other (it does not matter which). Linear polarizers will show huge changes in intensity and usually color changes as well. A few months ago I broke a polarizer (dropped it on a rock) and in a rush to get another picked up two Quantaray circular polarizers. They show extremely poor performance showing both intensity changes and color changes when the two polarizers are rotated. I replaced them with Hoyas and again they show no changes. The Quantaray's are fine as linear polarizers so I'll use them with my 4x5. > BTW, if you screw the 2 polarizers together, there are 3 rings that can > be rotated. Rotating the last ring does not rotate any filter. You need > to rotate the middle ring to rotate the rear filter. I know how to rotate a polarizer. Roger > >>Regarding the two circular polarizer experiment: if you take > >>two circular polarizers and stack them then rotate one, [quoted text clipped - 9 lines] > Do you have the two polarizers in the same direction? For example, > one threaded into the other, Yes. > and both are circular polarizers? No. The rear is a B+W linear polarizer. The color change effect is caused by the front circular polarizer. If the linear polarizer is used alone there is no such color change on non-polarized scenes. Of course on polarized scenes, color change is the expected effect of the polarizer, e.g., on grass, leaf, flower, sky, etc. > My Hoya multicoated filters do not show intensity or color changes as > one polarizer is rotated with respect to the other (it does > not matter which). If the scene is polarized, rotating the front polarizer may change the color. This color change will mix with the color change caused by the relative orientation of the 2 polarizers, sometimes the combined effect is reduced. > Linear polarizers will show huge changes > in intensity and usually color changes as well. If the linear polarizer is in front of the circular polarizer, they can almost black out the scene when the 2 are 90deg to each other. But there is no color change like those when the circular polarizer is in the front. >>>>Regarding the two circular polarizer experiment: if you take >>>>two circular polarizers and stack them then rotate one, [quoted text clipped - 19 lines] > on polarized scenes, color change is the expected effect of the > polarizer, e.g., on grass, leaf, flower, sky, etc. Then you are not doing the test that checks the quality of the circular polarizer. While you can consider rotating a polarizer a color change what is in fact happening, e.g. 1) reflections from surfaces: remove the specular component, or 2) remove haze scattering in the sky, or 3) because the blue sky is polarized 90 degrees from the sun, block that sky component from entering the camera. So it is like selectively subtracting some light from the scene, and in general because it is scattered light (e.g. the haze) the effect is to increase contrast which gives better color (more saturated). The color hasn't changed (e.g. blue sky is still blue sky simply a more saturated blue). >>My Hoya multicoated filters do not show intensity or color changes as >>one polarizer is rotated with respect to the other (it does [quoted text clipped - 12 lines] > there is no color change like those when the circular polarizer is in > the front. I think we are describing 2 different things. I am talking about the transmission profile of the polarizers, not a typical scene they includes polarized light. For example, polarizers are usually gray to greenish-gray when you look through them. Example: look through the polarizers at some white paper. Cross 2 linear polarizers while looking at the white paper and the color will usually change (as the light intensity is reduced a lot) to a purple color. The better the linear polarizers, the less the color change. With TWO CIRCULAR polarizers, as you rotate them while looking at the white paper, there should be no intensity or color change (for quality circular polarizers). Roger >No. The rear is a B+W linear polarizer. The color change effect is >caused by the front circular polarizer. This is because the retardation effect of the quarter wave plate varies with the wavelength of the light. Usually these things are designed so that green light has its extraordinary ray retarded by exactly 1/4 of a wavelength relative to the ordinary ray, making the light perfectly circular in polarisation *. Since the blue and red light have the same velocity but different wavelengths, they are retarded respectively slightly more, and slightly less, than 1/4 of a wavelength; instead of being perfectly circularly polarised they come out with slightly elliptical polarisation. When the CP is used alone this effect is not visible, but the second polariser effectively "detects" the lack of perfectly circular polarisation and, in certain orientations, will slightly attenuate the red and blue light. Thus the effect is to enhance the green a little. *See texts on birefringence and wave plates for full explanation, which really requires pictures to make it clear. David SignatureDavid Littlewood SNIP (for the essence, not for quality) > Since the blue and red light have the same velocity but different > wavelengths, they are retarded respectively slightly more, and > slightly less, than 1/4 of a wavelength; instead of being perfectly > circularly polarised they come out with slightly elliptical > polarisation. This is, IMHO, the real issue with 'quarter wave retarders', they are wavelength dependent. Even my (considered as mechanical *and* optical quality) B+W circular polarizers will produce a Magenta cast when used to cancel each other out. Mine are apparently optimized for cancelling Green polarized wavelengths. > When the CP is used alone this effect is not visible, but the second > polariser effectively "detects" the lack of perfectly circular > polarisation and, in certain orientations, will slightly attenuate > the red and blue light. Thus the effect is to enhance the green a > little. Wouldn't the second polarizer, assuming similar characteristics as the first, 'enhance' (in fact again under-correct) Magenta transmission? Bart |
Enable EMail Alerts
Start New Thread
Reply to this Message