Changes

[[Image:01 Apparatus with Red Laser.png|thumb|The Michelson interferometer to be used. The diamond is not to scale.]]

[[Image:01 Apparatus with Red Laser.png|thumb|The Michelson interferometer to be used. The diamond is not to scale.]]

We must be able to determine the precise thickness and shape of the diamond chip. Because it is tiny (on the order of 50 microns thick), conventional measurements are impossible. Instead, we will use a modified [http://en.wikipedia.org/wiki/Michelson_interferometer Michelson interferometer]. In our design, we will replace one of the mirrors with the target diamond chip. In this modified design, the plane wave returned to the detector will be a combination of three waves (neglecting internal reflection): one reflected off the front of the diamond, one reflected off the back of the diamond, and one reflected off the remaining mirror. However, all that the detector can record is the wave's amplitude. From this, we need to extract thickness and shape.

We must be able to determine the precise thickness and shape of the diamond chip. Because it is tiny (on the order of 50 microns thick), conventional measurements are impossible. Instead, we will use a modified [http://en.wikipedia.org/wiki/Michelson_interferometer Michelson interferometer]. In our design, we will replace one of the mirrors with the target diamond chip. In this modified design, the plane wave returned to the detector will be a combination of three waves (neglecting internal reflection): one reflected off the front of the diamond, one reflected off the back of the diamond, and one reflected off the remaining mirror. However, all that the detector can record is the wave's intensity. From this, we need to extract thickness and shape.

== The Diamond and Light Waves ==

== The Diamond and Light Waves ==

Both the front and back planes of the diamond are two-dimensional surfaces in three-dimensional space. The recorded amplitudes will form a two-dimensional graph and record amplitude at points across the diamond's surface. Basically, the light wave can be treated as a massive grid of one-dimensional waves normal to the diamond. All of the following calculations are applied to the recorded amplitude of one of these waves, which is the amplitude at one specific point on the diamond.

Both the front and back planes of the diamond are two-dimensional surfaces in three-dimensional space. The recorded intensities will form a two-dimensional graph and record intensity at points across the diamond's surface. Basically, the light wave can be treated as a massive grid of one-dimensional waves normal to the diamond. All of the following calculations are applied to the recorded intensity of one of these waves, which is the intensity at one specific point on the diamond.

We will be neglecting all sources of error during these calculations. They will be reintroduced after we have obtained our basic calculations.

We will be neglecting all sources of error during these calculations. They will be reintroduced after we have obtained our basic calculations.

<math>\Psi = A _{laser} \sin ( \omega t + d ) \,</math>

<math>\Psi = A _{laser} \sin ( \omega t + d ) \,</math>

where <math> A _{laser} </math> is the amplitude of the initial laser beam, <math>\omega</math> is the frequency, t is time, d is the phase-shift, and C is a constant dependent on the reflectivity of all surfaces the laser intercepts.

where <math> A _{laser} </math> is the intensity of the initial laser beam, <math>\omega</math> is the frequency, t is time, d is the phase-shift, and C is a constant dependent on the reflectivity of all surfaces the laser intercepts.

We have a sum of three waves, which can be expressed as

We have a sum of three waves, which can be expressed as

(For simplicity, we will say that the wave leaving the mirror has not been phase-shifted, as above.)

(For simplicity, we will say that the wave leaving the mirror has not been phase-shifted, as above.)

== Ideal Amplitude Calculation ==

== Ideal Intensity Calculation ==

[[Image:03 Apparatus for Beam 00.png|thumb|The zeroth wave path. Although some light passes through the splitter, it is lost, cannot affect the experiment, and therefore is not illustrated.]]

[[Image:03 Apparatus for Beam 00.png|thumb|The zeroth wave path. Although some light passes through the splitter, it is lost, cannot affect the experiment, and therefore is not illustrated.]]

Because all three waves are reflections of the same original wave, they all have the same wavelength. However, the processes of reflection and transmission will modify the ''amplitude'' of each wave. By removing the diamond and reflecting the laser solely off of the mirror, we will be able to calculate the amplitude of the initial light after it has reflected off the mirror and beam splitter once and been transmitted through the splitter once.  

Because all three waves are reflections of the same original wave, they all have the same wavelength. However, the processes of reflection and transmission will modify the ''intensity'' of each wave. By removing the diamond and reflecting the laser solely off of the mirror, we will be able to calculate the intensity of the initial light after it has reflected off the mirror and beam splitter once and been transmitted through the splitter once.  

