User:AubreyJack11/sandbox

Definitions

 * $$a$$ is the radius of the circular aperture,
 * $$\lambda$$ is the wavelength,
 * $$k = \frac{2\pi}{\lambda} $$ is the wavenumber,
 * $$w$$ is the angular distance from the center of the aperture in radians.

Diffraction through a circular aperture
The intensity from diffraction through a circular aperture at a distance $$w$$ from the center of the aperture is given by the following equation for the Airy disk
 * $$I(w) = I_0 \left ( \frac{2 J_1(kaw)}{kaw} \right )^2$$.

Where $$J_1(\cdot)$$ is the Bessel function of the first kind of order one.

Energy
The fraction of the total energy contained within a circle of radius $$w_0$$, denoted as $$L(w_0)$$ is given by

\begin{align} L(w_0) & = \frac{1}{E}\int\limits_{0}^{2\pi}\int\limits_{0}^{w_0}I(w)w dw d\psi\\ & = \frac{D}{\lambda^2}\int\limits_{0}^{2\pi}\int\limits_{0}^{w_0} \left [ \frac{2J_1\left ( kaw \right )}{kaw} \right ]^2 w dw d\psi \\ & = 2\int\limits_{0}^{kaw_0}\frac{J_1^2(x)}{x} dx\\ \end{align} $$

Given the following relations
 * $$ \frac{d}{dx} \left \{ x^{n+1} J_{n+1} (x) \right \} = x^{n+1} J_n(x) $$

and
 * $$ \frac{d}{dx} \left [ x^{-n} J_{n} (x) \right ] = -x^{-n} J_{n+1}(x) $$

This yields
 * $$ L(w_0) = 1 - J_0^2(kaw_0) - J_1^2(kaw_0) $$

At any $$w_0$$ corresponding to a dark ring $$J_1^2(kaw_0) = 0$$, so $$ L(w_0) = 1 - J_0^2(kaw_0) $$. For the first ring $$ kaw_0 = 1.22\pi $$, so $$ w_0 = \frac{1.22\pi}{ka} $$.

In this case $$ L(w_0) = 1 - J_0^2(kaw_0) = 1 - J_0^2(1.22\pi) \approx 0.838 $$. Thus approximately 84% of the total energy is contained within the first zero of the Airy disk.

Gaussian Approximation
The intensity at a distance $$w$$ from the center is given by

I(w) = \frac{2 I_0}{\sigma^2} e^{ -(\frac{(kaw)^2}{2\sigma^2}) }. $$

Energy


\begin{align} L(w_0) & = \frac{1}{E}\int\limits_{0}^{w_0} 2 \pi w I(w) dw \\ & = \frac{1}{E}\int\limits_{0}^{w_0} 2 \pi w \frac{2 I_0}{\sigma^2} e^{ -\frac{(kaw)^2}{2\sigma^2} } dw \\ & = \frac{2\pi 2 I_0}{E \sigma^2} \int\limits_{0}^{w_0} w e^{ -\frac{(kaw)^2}{2\sigma^2} } dw\\ \end{align} $$ From a table of integrals we have
 * $$\int x e^{-ux^2}dx = \frac{-1}{2u}e^{-ux^2}$$

In this case $$u = \frac{k^2a^2}{2\sigma^2}$$, so we have

\begin{align} L(w_0) & = \frac{2\pi 2 I_0}{E \sigma^2} \int\limits_{0}^{w_0} w e^{ \frac{-(kaw)^2}{2\sigma^2} } dw\\ & = \frac{2\pi 2 I_0}{E \sigma^2} \left [ \frac{-2\sigma^2}{2k^2a^2} e^{ \frac{-(kaw)^2}{2\sigma^2 } }\right ]_0^{w_0} \\ & = \frac{4\pi I_0}{E \sigma^2} \frac{2\sigma^2}{2k^2a^2} \left [ 1 - e^{ \frac{-(kaw_0)^2}{2\sigma^2 } } \right ] \\ & = \frac{4\pi D}{\lambda^2 \sigma^2} \frac{2\sigma^2}{2k^2a^2} \left [ 1 - e^{ \frac{-(kaw_0)^2}{2\sigma^2 } } \right ] \\ & = \frac{4\pi \pi a^2}{\lambda^2 \sigma^2} \frac{2\sigma^2}{2k^2a^2} \left [ 1 - e^{ \frac{-(kaw_0)^2}{2\sigma^2 } } \right ] \\ & = \frac{4\pi \pi a^2 k^2}{4 \pi^2 \sigma^2} \frac{2\sigma^2}{2k^2a^2} \left [ 1 - e^{ \frac{-(kaw_0)^2}{2\sigma^2 } } \right ] \\ & = 1 - e^{ -\frac{(kaw_0)^2}{2\sigma^2 } } \\ \end{align} $$ At the first dark ring we have $$kaw_0=1.22\pi$$, where for the Airy case the fractional energy is approximately 84%, so we have,
 * $$L(w_0) = 0.84 = 1 - e^{ \frac{-(1.22\pi)^2}{2\sigma^2 } } $$
 * $$ln( 1 - 0.84) = \frac{-(1.22\pi)^2}{2\sigma^2 } $$
 * $$\sigma^2 = \frac{-(1.22\pi)^2}{2 ln( 1 - 0.84) } $$
 * $$\sigma = \sqrt{\frac{-(1.22\pi)^2}{2 ln( 1 - 0.84) } }$$