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Homework Help: Polarization tanB = N1/N2

  1. May 9, 2008 #1
    Two sheets of polaroid are oriented so that there is a maximum transmission of light. One sheet is now rotated by 30 degrees, by what factor does the light intensity drop?

    OK, the only equation I could think to use is tanB = N1/N2 but it doesn't seem to work.

    the answer is 0.75 I just can't seem to get there.
    Any help would be GREATLY appreciated.
    Last edited by a moderator: Mar 7, 2013
  2. jcsd
  3. May 9, 2008 #2
    Define the terms in this equation: what exactly are B, N1, N2, and how do they relate to the problem you've quoted?
  4. May 9, 2008 #3
    Got it!!

    I feel silly! I realise equation I was trying to use was wrong. I solved it using Malus' Law

    intensity after = Intensity before*cos^30

    Thanks for the quick reply tho.
  5. May 9, 2008 #4
    Malus' Law is spot on. The intensity is greatest when angle=0deg, zero when angle=90 deg. The cos(angle) function fits the bill, especially as the situation is a rotation.
  6. May 10, 2008 #5
    Sorry, in the above post I should have said amplitude not intensity.
    In the above did you mean [tex]I = I_0 \sin 30^o[/tex] or [tex]I = I_0 \sin^2 30^o[/tex] ?
    Last edited: May 10, 2008
  7. May 11, 2008 #6
    I = I_0 \cos^2 30^o

    From Cutnell, Physics
    Last edited: May 11, 2008
  8. May 11, 2008 #7

    [itex]E=E_0 cos(30)[/itex] where [itex]E[/itex] is the field in Volts/m.

    Power (intensity) is proportional to [itex]E^2[/itex], which in turn is proportional to [itex]I^2[/itex] (current squared).

    [tex]\frac{I}{I_0}=\frac{E^2}{E_0^2}=\frac{(E_0 cos(30))^2}{E_0^2}=cos^2(30)=0.75[/tex]

    [n.b. "I" for intensity is not current - I hope that isn't too confusing]


    Last edited: May 11, 2008
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