The mirror has a coefficient of reflection r = 1 (it reflects all the light and does not transmit any), so we are only concerned with the half-silvered mirror. The recorded amplitude will be equal to <math> C _0 A _{laser} </math>. Because all beams that the detector will recieve will reflect off of and pass through the splitter once, we can create a new amplitude variable A such that

The mirror has a coefficient of reflection r = 1 (it reflects all the light and does not transmit any), so we are only concerned with the half-silvered mirror. The recorded intensity will be equal to <math> C _0 A _{laser} </math>. Because all beams that the detector will recieve will reflect off of and pass through the splitter once, we can create a new intensity variable A such that

<math> A = A _{laser} C _0 \,</math>

<math> A = A _{laser} C _0 \,</math>

Because of this, we will never be dealing with the amplitude of the original laser. This will have no effect on the calculations, but it is important to remember that our value <math> A </math> is ''not'' the same as <math> A _{laser} </math>.

Because of this, we will never be dealing with the intensity of the original laser. This will have no effect on the calculations, but it is important to remember that our value <math> A </math> is ''not'' the same as <math> A _{laser} </math>.

== Ideal Thickness Calculation ==

== Ideal Thickness Calculation ==

To find the thickness of the diamond, we ideally only need the first two waves. To remove the third wave, which reflects from the mirror, we can simply remove the mirror.

To find the thickness of the diamond, we ideally only need the first two waves. To remove the third wave, which reflects from the mirror, we can simply remove the mirror.

The combined wave equation is unimportant, since we only record its amplitude, which is

The combined wave equation is unimportant, since we only record its intensity, which is

<math>A^2 _{12} = C _1 A^2 + C _2 A^2 + 2 C _1 C _2 A^2 \cos ( d _2 - d _1 ) </math>

<math>A^2 _{12} = C _1 ^2 A^2 + C _2 ^2 A^2 + 2 C _1 C _2 A^2 \cos ( d _2 - d _1 ) </math>

<math>A^2 _{12} / A^2 = C _1 + C _2 + 2 C _1 C _2 \cos ( d _2 - d _1 ) </math>

<math>A^2 _{12} / A^2 = C _1 ^2 + C _2 ^2 + 2 C _1 C _2 \cos ( d _2 - d _1 ) </math>

Because the wave reflecting off the back of the diamond travels through the diamond twice, the term <math> d _2 - d _1 </math> is twice the thickness of the diamond, in radians. Because this measurement is in unhelpful units, we can multiply it by the wavelength <math> \lambda </math> and divide by <math> 2 \pi </math> for the thickness <math> \tau </math> in meters.

Because the wave reflecting off the back of the diamond travels through the diamond twice, the term <math> d _2 - d _1 </math> is twice the thickness of the diamond, in radians. Because this measurement is in unhelpful units, we can multiply it by the wavelength <math> \lambda </math> and divide by <math> 2 \pi </math> for the thickness <math> \tau </math> in meters.

Therefore, we can rewrite the earlier equation in terms of <math> \tau </math>:

Therefore, we can rewrite the earlier equation in terms of <math> \tau </math>:

<math>A^2 _{12} / A^2 = C _1 + C _2 + 2 C _1 C _2 \cos ( \frac{4 \pi \tau}{\lambda} ) </math>

<math>A^2 _{12} / A^2 = C _1 ^2 + C _2 ^2 + 2 C _1 C _2 \cos ( \frac{4 \pi \tau}{\lambda} ) </math>

== Ideal Shape Calculation ==

== Ideal Shape Calculation ==

Of course, thickness is not the only thing we need. After calculating <math> \tau </math> and uncovering the mirror, we will also detect the third laser reflection. We can calculate that the amplitude of this new combined wave will be

Of course, thickness is not the only thing we need. After calculating <math> \tau </math> and uncovering the mirror, we will also detect the third laser reflection. We can calculate that the intensity of this new combined wave will be

<math>A^2 _{012} = C^2 _0 A^2 + C^2 _1 A^2 + C^2 _2 A^2 + 2 C _1 C _2 A^2 \cos ( d _2 - d _1 ) + 2 C _0 C _1 A^2 \cos ( - d _1 ) + 2 C _0 C _2 A^2 \cos ( - d _2 ) </math>

