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I thought that modern 3d glasses work by having one lens filter horizontally polarized light, and the other filter vertically polarized light.

However, I found this pair of 3d glasses at my parents' house, and looked at the reflection from the floor at different angles:

What's confusing me is that turning the glasses 90 degreed changed the color of the light from yellow to blue, but it did that on both lenses simultaneously. I expected one to be yellow while the other is blue, and vice versa, since they should be polarized at a 90 degrees difference.

Can someone explain this?

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  • $\begingroup$ It's possible that one is slanted to the right (45°) and the other to the left. $\endgroup$
    – LDC3
    Aug 17, 2014 at 21:08
  • $\begingroup$ @LDC3 I saw the same color during the rotation, this would rule this out, no? $\endgroup$
    – Ram Rachum
    Aug 17, 2014 at 21:42
  • $\begingroup$ Most likely. There are 3D glasses that work with circular polarization (I did look it up). $\endgroup$
    – LDC3
    Aug 17, 2014 at 22:14

1 Answer 1

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Many 3d glasses use circular polarization, where one lens uses left-hand polarization while the other uses right-hand polarization. This lets the viewer tilt their head a bit without losing the 3D effect, where linear polarization would let the image bleed through into the other eye.

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    $\begingroup$ A fun fact I would like to note is that when looking at a mirror, while wearing these glasses, your right eye will not be able to see through the glass infront of your right eye in the reflection, but can through the glass of your left eye. This is because the mirror reverses the polarity of the light. $\endgroup$
    – fibonatic
    Aug 18, 2014 at 0:24
  • $\begingroup$ We verified this circular polarization in the lab a few years ago. (someone snitched a pair of 3D glasses from the theatre). BTW, the correct term is "stereoscopic," since all movies are (or aren't, depending on your point of view) 3-dimensional images. The so-called "3D" just provides artificially magnified depth. $\endgroup$ Aug 18, 2014 at 11:34

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