Huygens Principle for a Planar Source - Revision history

Pe1505686: /* Special Case */

2009-07-09T20:50:00Z

Special Case

Pe1505686: /* Special Case */

2009-07-09T20:09:37Z

Special Case

Pe1505686: /* Special Case */

2009-07-09T20:09:12Z

Special Case

Pe1505686: /* Special Case */

2009-07-09T20:08:48Z

Special Case

Pe1505686: /* Special Case */

2009-07-09T20:06:05Z

Special Case

Pe1505686: /* Special Case */

2009-07-09T20:03:37Z

Special Case

Pe1505686: /* Special Case */

2009-07-09T16:33:56Z

Special Case

Pe1505686: /* Special Case */

2009-07-09T15:52:06Z

Special Case

Pe1505686: /* Special Case */

2009-07-08T21:52:52Z

Special Case

Pe1505686: /* Special Case */

2009-07-08T20:35:42Z

Special Case

@@ Line 75: / Line 75: @@
 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>
+:<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>
+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>
 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}')=\frac{d}{k\sin{\theta}'dr} J_1(kr\sin{\theta}')</math><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 \frac{d}{k\sin{\theta}'dr} J_1(kr\sin{\theta}')</math><br><br>
+This gives us the equation
-<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^{-ikct'} 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'-ikct'}}{k\sin{\theta}'r'^2}J_1(ka\sin{\theta}')</math><br><br>
+:<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>
 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>&alpha;</sub>(x), for integer orders &alpha;=0,1,2.]]

@@ Line 85: / Line 85: @@
 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 tamplitude 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>
+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>

@@ Line 85: / Line 85: @@
 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> sould 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 tamplitude 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>
+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 tamplitude 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>

@@ Line 84: / Line 84: @@
 [[Image:Bessels_J0.svg|thumb|300px|right|Plot of Bessel function of the first kind, J<sub>&alpha;</sub>(x), for integer orders &alpha;=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>.  It it is shown, the first zero of <math>J_1(x) \quad</math> will <br>
+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> sould 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 tamplitude 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>

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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}')=\frac{d}{k\sin{\theta}'dr} 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 \frac{d}{k\sin{\theta}'dr} J_1(kr\sin{\theta}')</math><br><br>
-We invoke the principle that the integral of a derivative is the function evaluated at the end points to give us the equation<br><br><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^{-ikct'} 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'-ikct'}}{k\sin{\theta}'r'^2}J_1(ka\sin{\theta}')</math><br><br>
+<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^{-ikct'} 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'-ikct'}}{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>&alpha;</sub>(x), for integer orders &alpha;=0,1,2.]]

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 We invoke the principle that the integral of a derivative is the function evaluated at the end points to give us the equation<br><br><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^{-ikct'} 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'-ikct'}}{k\sin{\theta}'r'^2}J_1(ka\sin{\theta}')</math><br><br>
 To find the angle to the first diffraction minimum, we must find the first zero of this amplitude function.  This will occur when <math>J_1(ka\sin{\theta}')=0 \quad</math>
-[[Image:Bessels_J0J1J2.svg|thumb|300px|right|Plot of Bessel function of the first kind, J<sub>&alpha;</sub>(x), for integer orders &alpha;=0,1,2.]]
+[[Image:Bessels_J0.svg|thumb|300px|right|Plot of Bessel function of the first kind, J<sub>&alpha;</sub>(x), for integer orders &alpha;=0,1,2.]]

← Older revision		Revision as of 20:35, 8 July 2009
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	We invoke the principle that the integral of a derivative is the function evaluated at the end points to give us the equation<br><br><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^{-ikct'} 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'-ikct'}}{k\sin{\theta}'r'^2}J_1(ka\sin{\theta}')</math><br><br>		We invoke the principle that the integral of a derivative is the function evaluated at the end points to give us the equation<br><br><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^{-ikct'} 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'-ikct'}}{k\sin{\theta}'r'^2}J_1(ka\sin{\theta}')</math><br><br>
	To find the angle to the first diffraction minimum, we must find the first zero of this amplitude function. This will occur when <math>J_1(ka\sin{\theta}')=0 \quad</math>		To find the angle to the first diffraction minimum, we must find the first zero of this amplitude function. This will occur when <math>J_1(ka\sin{\theta}')=0 \quad</math>
		+	[[Image:Bessels_J0J1J2.svg\|thumb\|300px\|right\|Plot of Bessel function of the first kind, J<sub>α</sub>(x), for integer orders α=0,1,2.]]