Intensity of the incident light

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The discussion revolves around calculating the intensity of incident light on a system of three polarizers. The user initially applies the formula Iout = Iincos²θ but miscalculates the angles between the polarizers. After passing through the first polarizer, the intensity is halved, and the user attempts to apply the angles incorrectly for the second and third polarizers. Clarification is provided that the angle between the second and third polarizers should be used for the final calculation. The correct approach leads to a different intensity value for the incident light than initially calculated.
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Homework Statement



Unpolarized light is incident on a system of three polarizers. The second polarizer is oriented at an angle of 33.0° with respect to the first and the third is oriented at an angle of 45.0° with respect to the first. If the light that emerges from the system has an intensity of 2.3 W/m2, what is the intensity of the incident light?

Homework Equations



Iout=Iincos2\theta

The Attempt at a Solution



I know that after I0 passes through the first polarizer, it is 1/2I0. After it passes through the second, I thought it would be 1/2I0cos2(33), and after it passes through the third, 1/2I0cos2(33)cos2(45).
This would all equal 2.3, which I was given.
So I set up my final equation as:
1/2I0cos2(33)cos2(45) = 2.3, or
I0= 2.3 / (1/2cos2(33)cos2(45).

I solved and got 13.08, but that is incorrect. Can anyone offer any insight? Thanks.
 
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For the third polarizer, use the angle between it and the second polarizer.

45° is the angle between the third and first polarizer.
 

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