When you "cross" (90 degrees) two high-quality polarizing lenses, what happens?

1. 83

   Laura Schneiderposted 11 years ago

   When you "cross" (90 degrees) two high-quality polarizing lenses, what happens?
   
   Now, what happens when you put a third polarizing lens between the first two at a random (but unequal) angle with respect to the first to lenses? And can you explain why?

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2. 67

   Larry Fieldsposted 11 years ago

   With the first two lenses, you'd filter out essentially all of the incoming light. Adding the intermediate third lens would allow some of the light to pass through. Sorry; I can visualize it, but I can't explain it.

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      Laura Schneiderposted 11 years agoin reply to this

      Ding ding ding! You are correct, Sir! I'm pleased that someone else was awake during physics class, too.
      
      The last part of the question, how and why it works that way, we'll leave to someone else: you've gotten us off the ground.
      
      Anyone? Guesses?

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   calculus-geometryposted 11 years ago

   The explanation of the paradox is due to the fact that polarizing lenses don't "filter" in the normal sense of the word, they reduce the strength of light that passes through according to the angle of the light source and the angle of orientation of the lens.
   
   To illustrate, suppose the first two lenses are at angles of 0 degrees and 90 degrees, and that the middle lens is oriented at x degrees with x between 0 and 90.
   
   In the two-lens case, when light at an angle of z passes through the first lens (0 degrees, horizontal) the vertical component of the light is flattened. When that now horizontal light passes through the second lens (90 degrees, vertical) the horizontal component of the light is flattened and only the magnitude of the vertical component matters.  But at this point, after having passed through the first lens, there is no vertical component, therefore the magnitude of the output light is 0, no light at all.
   
   In the three-lens case, when the light passes through the lens oriented 0 degress and then the lens oriented x degrees (in the middle), the intensity of the light reduces to cos(x) of its original intensity. After it passes through the middle lens, the light still has some horizontal and vertical component.  Finally, when it passes through the last lens (90 degrees) the intensity is reduced by sin(x).  So the net reduction in strength is cos(x)*sin(x). 
   
   If you choose x = 45 degrees for the middle lens, then the reduction is cos(45)*sin(45) = 0.5.  So the light is reduced to 50% of its original strength.

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      Laura Schneiderposted 11 years agoin reply to this

      Excellent proof! You must be a professor, genius, over-educated unit of the general public, or all of these. :-)  Larry's answer got us off to a great start, and yours fleshed out the details. Congrats and thanks! I selected this as the Best Answer!

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      SidKempposted 11 years agoin reply to this

      Fascinating - I did not know abut the effect of adding the third lens. This means that the light leaving the 1st lens has its horizontal component eliminated at the lens, but it is somewhat restored before it reaches the intermediate lens,  I think.

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   3. 83

      Laura Schneiderposted 11 years agoin reply to this

      SidKemp--It boggles the mind, doesn't it? The 3rd, odd-angled filter introduces a perpendicular component back in to the mix because photons are not just waves or just particles, but rather hybrids of both: waveicles. Cool!

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