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Archived General question on Malus' Law

No specific question, just studying for the physics GRE and making sure that I remember all this correctly. Can someone verify or deny my rational? Here I go:

Say we have three linear polarizers in series, where [tex]\theta_1=0[/tex], and [tex]\theta_2=45^o[/tex] and [tex]\theta_3=90^o[/tex]. An incident beam of light (with intensity [tex]I_o[/tex]) goes through the first polarizer and loses half it's intensity, since the time average of Malus' law is equal to [tex]\frac{I_o}{2}[/tex]. Now, after the now diminished light passes through the second polarizer (oriented [tex]45^o[/tex] with respect to the first), the intensity is given by

[tex]I=\frac{I_o}{2}Cos(\theta_2)^2=\frac{I_o}{2}\left(\frac{\sqrt{2}}{2}\right)^2=\frac{I_o}{4}[/tex]

So, when it passes through the third polarizer, the incident intensity is equal to [tex]\frac{I_o}{4}[/tex] and [tex]\theta_3=45^o[/tex]. This is where I'm confused. [tex]\theta_3=45^o[/tex] is the angle we choose because it is always with respect to the previous filter, not the original, since in that case we have [tex]\theta_3=90^o[/tex] and Malus' law warrants a big fat zero. So, the resultant intensity is equal to [tex]\frac{I_o}{8}[/tex].

Also: can an analysis of this sort be done with polarizers of different types (circular, elliptical)? I would assume that the equation would be somewhat different, but I would think that the general idea could be extended. Thanks yall

IHateMayonnaise
1. Homework Statement



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3. The Attempt at a Solution
 
You did it correct.it is always with respect to the previous filter.
 

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