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Intensity after different polarizers

  1. Jun 3, 2013 #1
    1. The problem statement, all variables and given/known data
    1.An unpolarized beam of light is incident on a stack of four ideal polarizing filters. Each of the second, third and fourth polarizers is rotated by 30.0o relative to the preceding polarizer. In this way, the axis of the first polarizer is perpendicular to the axis of the fourth polarizer in the stack. Calculate the fraction by which the incident intensity is reduced after going through the polarizers.

    The first polarizer will be vertical which will 1/2 the intensity; the second will be at 60° so
    1/2(cos60°squared)=1/8; the third will be at 30° so 1/8(cos30°squared)=3/32; the last polarizer is horizontal and am not sure as to how much the intensity will be reduced. cos(0°)=1, so do I multiply 3/32 by 1 to get the final intensity of the light?
     
  2. jcsd
  3. Jun 3, 2013 #2

    mfb

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    60° relative to what? Is that important?
    You see this problem at the last polarizer: it will reduce the intensity...
     
  4. Jun 3, 2013 #3
    The angle for Malus' Law is the angle between the incident light and the polarizer axis.

    It is not the angle with respect to some arbitrary coordinate axis (such as the horizontal axis, like you seem to have used).

    Every time linearly polarized light passes through a linear polarizer, the light that comes out is polarized along the angle of polarizer, regardless of what the initial angle was.

    To exemplify this, consider a vertical polarizer, followed by a polarizer at 45 degrees, and finally a horizontal polarizer.

    Suppose after the first polarizer we have intensity [itex]I_0[/itex] polarized vertically.

    After the second polarizer we'd have intensity [itex]I_0(\cos 45^\circ)^2[/itex] polarized at 45 degrees.

    Now the angle from 45 degrees to the final horizontal polarizer is just 45 degrees, so the the final intensity is [itex]I_0(\cos 45^\circ)^2(\cos 45^\circ)^2[/itex] polarized horizontally.

    Notice that the original polarization does not directly matter for the final calculation; only the intensity and polarization of light coming out of the second filter matters.
     
  5. Jun 3, 2013 #4
    The polarizers are changing in 30 degree increments relative to the preceding polarizer; verticle (90),60,30,horizontal. I am sure that the last polarizer (horizontal) will eliminate any light from getting through, am I correct?
     
  6. Jun 3, 2013 #5
    No, you're not correct. See what I wrote above. Notice that even though there is a 90 degree difference between the first and last polarizer, some light still gets through. You might think of it like polarizer in the middle rotating the light so that some gets through the final filter. But if the one in the middle wasn't there, then none would get through the second filter.
     
  7. Jun 3, 2013 #6
    ok so essentially the intensity is reduced by cos30^2 every polarizer... so (1/2)(3/4)(3/4)(3/4) = 27/128?
     
  8. Jun 3, 2013 #7

    mfb

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    Right.
     
  9. Jun 3, 2013 #8
    Thank you so much mfb and misterx, very much appreciated.
     
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