# Fresnel–Arago laws

One may understand this more clearly when considering two waves, given by the form $\mathbf {E_{1}} (\mathbf {r} ,t)=\mathbf {E} _{01}\cos(\mathbf {k_{1}\cdot r} -\omega t+\epsilon _{1})$ and $\mathbf {E_{2}} (\mathbf {r} ,t)=\mathbf {E} _{02}\cos(\mathbf {k_{2}\cdot r} -\omega t+\epsilon _{2})$ , where the boldface indicates that the relevant quantity is a vector, interfering. We know that the intensity of light goes as the electric field squared (in fact, $I=\epsilon v\langle \mathbf {E} ^{2}\rangle _{T}$ , where the angled brackets denote a time average), and so we just add the fields before squaring them. Extensive algebra  yields an interference term in the intensity of the resultant wave, namely: $I_{12}=\epsilon v\mathbf {E_{01}\cdot E_{02}} \cos \delta$ , where $\delta =(\mathbf {k_{1}\cdot r-k_{2}\cdot r} +\epsilon _{1}-\epsilon _{2})$ represents the phase difference arising from a combined path length and initial phase-angle difference.
Now it can be seen that if $\mathbf {E_{01}}$ is perpendicular to $\mathbf {E_{02}}$ (as in the case of the first Fresnel–Arago law), $I_{12}=0$ and there is no interference. On the other hand, if $\mathbf {E_{01}}$ is parallel to $\mathbf {E_{02}}$ (as in the case of the second Fresnel–Arago law), the interference term produces a variation in the light intensity corresponding to $\cos \delta$ . Finally, if natural light is decomposed into orthogonal linear polarizations (as in the third Fresnel–Arago law), these states are incoherent, meaning that the phase difference $\delta$ will be fluctuating so quickly and randomly that after time-averaging we have $\langle \cos \delta \rangle _{T}=0$ , so again $I_{12}=0$ and there is no interference (even if $\mathbf {E_{01}}$ is rotated so that it is parallel to $\mathbf {E_{02}}$ ).