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George

First a primer for those without your scientific/engineering background:

A polarizing filter is akin to a picket fence.  A series of fine lines that
allow light vibrating on one plane to pass through.  Using the fence
analogy, a telephone pole (vertical) aligned with the fence can be seen in
its entirety through the gaps in the fence (given sufficient distance).
With a curb (horizontal) only the portions not blocked by the fence can be
observed.

How much affect a polarizing filter can have depends on how "scattered" the
light source is.  A north sky (in the northern hemisphere) is reflected
light, so very scattered. In this case the filter may reduce the sky by 2 f
stops (1/4 of light transmitted). In other cases, where the light operates
closer to one plane, only 1 f stop loss (1/2 the light).

NOTE: f-stops are based on the square root of 2.  Hence 1, 1.4, 2, 2.8, 4,
5.6, 8, 11.2, 16, 22, 32 ....  Each stop admits 1/2 the light of the
previous stop.

Photo texts will tell you (in theory) that a polarizing filter will not help
reduce glare from a mirror or other highly-reflective surface.  In practice
this is not true.  A lot depends on what the surface is reflecting.

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.

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.

From a practitioner's viewpoint, I have stacked polarizing filters to
photograph a waterfall of rocky stream bed for a "dream like" quality where
the moving water becomes "fog."  Adjusted filters for approx. 1/30 second
exposure. Then adjust for reciprocity failure (which truly is a dog chasing
its own tail). Then bracket, bracket, bracket exposures. Or for copy-stand
work of glossy flats, polarize each light source, and add a polarizing
filter over the lens.

Speaking to Mikes point on IE change, "It is important to note however that
even good polarizers change the index error when used. I know that the
manuals say that this can happen with any shades but I have only ever
measured it with polarisers - maybe it has to do with the two pieces of
glass in each filter. In any event IE must be measured with the filters in
place - orientation does not seem to matter."

Yes, the pragmatic problems of stacking filters. Myriad. Are the filters
exactly parallel to each other and the ray path?  If not, different
refraction indexes of colored glasses.  All but the new "ideal glass"
transmits and reflects, so ghosts (flare) from light bouncing between
filters. Stacked filters allow the air between the filters to heat up...

Hope that helps,

Bill

> 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.
>
> 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.
>
> But then, if you add a second filter, that's polarised in the same direction
> as the first, 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.

   

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