Amnesia's Dream: Not Exactly A Fluorescence Polarizer

Tuesday, July 28, 2015

t a n (θ)

$tan (\theta)$ has a Cauchy distribution,

f (x) = \frac{1}{π (1 + x^{2})}

$f(x)=\cfrac{1}{\pi(1+x^2)}$

for

- \frac{π}{2} < θ < \frac{π}{2}

$-\cfrac{\pi}{2}\lt\theta\lt\cfrac{\pi}{2}$

then

\sqrt{| t a n (θ) |}

$\sqrt{|tan(\theta)|}$ has a distribution

f (x) = \frac{1}{π (1 + x^{4})} | 2 x |

$f(x)=\cfrac{1}{\pi(1+x^4)}|2x|$

for

- \frac{π}{2} < θ < \frac{π}{2}

$-\cfrac{\pi}{2}\lt\theta\lt\cfrac{\pi}{2}$ in blue below.

These set of peaks for the probability density of

\sqrt{t a n (θ)}

$\sqrt{tan(\theta)}$ are not at

\frac{π}{2}

$\cfrac{\pi}{2}$ apart but

\frac{2}{\sqrt[4]{3}}

$\cfrac{2}{\sqrt[4]{3}}$ apart.

\frac{2}{\sqrt[4]{3}} = 1.5197 r a d = {87.071}^{o} \neq 90^{o}

$\cfrac{2}{\sqrt[4]{3}}=1.5197\,\, \small{rad}=87.071^o\ne90^o$

From,

\frac{E_{p v}}{ℏ} = {1 - \sqrt{t a n (θ_{v})}} \frac{c}{x_{v}}

$\cfrac{E_{p\,v}}{\hbar}=\left\{1-\sqrt{tan(\theta_v)}\right\}\cfrac{c}{x_v}$

The probability density of

1 - \sqrt{t a n (θ_{v})}

$1-\sqrt{tan(\theta_v)}$ has been right shifted by

1

$1$ . The peak intensity of

E_{p}

$E_p$ occurs not at

θ = 0 r a d

$\theta=0\,\,\small{rad}$ but at

θ = 1 r a d = {57.296}^{o}

$\theta=1\,\, \small{rad}=57.296^o$ .

a = \frac{c}{x_{v}}

$a=\cfrac{c}{x_v}$

is assumed constant.

Note: Also,

\frac{E_{p v}}{ℏ} = {\sqrt{t a n (θ_{v})} - 1} \frac{c}{x_{v}}

$\cfrac{E_{p\,v}}{\hbar}=\left\{\sqrt{tan(\theta_v)}-1\right\}\cfrac{c}{x_v}$

in which case two peaks occurs centered about

θ = - 1 r a d

$\theta=-1\,\, \small{rad}$ ,

θ = 1.5197 r a d

$\theta=1.5197\,\, \small{rad}$ apart. A notch occurs at

θ = - 1 r a d

$\theta=-1\,\, \small{rad}$ .

Tuesday, July 28, 2015