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3,521 bytes added , 20:50, 9 July 2009

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:<math>A(r')=-\int d^2r\int cdt\left[A(r)\frac{\part}{\part z}G_1(r,r')-G_1(r,r')\frac{\part}{\part z}A(r)\right]</math>, where the last term is zero by the condition of<math>G_1(z=0,r')=0 \quad</math>

:<math>A(r')=-c\int dt\int d^2rA(r)\frac{\part}{\part z}G_1(r,r')</math>

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To do the t integral, I need to bring out the z derivative. To do this, I first turn it into a z' derivative, using the relation~~: ~~

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:<math>G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}''|-c(t-t'))}{|\mathbf{r}-\mathbf{r}''|}\right)</math>, where :<math>\mathbf{r}''=\mathbf{r}'-2z'\hat{e_3}</math>

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To do the t integral, I need to bring out the z derivative. To do this, I first turn it into a z' derivative, using the relation

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:<math>G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}''|-c(t-t'))}{|\mathbf{r}-\mathbf{r}''|}\right)</math>

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where <math>\mathbf{r}''=\mathbf{r}'-2z'\hat{e_3}</math>

:<math>\frac{\part}{\part z}G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\part}{\part z}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}''|-c(t-t'))}{|\mathbf{r}-\mathbf{r}''|}\right)\right)</math>

&there4; <math>A(r')=\frac{-1}{4\pi}\frac{\part}{\part z'}\int_{z=0} d^2r\left(2\frac{A(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c})}{|\mathbf{r}-\mathbf{r}'|}\right)</math>

At <math>z=0 \quad </math>, <math>|\mathbf{r}-\mathbf{r}'|=\sqrt{r^2+z'^2}=S, dS=\frac{rdr}{\sqrt{r^2+z'^2}}</math>

If<math>A(\mathbf{r},t) \quad</math> is independent of position, as in a plane wave propagating along the z axis, then:

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<math>A(r')=\frac{-\part}{\part z'}\int_{z'}^\infin dS A\left(\mathbf{0},t-\frac{S}{c}\right)=A\left(\mathbf{\mathbf{0}},t'-\frac{z'}{c}\right)</math>

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:<math>A(r')=\frac{-\part}{\part z'}\int_{z'}^\infin dS A\left(\mathbf{0},t-\frac{S}{c}\right)=A\left(\mathbf{\mathbf{0}},t'-\frac{z'}{c}\right)</math>

This gives us uniform translation of waves at velocity c. More generally:

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<math>A(r')=\frac{-1}{2\pi}\int_{z=0} d^2r\frac{\part}{\part z'}\left(\frac{A\left(\mathbf{r}, t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|}\right)</math>

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:<math>A(r')=\frac{-1}{2\pi}\int_{z=0} d^2r\frac{\part}{\part z'}\left(\frac{A\left(\mathbf{r}, t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|}\right)</math>

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<math>=\frac{-1}{2\pi}\int_{z=0} d^2r\left(\frac{A\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^3}(-z')+\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|c}\frac{-z'}{|\mathbf{r}-\mathbf{r}'|}\right)</math>

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:<math>=\frac{-1}{2\pi}\int_{z=0} d^2r\left(\frac{A\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^3}(-z')+\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|c}\frac{-z'}{|\mathbf{r}-\mathbf{r}'|}\right)</math>

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<math>A(r')=\frac{-1}{2\pi}\int_{z=0} d^2r\left(\frac{A\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^3}(-z')+\frac{1}{c}\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^2}(-z')\right)</math>

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:<math>A(r')=\frac{-1}{2\pi}\int_{z=0} d^2r\left(\frac{A\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^3}(-z')+\frac{1}{c}\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^2}(-z')\right)</math>

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In our case, we consider only those waves which drop off as <math>\frac{1}{r'} \quad</math>, so~~: ~~

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~~<math>A(r')=\frac{1}{2\pi}\int_{z=0} d^2r\left(\frac{1}{c}\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^2}(z')\right)</math> ~~

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In our case, we consider only those waves which drop off as <math>\frac{1}{r'} \quad</math>, so

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<math>A(r')=\frac{z'}{2\pi c}\int_{z=0} d^2r\left(\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^2}\right)</math> ~~ ~~

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:<math>A(r')=\frac{z'}{2\pi c}\int_{z=0} d^2r\left(\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^2}\right)</math>

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In cylindrical coordinates, <math>d^2r=rdrd\phi \quad</math>. ~~Also~~, <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>~~. So:~~

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<math>A(r')=\frac{z'~~\dot{A_0}~~}{2\pi c}\int_{z=0} rdrd\phi \frac{e^{-ik~~(t'c-~~|\mathbf{r}-\mathbf{r}'|)}}{|\mathbf{r}-\mathbf{r}'|^2}</math>~~ ~~

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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.

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:<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>

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:<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>

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== Special Case ==

Picture an opaque screen with a circular aperture of radius a.

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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>

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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>

Then <math>A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')</math>

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<math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}</math~~><br~~>

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But <math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}</math>

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<math>=r'-\frac{2r\rho'\cos\phi}{2r'}, \frac{\rho'}{r'}=\sin\theta'</math>

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::<math>=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}</math>

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<math>\frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}\left(1+\frac{2r\sin\theta'\cos\phi}{r'}\right)</math>

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::<math>=r'-\frac{2r\rho'\cos\phi}{2r'}, \frac{\rho'}{r'}=\sin{\theta}'</math>

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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>

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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>

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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>

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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:

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:<math>\mathcal{J}(r')=\frac{e^{ikr'}}{r'^2}\int_0^a rdr 2\pi J_0(kr \sin{\theta}') </math>

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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

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:<math>z^{v-k}J_{v-k}(z)=\left(\frac{1}{z}\frac{\part}{\part z}\right)^kz^vJ_v(z)</math>

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In this case, we take <math>v=k=1 \quad</math> and <math>z=kr\sin{\theta}' \quad</math>. So

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:<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>

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This gives us the equation

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:<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>

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Let <math>x=kr\sin{\theta'} \quad </math> so that

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:<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>

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<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> 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>

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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>

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[[Image:Bessels_J0.svg|thumb|300px|right|Plot of Bessel function of the first kind, Jα(x), for integer orders α=0,1,2.]]

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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

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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>

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<math>\frac{2\pi D\sin{\theta}'}{2\lambda}=3.832\rightarrow \sin{\theta}'= \frac{1.22\lambda}{D}</math>

Pe1505686

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Huygens Principle for a Planar Source (view source)

Revision as of 20:50, 9 July 2009

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