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The Cokin "Varicolor" 173 filter is a polarization filter of sorts that they sell for their various filter systems. Apparently its purpose is to provide some sort of polarization-dependent blue/yellow color shift, or to intensify certain colors from polarized sources (from what I've read, landscape photographers used them to intensify blue water, for instance).

Here's the thing, though: I have a cheap regular CPL filter (Altura branding) which visually exhibits a normal polarizing effect when looked through from the lens side, but, when looked through "backwards," shifts from blue through orange as it's rotated. This got me to thinking - is that all the Cokin filter is (linear polarizer + quarter-wave plate)? And if it is, is there some way I can fabri-cobble something to mount this on my lens backwards?

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Is the Cokin 173 filter just a backwards CPL?

No. The effect is similar to mounting your polarizer backwards, but much more pronounced on the Varicolor filter.

Here is what my Hoya HD CPL looks like at 0° and 90° polarization angles, when oriented for correct mounting (male threads towards camera):

  • Note that this is NOT a variable-ND filter. In the right image, the filter blacks out the light because the light source itself is polarized (coming from an LCD, which uses polarization to control light output).

Circular polarizer, 0° rotation, correctly mounted Circular polarizer, 90° rotation, correctly mounted
Hoya circular polarizer, normal orientation, at 0° (left) and 90° (right) polarization

When I turn the polarizer backwards, here's what it looks like, again at 0° and 90° polarization angles.

  • Note that my iPhone's auto-white balance tried to adjust for the color shift in both images. The left, bluish-cast of the polarizer is actually more pronounced than shown. You can see the yellow-shifted LCD monitor outside of the CPL. The left image should actually have a cooler color-temperature. Similarly, the yellow-ish cast of the right image should be warmed up slightly (the monitor is slightly bluer than normal outside of the CPL view).

Circular polarizer, 0° rotation, reverse mounted Circular polarizer, 90° rotation, reverse mounted
Hoya circular polarizer, reverse-mount orientation, at 0° (left) and 90° (right) polarization

I don't own the Cokin Varicolor, but I have Singh-Ray's Gold-N-Blue polarizer (same effect). Here's the Gold-N-Blue filter at 0° and 90°, oriented for correct mounting:

Singh-Ray Gold-N-Blue polarizer, 0° rotation, correctly mounted Singh-Ray Gold-N-Blue polarizer, 90° rotation, correctly mounted
Singh-Ray Gold-N-Blue polarizer, normal orientation, at 0° (left) and 90° (right) polarization

Here's the Gold-N-Blue oriented for reverse-mounting:

Singh-Ray Gold-N-Blue polarizer, 0° rotation, reverse mounted Singh-Ray Gold-N-Blue polarizer, 90° rotation, reverse mounted
Singh-Ray Gold-N-Blue polarizer, reverse-mount orientation, at 0° (left) and 90° (right) polarization

I couldn't tell a difference the obverse- and reverse-mounted Singh-Ray Gold-N-Blue filter. But the degree of the effect between the the Gold-N-Blue and a reverse-mounted CPL is significant.

  • Also note, that in real life, the effect of the Gold-N-Blue (and Varicolor, I presume) is not so pronounced to turn the world into Denver Broncos colors, or to make it look like a Michel Bay color-graded film. The blue and gold/orange is extremely saturated here because the color-separated light from the LCD is already polarized (by definition of being an LCD monitor).
  • Because of the way a circular polarizer is constructed it doesn't polarize when mounted in the reverse position. You can check this by looking at a reflection from each side and rotating it. You'll see the reflection (assuming it's from a non-conducting surface) change in the normal direction, and not change in the reverse position. – BobT Dec 21 '17 at 3:17
  • @BobT Correct. So, open question: how is the Varicolor (and Singh-Ray) gold/blue CPL constructed? Is it 2 quarter-wave plates sandwiching a linear polarizer, so that no matter the orientation it looks the same? – scottbb Dec 21 '17 at 3:19
  • Good question... Beats me, but I'll try to find an answer. Apologize for restating part of your question- I'm evidently not a careful enough reader... – BobT Dec 21 '17 at 3:22

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