If water is magnetic, how does it work? Water being magnetic will help explain the formation of water droplets, but that would be jumping the gun.
What else is in optics? Plenty!
In the case of Lambert's cosine law, where intensity is directly proportional to the incident angle, but as noted in the post "No \(B\), Speed Alone" dated 25 Aug 2015, since the two loops makes an angle \(\theta\) with the ray,
\(cos(\alpha)\to cos(\alpha_{adj})=cos(\alpha+90^o-\theta)=-sin(\alpha-\theta)=sin(\theta-\alpha)\)
and
\(cos(\alpha)\to cos(\alpha_{adj})=cos(\alpha-90^o+\theta)=sin(\alpha+\theta)\)
as we adjust for the angle of incident between the loops and the surface normal at the point of reflection, we have an adjusted Lambert, where the intensity of an ideal diffusely reflecting surface is proportional to,
\(sin(\theta-\alpha)+sin(\theta+\alpha)=sin(\theta)cos(\alpha)\)
\(\alpha\) is the incident angle of the ray, and \(\theta\) is the angle the loops make with the ray, polarization.
Since \(\theta\) is distributed over a range of values, \(0\lt\theta\lt\pi/2\), intensity \(I\),
\(I\propto\int^{\pi/2}_0{f(\theta)sin(\theta)}d\theta.cos(\alpha)\)
where
\(f(\theta)=\cfrac{2\theta}{\pi(1+\theta^4)}\)
from the post "Not Exactly A Fluorescence Polarizer" dated 28 Jul 2015. The problem is with the abrupt cutoff at \(\pi/2\), may be
\(I\propto\int^{\theta\to\infty}_0{f(\theta)sin(\theta)}d\theta.cos(\alpha)\)
or even,
\(I\propto\int^{\theta\to\infty}_{\theta\to -\infty}{f(\theta)sin(\theta)}d\theta.cos(\alpha)\)
Only after \(\theta\) has been accounted for, (for example, \(\theta=90^o\)) is it possible to have a constant \(A\) such that,
\(I=AF(\theta).cos(\alpha)\)
where \(F(\theta)\) is a function of \(\theta\) only. In the case \(\theta=90^o\), a laser source,
\(I=A.cos(\alpha)\)
where \(A\) is a constant.
