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Huygens Principle for a Planar Source (view source)
Revision as of 20:50, 9 July 2009
, 20:50, 9 July 2009→Special Case
In cylindrical coordinates, <math>d^2r=rdrd\phi \quad</math>. Without loss of generality, we consider a harmonic solution with a particular frequency ω = kc.
In cylindrical coordinates, <math>d^2r=rdrd\phi \quad</math>. Without loss of generality, we consider a harmonic solution with a particular frequency ω = kc.
:<math>\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)=\dot{A}(\mathbf{r},0)e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}</math><br><br>
:<math>\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)=\dot{A}(\mathbf{r},0)e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}</math><br><br>
:<math>A(r')=\frac{-ikz'A_0}{2\pi}\,e^{-i\omega t'} \int_{z=0} rdrd\phi\, \frac{e^{ik|\mathbf{r}-\mathbf{r}'|)}}{|\mathbf{r}-\mathbf{r}'|^2}</math>
:<math>A(r')=\frac{ikz'}{2\pi}\,e^{-i\omega t'} \int_{z=0} rdrd\phi\, \frac{e^{ik|\mathbf{r}-\mathbf{r}'|)}}{|\mathbf{r}-\mathbf{r}'|^2}A_0(\mathbf{r},0)</math>
== Special Case ==
== Special Case ==
Picture an opaque screen with a circular aperture of radius a.<br><br>
Picture an opaque screen with a circular aperture of radius a.<br><br>
Let<math>\mathcal{J}(r')=\int_0^a rdr\int_0^{2\pi} d\phi\, \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}</math><br><br>
Let<math>\mathcal{J}(r')=\int_0^a rdr\int_0^{2\pi} d\phi\, \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|^2}</math><br><br>
Then <math>A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')</math><br><br>
Then <math>A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')</math><br><br>
But <math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}</math>
But <math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}</math>
::<math>=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}</math><br>
::<math>=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}</math><br>
::<math>=r'-\frac{2r\rho'\cos\phi}{2r'}\frac{\rho'}{r'}\sin\theta'</math><br><br>
::<math>=r'-\frac{2r\rho'\cos\phi}{2r'}, \frac{\rho'}{r'}=\sin{\theta}'</math><br><br>
so that <math>\frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}\left(1+\frac{2r\sin\theta'\cos\phi}{r'}\right)</math>
so that <math>|\mathbf{r}-\mathbf{r}'|=r'-r\cos{\phi}\sin{\theta}'</math> and <math> \frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}\left(1+\frac{2r\sin\theta'\cos\phi}{r'}\right)</math><br><br>
In this particular case, we are dealing with far-field effects only, so <math>\frac{2r\sin\theta'\cos\phi}{r'}\rightarrow 0 </math> and <math>\frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}</math><br><br>
So, <math> \mathcal{J}(r')=\int_0^a rdr\int_0^{2\pi} d\phi\, \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}=\frac{e^{ikr'}}{r'^2}\int_0^a rdr\int_0^{2\pi} d\phi\, e^{-ikr\sin{\theta}'\cos{\phi}}</math><br><br>
The integral <math>\int_0^{2\pi} d\phi\, e^{-ikr\sin{\theta}'\cos{\phi}}</math> is the integral representation of the zero order Bessel function of the first kind with <math> kr\sin{\theta}' \quad</math> as the argument. This gives us the equation:<br><br>
:<math>\mathcal{J}(r')=\frac{e^{ikr'}}{r'^2}\int_0^a rdr 2\pi J_0(kr \sin{\theta}') </math><br><br>
To simplify the math, we make use of the fact that we can represent this Bessel functions as the derivative of a Bessel function of a different order. In general, the formula to compute this derivative is <br><br>
:<math>z^{v-k}J_{v-k}(z)=\left(\frac{1}{z}\frac{\part}{\part z}\right)^kz^vJ_v(z)</math><br><br>
In this case, we take <math>v=k=1 \quad</math> and <math>z=kr\sin{\theta}' \quad</math>. So<br><br>
:<math>J_0(kr\sin{\theta}')=\left(\frac{1}{kr\sin{\theta}'}\frac{\part}{\part (kr\sin{\theta}')}\right)(kr\sin{\theta}')J_1(kr\sin{\theta}')</math><br><br>
This gives us the equation
:<math>\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{r'^2}\int_0^a rdr \left(\frac{1}{kr\sin{\theta}'}\frac{\part}{\part (kr\sin{\theta}')}\right)(kr\sin{\theta}')J_1(kr\sin{\theta}')</math><br><br>
Let <math>x=kr\sin{\theta'} \quad </math> so that
:<math>\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{k^2\sin^2{\theta}'r'^2}\int_0^{ka\sin{\theta'}} dx \frac{d}{dx}xJ_1(x)</math>
<math>\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}\left[aJ_1(ka\sin{\theta}')-0J_1(0k\sin{\theta}')\right]=2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}aJ_1(ka\sin{\theta}')</math><br><br> and <math>A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-i\omega t'} 2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}aJ_1(ka\sin{\theta}')=\frac{z'\dot{A_0}a}{c}\frac{e^{ikr'-i\omega t'}}{k\sin{\theta}'r'^2}J_1(ka\sin{\theta}')</math><br><br>
To find the angle to the diffraction minimum, we must find the zeroes of this amplitude function. This will occur when <math>J_1(ka\sin{\theta}')=0 \quad</math><br><br>
[[Image:Bessels_J0.svg|thumb|300px|right|Plot of Bessel function of the first kind, J<sub>α</sub>(x), for integer orders α=0,1,2.]]
To the right is a graph of three Bessel functions of the first order, specifically <math> J_0(x), J_1(x), and J_2(x) \quad</math>. As it is shown, the first zero of <math>J_1(x) \quad</math> will <br>
occur at <math>x=0 \quad</math>. This will correspond to the center of the pattern, at <math>\theta=0 \quad</math>. Here, we would expect a bright spot, so <math>A(r') \quad</math> should be positive and finite. At <math>\theta=0 \quad</math> the term <math>\frac{J_1(ka\sin{\theta}')}{\sin{\theta}'}</math> is positive and finite, so this expression gives the correct amplitude at <math>\theta=0 \quad</math>. The next zero of <math> J_1(x) \quad</math> corresponds to the first minumum of the diffraction pattern. In this case, this zero occurs at x=3.832. So, <math>ka\sin{\theta}'=3.832 \quad</math> . Since<math> k=\frac{2\pi}{\lambda}\quad</math> and <math> a=\frac{D}{2}\quad</math><br><br>
<math>\frac{2\pi D\sin{\theta}'}{2\lambda}=3.832\rightarrow \sin{\theta}'= \frac{1.22\lambda}{D}</math>