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    Re: Sextants with Polarizing filters
    From: George Huxtable
    Date: 2006 Jan 25, 16:02 -0000

    In response to Peter Fogg's recent message stating-
    "...On the certain range: each polarising filter
    reduces the brightness of light passing through by about one and a half f/
    stops, or 150% (this varies between manufacturers; 150/200%)."
    
    I had asked-
    
    | > Some things I don't understand about Peter Fogg's recent message are
    | >
    | > 1. How can you reduce the brightness of light by more than 100%? If you
    | > have reduced it by 100%,
    | > then you have already achieved complete darkness.
    
    To which he has replied-
    
    | An f/stop more is a doubling of light brightness, thus 100%, while an f/stop
    | less is a halving, thus indeed 50%.
    
    (which is correct). And then added, somewhat acrimoniously-
    
    |A somewhat pedantic point hardly worth
    | making, I would have thought.
    
    Well, if Peter would like others to follow his meaning, it needs to be expressed in terms others can
    understand. Reducing brightness by one f-stop is a reduction to one-half, or 50% (as Bill has
    explained). Reducing by two f-stops is a reduction to a quarter, or 25% (or, if you really want to
    express it in such an awkward way, a reduction BY 75%). Neither percentage increases nor percentage
    reductions can be added in the way that Peter has done, they multiply together. A reduction by one
    and a half f-stops is a reduction to 1/2.8 of the original, or to about 35%, or a reduction BY 65%.
    It is of course quite impossible to reduce light intensity BY more than 100%.
    
    If light intensity is reduced to complete darkness, I ask Peter by what percentage the light has
    been reduced, in his reckoning?
    
    ==========================================
    
    Then I added-
    
    | > 2. Yes, if you introduce one polarising filter, you will approximately
    | > halve the light intensity,
    | > for initially unpolarised light, because you are excluding the 50% of the
    | > light that is polarised at
    | > right angles to the direction of the filter that lets light through. Plus
    | > a slight further
    | > reduction, to allow for losses in the plastic material and in surface
    | > reflections.
    | >
    | , that
    | > won't halve the light intensity again, because now all the light entering
    | > it is polarised in the
    | > preferred direction. This second filter will therefore cause a slight
    | > reduction in intensity, as
    | > before, because of further losses in that extra layer of plastic and
    | > further reflections, but
    | > nothing like the reduction that the first filter caused. That's a
    | > deduction based on theoretical
    | > reasoning, not on any practical tests by me, so it may possibly be
    | > contradicted by experiment. I
    | > would be interested to learn if it is.
    
    And Peter responded-
    |
    | Take two polarising filters and try it out, George, as suggested by me
    | earlier (I do hope it was included in the part you did understand).
    
    [do I detect another sour note there, perhaps? - George]
    
    |Once you
    | see it working, more convoluted reasoning may be introduced to explain how
    | it does, perhaps..
    |
    | > ... if you add a second filter, that's polarised in the same
    | > direction as the first ...
    |
    | By turning the second filter the remaining light polarised in the other
    | direction is gradually choked off until no light passes.
    
    Well, of course, there's no argument about that last sentence. That's the way crossed Polaroids
    operate. What I was pointing out was something rather different. Here it is again.
    
    If you look at a scene through a single polaroid, its brightess is reduced, usually to about half
    the original, but it depends on how polarised the light was in the original scene. Now add another
    similar polaroid. First find the angle at which the two polarisation planes are crossed, by turning
    one until you get blackout. Now turn one through 90 degrees, until they are aligned, and are
    transnitting maximal light. Now remove one of the polaroids. Does the brightness increase
    significantly when you do that? I predict that there won't be a noticeable increase. Because the
    second polaroid, when its axis is aligned with the first, didn't give rise to a significant
    reduction in brightness, compared with what is seen through a single polaroid, the light already
    being polarised in the right plane. Bill may even be able to measure it with a light-meter. Of
    course, some reduction, perhaps 10% or so, is expected, simply because any optical component you put
    into a light path, even clear glass, loses some light by surface reflections.
    
    Has Peter tried that test, I wonder?
    
    There's nothing like a bit of practical observation, but I don't intend to sacrifice a good pair of
    polaroid sunglasses by sawing them across the bridge, just to try that out. But no doubt many list
    members (perhaps even Peter) can find a second pair in their bottom drawer, and will be able to do
    so. It would be interesting to compare results.
    
    Bill wrote-
    
    "On average, a polarizing filter will reduce light by 1.5 f-stops (1 to 2 stops range depending on
    light source.)
    
    Now, if we stack 2 filters (picket fences) together, with each axis aligned, we have a combined loss
    of 1-2 f stops + 1-2 f stops (depending on light source).  Therefore a loss of 2-4 f stops."
    
    That's where we disagree, I suggest that adding a second polaroid filter, aligned with the first in
    its direction of polarisation, will give rise to only a tiny reduction in brightness. That may be
    surprising (to Bill and to Peter), but that doesn't mean that it's wrong..
    
    Bill added-
    
    "Given your analytical mind, and to paraphrase your observation, "If the sun's rays are assumed as
    parallel, and I nail 2 identical fences together, why would the second fence cause anymore light
    loss than the first fence." In a perfect world, you would be correct.  But alas...
    
    Back to the pragmatic world. As we begin to rotate one filter (picket fence) relative to the other,
    the "apertures" that allows light to pass through get smaller and smaller.  Finally, at 90 degrees
    rotation, all we are left with are square areas whose sides are equal to the distance between the
    "slats" transmitting light.  Not a bunch of aligned photons given the spacing between "slats" on a
    polarizing filter.  Nonetheless, not 0 light transmission."
    
    First, I didn't say any such words as Bill attributes to me. Second, Bill is pursuing his
    picket-fence analogy rather further than it will usefully go. Bill describes the effect of rotation
    of the second polaroid in a way that I would concur with, even as far as "not 0 light transmission",
    though a good pair of polaroids gets pretty damn close to that ideal. No doubt Bill has plenty of
    relevant kit in his locker for making just such a test as I have suggested, in his "pragmatic
    world", and getting numerical answers. I ask him to do so, and tell us what he finds.
    
    George.
    
    contact George Huxtable at george@huxtable.u-net.com
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    
    
    

       
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