<math>A^2 _{012} = C^2 _0 A^2 + C^2 _1 A^2 + C^2 _2 A^2 + 2 C _1 C _2 A^2 \cos ( d _2 - d _1 ) + 2 C _0 C _1 A^2 \cos ( - d _1 ) + 2 C _0 C _2 A^2 \cos ( - d _2 ) </math>

== Compensating for Internal Reflection ==

== Compensating for Internal Reflection ==

[[Image:02 Internal Reflection with Labels.png|thumb|Internal reflection of the diamond, where C is a coefficient to be multiplied by the incoming amplitude. The angles have been exaggerated from zero degrees. The diagram also shows that all error comes from about two percent of the amplitude.]]

[[Image:02 Internal Reflection with Labels.png|thumb|Internal reflection of the diamond, where C is a coefficient to be multiplied by the incoming intensity. The angles have been exaggerated from zero degrees. The diagram also shows that all error comes from about two percent of the intensity.]]

Realistically, the laser will not miraculously split in two upon reaching the diamond, creating one wave that reflects back and a second that reflects off of the back of the diamond and then passes perfectly through the front. Internal reflection will occur; we must calculate how much there will be and whether or not we must compensate for it.

Realistically, the laser will not miraculously split in two upon reaching the diamond, creating one wave that reflects back and a second that reflects off of the back of the diamond and then passes perfectly through the front. Internal reflection will occur; we must calculate how much there will be and whether or not we must compensate for it.

Once again, the terms become very small very quickly. Because even the largest internal-reflection induced term (<math> 2 C _2 C _3 </math>) is more than an order of magnitude smaller than the needed terms, we can treat all internal reflection as error and ignore it.

Once again, the terms become very small very quickly. Because even the largest internal-reflection induced term (<math> 2 C _2 C _3 </math>) is more than an order of magnitude smaller than the needed terms, we can treat all internal reflection as error and ignore it.

This rapid decrease in amplitude is primarily because much of the initial amplitude is completely lost during the reflection process, since 83% of any remaining amplitude is lost each time the wave reflects off of the back of the diamond. This decrease is ''very'' significant. For example, when the wave <math> \Psi _2 </math> exits the diamond, the light reflected back into the diamond has an amplitude less than three percent of the original wave, and 83% of ''this'' is lost when the wave reflects off of the back of the diamond, leaving less than half a percent of the initial amplitude to make up all errors.

This rapid decrease in intensity is primarily because much of the initial intensity is completely lost during the reflection process, since 83% of any remaining intensity is lost each time the wave reflects off of the back of the diamond. This decrease is ''very'' significant. For example, when the wave <math> \Psi _2 </math> exits the diamond, the light reflected back into the diamond has an intensity less than three percent of the original wave, and 83% of ''this'' is lost when the wave reflects off of the back of the diamond, leaving less than half a percent of the initial intensity to make up all errors.

== Interference ==

== Interference ==

[[Image:06 Spherical Propogation.png|thumb|Each point on the wavefront generates spherical wavelets. Note that the waveform is not uniform.]]

[[Image:06 Spherical Propogation.png|thumb|Each point on the wavefront generates spherical wavelets. Note that the waveform is not uniform.]]

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.

The propagating 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 propagator g.

<math> g( x_i, t_i, x_f, t_f)\,</math>

<math> g( x_i, t_i, x_f, t_f)\,</math>

Because the propogator is actually in terms of the <i>differences</i> between the x and t values, we will write the difference between the x-vectors as '''&Delta;x''' and the difference between the times as <math>\Delta t</math>.

Because the propagator is actually in terms of the <i>differences</i> between the x and t values, we will write the difference between the x-vectors as '''&Delta;x''' and the difference between the times as <math>\Delta t</math>.

The integral is therefore

The integral is therefore

<math>\int{dt_i} \int{f(x_i,t_i)g(\Delta \mathbf{x}, \Delta t) dx_i dy_i}</math>  

<math>f (x_f,t_f) = \int{dt_i} \int{f(x_i,t_i)g(\Delta \mathbf{x}, \Delta t) dx_i dy_i}</math>  

where <math>f(x_i,t_i)\,</math> is nothing more than the product of the phases of the incoming and outgoing waves and the intensity function. Because this equation only takes into account one incoming and one outgoing wave, we will need to recalculate the later equations using different f-functions for all outgoing waves and sum these results. Fortunately, as we discovered above, only the first and second outgoing waves are relevant.

