= Malus's law
{c}
{title2=$I_{\mathrm{out}}=I_{\mathrm{in}}\cos^2\theta$}
For incident <light> with <linear polarization>, an ideal linear <polarizer> transmits
$$
I_{\mathrm{out}}=I_{\mathrm{in}}\cos^2\theta,
$$
where $I$ is <light intensity> and $\theta$ is the <angle> between the incident <electric field>'s line of <oscillation> and the <polarizer>'s pass axis. The input and output are compared in the same medium. The classical explanation is an <orthogonal projection>: for a <unit vector> $\mathbf u$ along the pass axis,
$$
\mathbf E_{\mathrm{out}}=(\mathbf E_{\mathrm{in}}\mathbin{\cdot}\mathbf u)\mathbf u.
$$
The <dot product> reduces the transmitted <wave amplitude> by $\cos\theta$. For a <plane electromagnetic wave> in the same medium, <light intensity> is proportional to squared <electric field> <wave amplitude>, giving <Malus's law>. For example, $\theta=60^\circ$ transmits one quarter of the incident <light intensity>, although the transmitted <wave amplitude> is one half. https://www.feynmanlectures.caltech.edu/I_33.html#Ch33-S4[Feynman Lectures, I §33–4: polarizers]{ref}.
Start with <light> already in <linear polarization> along the horizontal $0^\circ$ axis and let its <light intensity> be $I_0$. Two ideal absorbing linear <polarizers> with pass axes $0^\circ$ and $90^\circ$ give $I_0\cos^2 0^\circ\cos^2 90^\circ=0$. Insert a $45^\circ$ <polarizer> between them: the first leaves $I_0$ unchanged, the middle transmits $I_0/2$ with <linear polarization> along $45^\circ$, and the last transmits $I_0/4$ with <linear polarization> along $90^\circ$. Each <angle> in <Malus's law> is relative to the preceding output, so the final step uses $90^\circ-45^\circ=45^\circ$:
$$
I_{\mathrm{out}}=I_0\cos^2 0^\circ\cos^2 45^\circ\cos^2(90^\circ-45^\circ)=\frac{I_0}{4}.
$$
\Image[/polarization-through-three-filters.png]
{title=Two crossed <polarizers> transmit zero; inserting a middle <polarizer> transmits $I_0/4$ for initially horizontal <linear polarization>}
{description=Pass axes are measured from horizontal. Each blue segment represents transmitted <light intensity>, and each inset shows the pass axis in the transverse plane. The two successive $45^\circ$ changes each halve <light intensity> by <Malus's law>.}
For these ideal absorbing <polarizers>, the rejected <electric field> component transfers <energy> to the filter. In the three-filter arrangement the middle absorbs the fraction $I_0/2$, and the last absorbs $I_0/4$, leaving $I_0/4$ transmitted. <Conservation of energy> holds at each step. Inserting the middle <polarizer> changes the <linear polarization> arriving at the last one; it does not create <light>. The effect illustrates why successive <orthogonal projections> cannot generally be replaced by a single comparison of the first and last axes.
The initially horizontal assumption matters. For <unpolarized light> of <light intensity> $I_0$, the first ideal linear <polarizer> transmits $I_0/2$: averaging $\cos^2\theta$ over uniformly distributed incident orientations gives $1/2$. The same $0^\circ,45^\circ,90^\circ$ sequence then transmits $I_0/8$. This extra first-stage factor does not apply to the already horizontally polarized input in the figure.
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