Polarization and decreasing intensity of light to 10%

AI Thread Summary
To decrease the intensity of unpolarized light to 10% using two polarizers, the angle θ between their transmission axes must be calculated. The Law of Malus is applied, where the intensity after the first polarizer is halved, leading to the equation 1/10*I = 1/2*I*cos²θ. This simplifies to 1/5 = cos²θ, resulting in θ being approximately 63.4°. The discussion confirms the correct application of the Law of Malus for this problem.
aChordate
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Homework Statement



In order to decrease the intensity of a beam of unpolarized light to 10% of its original
intensity using two polarizers, what is the required value of θ, the angle between the
transmission axes of the two polarizers?


Homework Equations



tanθB=n2/n1


The Attempt at a Solution



tanθB=n2/n1

I'm not sure how to do this. I read the section on polarization and there are no examples in the book.
 
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aChordate said:

Homework Statement



In order to decrease the intensity of a beam of unpolarized light to 10% of its original
intensity using two polarizers, what is the required value of θ, the angle between the
transmission axes of the two polarizers?


Homework Equations



tanθB=n2/n1


The Attempt at a Solution



tanθB=n2/n1

I'm not sure how to do this. I read the section on polarization and there are no examples in the book.

You seem to have mixed up Brewster's law with the Law of Malus. Check you textbook about the Law of Malus.
 
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So my equation would change to S=So=cos2θ

Would it be:

1/10*I=1/2*I*cos2θ ?


1/5=cos2θ

θ=63.4°

?
 
aChordate said:
So my equation would change to S=So=cos2θ

Would it be:

1/10*I=1/2*I*cos2θ ?


1/5=cos2θ

θ=63.4°

?

Looks good to me.
 
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