# Homework Help: Archived General question on Malus' Law

1. Nov 4, 2008

### IHateMayonnaise

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
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2. Nov 7, 2016

### Wolf Legend

You did it correct.it is always with respect to the previous filter.