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== Interference ==
 
== Interference ==
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The propogating wave will not reflect back from the diamond coherently; it will instead form spherical wavelets. We must either account for these or find them insignificant.
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[[Image:06 Spherical Propogation.png|thumb|Each point on the wavefront generates spherical wavelets.]]
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We can treat the diamond as a single-slit experiment. In this analysis, the slit will be represented by some finite slice of the diamond, which will then be the resolution size. We will call this resolution size R. The distance the light travels, from the diamond to the detector, is L. From this projection, we know that there will be a first intensity of zero at a distance <math>\frac{b}{2}</math> from the center of the screen. Basic trigonometry tells us that
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The propogating wavefront generates spherical wavelets as it propogates. What we want to know is what the wave at the initial point <math> f(x,y,0,t) </math> will look like at <math>f(x,y,D,t)</math>. To do this, we can integrate over the product of this equation times a propogator g.
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<math>\theta = \tan^{-1} (\frac{R}{L})</math>
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<math> g( x_i, t_i, x_f, t_f)
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Because the irradiance, from the Fraunhofer approximation, is
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Because the propogator is actually in terms
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<math>I(\theta) = I(0) \frac{\sin(\beta)}{\beta}^2 </math>
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of the differences between these values, we will
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and <math>\beta</math> is defined as
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write the difference between the x-vectors as  
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<math>\beta = \frac{kR}{2}\sin(\theta)</math>
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{delta, hereafter d} dx and the difference between
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we can solve for I in terms of X.
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the times as dt.
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<math> I(x)= I (\frac{\sin(\frac{KR}{2}\sin^{-1}(\frac{R}{L}))}{\frac{KR}{2}\sin^{-1}(\frac{R}{L})})^2</math>
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[integral]dti [integral]g(stuff)f(stuffi)dxi dyi =
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Since
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f(xf, tf)
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<math>\sin(\lambda) = 0</math>
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We know that f(stuff) is nothing more than the
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our zeroes are at
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product of the phases of the incoming and outgoing
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<math> \frac{KR}{2}\sin^{-1}(\frac{R}{L}) = 0 </math>
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waves and the amplitude function.
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since this makes the sine function in the numerator zero.
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Calculation of g proves more challenging. We know
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This means that
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the wave equation
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<math> L = \frac{R}{2\tan(\sin^{-1}(\frac{\lambda}{R}))}</math>
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d2y/dt2 - c2d2y/dx2 = 0
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Unfortunately, this gives us absurd values for the necessary length of the optical path; for R to be plausibly small, L quickly approaches similar orders of magnitude. This approach is therefore invalid.
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However, this is for a uniform, sourceless wave.  
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If we use the double-slit approach, treating the width D of the diamond as the spacing between the slits, the size B of the slits becomes arbitrarily large. Under this approximation,
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Like most generalizations, this is an unrealistic
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<math>I(\theta) = 4I\frac{\sin^2(\beta)}{beta}\cos^2(\alpha)</math>
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situation. What we need is a function that can
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where
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generate a brief pulse. This sounds like a delta
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<math> \alpha = \frac{kD}{2}\sin(\theta)</math>
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function, so we will use this.
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and
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d2g/dt2 - c2d2g/dx2 = (d)(x-xi)(t-ti)
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<math> \beta = \frac{kB}{2}\sin(\theta)</math>
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This is not an easy equation to solve without
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Again, this approximation fails, here because B can be arbitrarily large. Because of this, <math>\beta</math> can be arbitrarily large, and I goes to zero.
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using a Fourier transform. Therefore, we'll do
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just that.
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G = (copy from notes)
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g = (copy)
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We will then need to plug this function into the
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earlier equation. We can change
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(copy equals-right-hand-side delta-function from
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notes)
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Now, we have an equation with integrals on both
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sides. By definition, these integrands are equal,
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so we can drop the integral signs.
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(eqn)
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After solving this,
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(G from Jones)
    
== Color of the Laser ==
 
== Color of the Laser ==
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* Figure out problems with interference
 
* Figure out problems with interference
 
* Compensate for other sources of error
 
* Compensate for other sources of error
* Calculate expected error
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* Calculate expected error</math>
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