Changes

3,549 bytes added , 20:50, 9 July 2009

→‎Special Case

Line 10: Line 10:

:<math>\square^2_rG(r,r')=\delta^4(r-r')</math>

and its Fourier transform

−

:<math> \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4r e^{-iq\cdot r} G(r,0)</math>

+

:<math> \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4r\, e^{-iq\cdot r} G(r,0)</math>

−

:<math> G(r,0)=\frac{1}{(2\pi)^2} \int d^~~4qe~~^{iq\cdot r} \tilde{G}(q)</math>

+

:<math> G(r,0)=\frac{1}{(2\pi)^2} \int d^4q\, e^{iq\cdot r} \tilde{G}(q)</math>

Translational symmetry implies

−

:<math>G(r-r',0)=G(r,r')</math>

+

:<math>G(r-r',0)=G(r,r')\quad</math>

−

~~&there4;~~<math> G(r,r')=\frac{1}{(2\pi)^2}\int d^4q e^{iq\cdot (r-r')} \tilde{G} (q)</math> <math>\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^~~4qe~~^{iq\cdot (r-r')}\tilde{G}(q)</math~~><br~~>

+

so that

−

<math>\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^~~4qe^~~{iq\cdot (r-r')}(-k^2+\frac{\omega^2}{c^2})</math>, where <math>q=(\mathbf{k},\frac{\omega}{c}) ~~\quad~~</math>~~ ~~

+

:<math> G(r,r')=\frac{1}{(2\pi)^2}\int d^4q\, e^{iq\cdot (r-r')} \tilde{G} (q)</math>

−

~~But,~~ <math>\square^2_rG(r,r')=\delta^4(r-r')=\frac{1}{(2\pi)^4}\int d^4q e^{iq\cdot (r-r')}</math~~><br~~>

+

:<math>\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^4q\,(-q^2)e^{iq\cdot (r-r')}\tilde{G}(q)</math>

−

~~&there4;~~<math>\tilde{G}(q)=\frac{(2\pi)^2}{(2\pi)^4}\frac{1}{-q^2}= \frac{-1}{(2\pi)^2q^2}</math~~><br~~>

+

:<math>\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^4q\, e^{iq\cdot (r-r')}(-k^2+\frac{\omega^2}{c^2})</math>

−

<math>G(r,r')=\frac{-1}{(2\pi)^4} \int d^~~4qe~~^{iq\cdot (r-r')} \frac{1}{(k+\frac{\omega}{c})(k-\frac{\omega}{c})}</math~~><br~~>

+

where <math>q=(\mathbf{k},\frac{\omega}{c})</math>. But

−

Chose the "retarded" solution, such that the function is zero unless t>t'~~ ~~

+

:<math>\square^2_rG(r,r')=\delta^4(r-r')=\frac{1}{(2\pi)^4}\int d^4q\, e^{iq\cdot (r-r')}</math>

−

<math>G(r,r')=\frac{1}{(2\pi)^4}\int d^~~3ke~~^{i\mathbf{k}\cdot (r-r')}\int d(\frac{\omega}{c}) \frac{e^{-i\omega(t-t')}}{(\frac{\omega}{c}-k)(\frac{\omega}{c}+k)}\Theta(t-t')</math>

+

:<math>\tilde{G}(q)=\frac{(2\pi)^2}{(2\pi)^4}\frac{1}{-q^2}= \frac{-1}{(2\pi)^2q^2}</math>

−

<math>=\frac{1}{(2\pi)^4}\int d^~~3ke~~^{i\mathbf{k}\cdot (r-r')}(2\pi i \frac{e^{ick(t-t')}-e^{-ick(t-t')}}{2k})\Theta</math>

+

:<math>G(r,r')=\frac{-1}{(2\pi)^4} \int d^4q\, e^{iq\cdot (r-r')} \frac{1}{(k+\frac{\omega}{c})(k-\frac{\omega}{c})}</math>

−

<math>=\frac{-2\pi}{(2\pi)^4}\int_0 \frac{k^2dk}{k} \sin\left({ck(t-t')}\right) 2\pi\int_{-1}^1 dze^{ik|\mathbf{r}-\mathbf{r'}|z}\Theta</math>

+

Chose the "retarded" solution, such that the function is zero unless t>t'.

