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Light Polarization

  1. Mar 6, 2007 #1
    1. The problem statement, all variables and given/known dataLight that is polarized along the vertical direction is incident on a sheet of polarizing material. Only 66% of the intensity of the light passes through the sheet and strikes a second sheet of polarizing material. No light passes through the second sheet. What angle does the transmission axis of the second sheet make with the vertical?

    2. Relevant equations
    If light intensity = S, then the light passing through a polarizer = (1/2)S
    Malus Law: Averge intensity of light leaving the analyzer = (.5*S)cos^2(theta)

    3. The attempt at a solution
    At 90 degress from the vertical, the polarizer would block all light from passing through. however the problem states that 66% of the light passes through the polarizer. I think that an angle that is 33% of 90 degrees would allow 66% of the light to pass through the polarizer. if i am correct in thinking that a polarizer must be at 90 degrees to block light, then the analyzer must be at a 90 with the polarizer which is at a 30 degree angle with respect to the vertical. 30 degrees + 90 degrees = 120 degrees from the vertical.
  2. jcsd
  3. Mar 6, 2007 #2
    You are right that the second sheet is at 90 degrees to the first. However your angle for the first sheet is wrong. You need to use Malus to find this angle.
    However you have not quoted Malus law correctly.

    Transmitted intensity = [tex]s*cos^2(/the)[/tex]

    (There should not be a half in front of this.)

    We do not know initial intensity but we know it is cut down by 66% after the first sheet. Therefore the transmitted intensity = 0.66*s.

    Put this into the Malus law and solve for theta.
  4. Mar 6, 2007 #3
    the more i thought about it the more confused i seemed to get. thanks..i finally got it
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