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Consider the Glan-Taylor prism. I have 2 questions about it:

  1. Is the second half of the Glan-Taylor prism redundant? In the sense that, without it, the light be still polarised.

  2. Generally the reflected light will only be partially polarised. Can I adjust it such that the p-mode contains no reflection so the the reflected light will be completely polarised?

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  • $\begingroup$ A somewhat pedantic answer to your second question is no, because you would have to have completely monochromatic incident light, and have the prism cut at exactly Brewster's angle for that wavelength. In reality, both of these things are unfeasible (there will always be some spread in the spectrum of your beam and some machining tolerance for your prism), so getting completely polarized light is impossible. But this probably misses the point of your second question, which is why I put it here in the comments. $\endgroup$ – probably_someone Apr 17 '18 at 0:40
  • $\begingroup$ @probably_someone Thanks for your comment. I'm considering the ideal case now. Assuming monochromatic light and exact cut, theoretically is it possible? $\endgroup$ – user148792 Apr 17 '18 at 0:50
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  1. Yes, the light would still be polarized. The second half makes the output beam direction less sensitive to the input angle, and makes the polarizer symmetric so it can be used in reverse as well.

  2. Yes. The transmitted beam is always perfectly polarized because $s-$polarized light must reflect (total internal reflection); so whatever light is transmitted must be $p-$polarized. If, in addition, the reflection occurs at Brewster's angle for the $p-$polarization refractive index, then $p-$polarized light must transmit, so whatever light is reflected must be $s-$polarized. However, as probably_someone pointed out above, this requires that the input beam be collimated and aligned such that the reflection takes place at Brewster's angle.

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  • $\begingroup$ Good answer I was just going to type this. " is always perfectly polarized" Should read "is always very well polarised" $\endgroup$ – MJC Apr 19 '18 at 16:17

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