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Polarization tanB = N1/N2

  • Thread starter fionamb83
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  • #1
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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.
 
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Answers and Replies

  • #2
the only equation I could think to use is tanB = N1/N2 but it doesn't seem to work.
Define the terms in this equation: what exactly are B, N1, N2, and how do they relate to the problem you've quoted?
 
  • #3
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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.
 
  • #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.
 
  • #5
Sorry, in the above post I should have said amplitude not intensity.
... Malus' Law

intensity after = Intensity before*cos^30
In the above did you mean [tex]I = I_0 \sin 30^o[/tex] or [tex]I = I_0 \sin^2 30^o[/tex] ?
 
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  • #6
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[tex]
I = I_0 \cos^2 30^o
[/tex]

From Cutnell, Physics
 
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  • #7
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[tex]
I = I_0 \cos^2 30^o
[/tex]

From Cutnell, Physics

[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]

Regards,

Bill
 
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