−

<math>=\frac{-1}{(2\pi)^2}\left(\frac{1}{|\mathbf{r}-\mathbf{r'|}}\right)2\int_0 dk \sin(ck(t-t')) \sin(k|\mathbf{r}-\mathbf{r'}|)\Theta</math>

+

:<math>G(r,r')=\frac{1}{(2\pi)^4}\int d^3k\, e^{i\mathbf{k}\cdot (r-r')}\int d(\frac{\omega}{c}) \frac{e^{-i\omega(t-t')}}{(\frac{\omega}{c}-k)(\frac{\omega}{c}+k)}\Theta(t-t')</math>

−

<math>=\frac{1}{(2\pi)^2}\frac{2}{|\mathbf{r}-\mathbf{r}'|}\frac{2\pi}{4} \left[\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))-\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))\right]\Theta</math>

+

::<math>=\frac{1}{(2\pi)^4}\int d^3k\, e^{i\mathbf{k}\cdot (r-r')}(2\pi i \frac{e^{ick(t-t')}-e^{-ick(t-t')}}{2k})\Theta</math>

−

But the term <math>\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'</math>~~ ~~

+

::<math>=\frac{-2\pi}{(2\pi)^4}\int_0 \frac{k^2dk}{k} \sin\left({ck(t-t')}\right) 2\pi\int_{-1}^1 dze^{ik|\mathbf{r}-\mathbf{r'}|z}\Theta</math>

−

~~&there4;~~<math> G(r,r')=\frac{-1}{4\pi}\quad \frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}</math> ~~ ~~

+

::<math>=\frac{-1}{(2\pi)^2}\left(\frac{1}{|\mathbf{r}-\mathbf{r'|}}\right)2\int_0 dk \sin(ck(t-t')) \sin(k|\mathbf{r}-\mathbf{r'}|)\Theta</math>

−

Now to get the <math>G_1(r,r')\quad </math> in the half-space with z>0 with the boundary condition <math>G_1\quad </math> at<math> r_3=z=0 \quad</math> we take the difference:

+

::<math>=\frac{1}{(2\pi)^2}\frac{2}{|\mathbf{r}-\mathbf{r}'|}\frac{2\pi}{4} \left[\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))-\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))\right]\Theta</math>

−

<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}'+2z'\hat{e_3}|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'+2z'\hat{e_3}|}\right)</math>

+

−

Now use Green's theorem:~~ ~~

+

But the term <math>\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'</math> so that

−

~~Let~~ <math>F^\mu=A(r)\part_\mu G_1(r,r')-G_1(r,r')\part_\mu A(r)</math>

+

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

−

+

−

~~But~~ <math>\part_\mu \part^\mu G_1(r,r')=\delta^4(r-r')</math>

+

Now to get the <math>G_1(r,r')\quad </math> in the half-space with z>0 with the boundary condition <math>G_1\quad </math> at<math> r_3=z=0 \quad</math>   we take the difference:

−

+

:<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}'+2z'\hat{e_3}|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'+2z'\hat{e_3}|}\right)</math>

−

+

Now use Green's theorem, with the generating function

+

:<math>F^\mu=A(r)\part_\mu G_1(r,r')-G_1(r,r')\part_\mu A(r)</math>

+

:<math>\int \part_\mu F_\mu d^4r= \int cdt \int d^3r[\part_\mu A \part^\mu G+A\part_\mu \part^\mu G_1-\part_\mu G \part^\mu A -G_1\part_\mu \part^\mu A]</math>

+

:<math>\part_\mu \part^\mu G_1(r,r')=\delta^4(r-r')</math>

+

:<math>\part_\mu \part^\mu A(r)= \mu j(r)</math>, let <math>j(r)=0 \quad</math>

+

:<math>\int \part_\mu F_\mu d^4r=A(r')</math>

Now invoke the divergence theorem on the half space <math>z>0 \quad</math>:

−

<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')=-\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>

−

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~~: ~~

+

−

<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>

+

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

−

<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>

+

:<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>

+

:<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:

−

<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>

+

:<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:

−

<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>

+

:<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>

−

<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>

+

:<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>

−

<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>

+

:<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>

−

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

+

−

~~<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> ~~

+

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

−

<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> ~~ ~~

+

:<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>

−

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:~~

+

−

<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>~~ ~~

+

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>

+

:<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 ==

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

−

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>

+

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>

−

<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~~>

+

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

−

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

+

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

−

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

+

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

+

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>

+

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>

+

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:

+

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

+

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

+

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

+

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

+

:<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>

+

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>

+

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> 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>

+

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>

+

[[Image:Bessels_J0.svg|thumb|300px|right|Plot of Bessel function of the first kind, Jα(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

+

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>

+

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

Pe1505686

196

edits

Changes

Huygens Principle for a Planar Source (view source)

Revision as of 20:50, 9 July 2009

Navigation menu

Search