Angle between polarizer and analyzer

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The discussion centers on calculating the angle between a polarizer and an analyzer based on the intensity reduction of light passing through them. The intensity is reduced by 67.6%, leading to a remaining intensity of 32.4% of the original. The correct equation to use is I = I_o * cos^2(∅), where I is the transmitted intensity. A misunderstanding arose regarding the use of the intensity reduction percentage, which was clarified in the conversation. The final realization corrected the approach to finding the angle using the appropriate formula.
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Homework Statement


Consider a polarizer-analyzer arrangement, as shown (here's a link http://i.imgur.com/APZRJ6J.png )

At what angle is the axis of the analyzer to the axis of the polarizer if, after passing through both sheets, the beam intensity is reduced by 67.6 percent?
Answer in units of ◦

Homework Equations


I=(1/2)I_o
I=I_o*cos^2(∅)

The Attempt at a Solution


Since the beam intensity is reduced by 67.7%, I knew I/I_o was .676, so I tried doing ∅=cos(\sqrt{.676}) but it wasn't right. I'm not sure what I did wrong and whether I'm supposed to use the first equation in this type of problem or not.
 
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It wasn't reduced to 67.6 percent. It was reduced by 67.6 percent
 
so that would mean that .324I=I_o*cos^2(∅)?
 
Last edited:
ok I got it. Thanks for pointing out my mistake
 

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