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Homework Help: Light through a polarizer

  1. Nov 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Unpolarized light passes through a polarizer oriented at 33 degrees from the horizontal axis.

    1) At what angle (relative to the horizontal axis) would you place a polarizer so that no light passes through it?
    2) At what angle (relative to the horizontal axis) would you place a polarizer so that only a third of the light from the

    2. Relevant equations

    Malus' law
    I/I0 = cos(θ)[itex]^{2}[/itex]

    3. The attempt at a solution

    1) Light into the second polarizer is polarized at an angle of 33 degrees. No light will pass through the second polarizer if the difference in angles is 90°, so the second polarizer should be at an angle of 90° + 33° = 123°.

    2)
    [itex]\frac{1}{3}= cos(θ)^{2}[/itex] where θ is the difference in angles between the two polarizers.
    [itex]θ difference = arccos(\sqrt{\frac{1}{3}})[/itex]
    [itex]θ_{final} - θ{_initial} = arccos(\sqrt{\frac{1}{3}})[/itex]
    [itex]θ_{final} - 33° = arccos(\sqrt{\frac{1}{3}})[/itex]
    θ = 54.7° + 33° = 87.7° from the horizontal axis

    But my instructor says that the left-hand side of the equation is [itex]θ_{final} - θ{_initial}[/itex], which results in θ = 54.7° - 33°, so I'm confused now... have I done these two correctly?
     
  2. jcsd
  3. Nov 26, 2011 #2

    HallsofIvy

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    Science Advisor

    This is a physics problem, not math so I am moving it.
     
  4. Nov 26, 2011 #3
    I'm sorry, I must have been looking at the wrong section when I posted it.
     
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