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IHateMayonnaise
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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 \theta_1=0, and \theta_2=45^o and \theta_3=90^o. An incident beam of light (with intensity I_o) goes through the first polarizer and loses half it's intensity, since the time average of Malus' law is equal to \frac{I_o}{2}. Now, after the now diminished light passes through the second polarizer (oriented 45^o with respect to the first), the intensity is given by
I=\frac{I_o}{2}Cos(\theta_2)^2=\frac{I_o}{2}\left(\frac{\sqrt{2}}{2}\right)^2=\frac{I_o}{4}
So, when it passes through the third polarizer, the incident intensity is equal to \frac{I_o}{4} and \theta_3=45^o. This is where I'm confused. \theta_3=45^o is the angle we choose because it is always with respect to the previous filter, not the original, since in that case we have \theta_3=90^o and Malus' law warrants a big fat zero. So, the resultant intensity is equal to \frac{I_o}{8}.
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
Say we have three linear polarizers in series, where \theta_1=0, and \theta_2=45^o and \theta_3=90^o. An incident beam of light (with intensity I_o) goes through the first polarizer and loses half it's intensity, since the time average of Malus' law is equal to \frac{I_o}{2}. Now, after the now diminished light passes through the second polarizer (oriented 45^o with respect to the first), the intensity is given by
I=\frac{I_o}{2}Cos(\theta_2)^2=\frac{I_o}{2}\left(\frac{\sqrt{2}}{2}\right)^2=\frac{I_o}{4}
So, when it passes through the third polarizer, the incident intensity is equal to \frac{I_o}{4} and \theta_3=45^o. This is where I'm confused. \theta_3=45^o is the angle we choose because it is always with respect to the previous filter, not the original, since in that case we have \theta_3=90^o and Malus' law warrants a big fat zero. So, the resultant intensity is equal to \frac{I_o}{8}.
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