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

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

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>

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