 

<math>f(x_i,t_i)\,</math>

 

is nothing more than the product of the phases of the incoming and outgoing waves and the amplitude function. Because this equation only takes into account one incoming and one outgoing wave, we will need to recalculate the later equations using different f-functions for all outgoing waves and sum these results. Fortunately, as we discovered above, only the first and second outgoing waves are relevant.

Calculation of g proves more challenging. We know the wave equation

Calculation of g proves more challenging. We know the wave equation

<math>\frac{d^2y}{dt^2} - c^2\frac{d^2y}{dx^2} = 0</math>

<math>\frac{d^2y}{dt^2} - c^2\nabla^2 = 0</math>

However, this is for a uniform, sourceless wave. Like most generalizations, this is an unrealistic situation in the real world. What we need is a function that can generate a brief pulse. This sounds like a delta function.

However, this is for a uniform, sourceless wave. Like most generalizations, this is an unrealistic situation in the real world. What we need is a function that can generate a brief pulse. This sounds like a delta function.

<math>\frac{d^2g}{dt^2} - c^2\frac{d^2g}{dx^2} = \delta^3(\Delta \mathbf{x})\delta(\Delta t)</math>

<math>\frac{d^2g}{dt^2} - c^2nabla^2 = \delta^3(\Delta \mathbf{x})\delta(\Delta t)</math>

This is not an easy equation to solve without using a Fourier transform. Therefore, we'll do just that, with G as the transformed function.

This is not an easy equation to solve without using a Fourier transform. Therefore, we'll do just that, with G as the transformed function.

Therefore,

Therefore,

<math>g(\Delta \mathbf{x},\Delta t) = \frac{1}{4\pi^2} \int{G(\mathbf{k},\omega)e^{-i\mathbf{k}\cdot\Delta x}e^{i\omega \Delta t}d^3k d\omega}</math>

<math>g(\Delta \mathbf{x},\Delta t) = \frac{1}{4\pi^2} \int{G(\mathbf{k},\omega)e^{i\mathbf{k}\cdot\Delta x}e^{-i\omega \Delta t}d^3k d\omega}</math>

We will then need to plug this function into the earlier equation. To make this easier, we know that the four-dimensional delta function on the right-hand side can be simplified.

We will then need to plug this function into the earlier equation. To make this easier, we know that the four-dimensional delta function on the right-hand side can be simplified.

Now, we have an equation with integrals on both sides. Since both of these integrals have the same limits and integrands, they must be integrals of equal functions. We can simply drop the integrals.

Now, we have an equation with integrals on both sides. Since both of these integrals have the same limits and integrands, they must be integrals of equal functions. We can simply drop the integrals.

<math> \frac{d^2}{dt^2} \left(\frac{1}{4\pi^2} G(\mathbf{k},\omega)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}\right)-c^2\frac{d^2}{dx^2}\left(\frac{1}{4\pi^2} G(\mathbf{k},\omega)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}\right) = \frac{1}{16\pi^4} e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}</math>

<math> \frac{d^2}{dt^2} \left(\frac{1}{4\pi^2} G(\mathbf{k},\omega)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}\right)-c^2nabla^2\left(\frac{1}{4\pi^2} G(\mathbf{k},\omega)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}\right) = \frac{1}{16\pi^4} e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}</math>

This is a complicated equation, but it can be solved for G.

This is a complicated equation, but it can be solved for G.

<math>G(\mathbf{k},\omega) = \frac{1}{4\pi^2}\left(\frac{1}{c^2k^2-\omega^2}\right)</math>

<math>G(\mathbf{k},\omega) = \frac{-1}{4\pi^2}\left(\frac{1}{c^2k^2-\omega^2}\right)</math>

Once G is calculated, we can apply an inverse Fourier transform and find g; we can then plug this into

Once G is calculated, we can apply an inverse Fourier transform and find g; we can then plug this into

== Color of the Laser ==

== Color of the Laser ==

The laser used must have a wavelength greater than the maximum expected defect of the diamond. Because any defects will be extremely small (on the order of one micron thick), we need light with a wavelength of around 500 nm. We could use either a green or red laser. Once expected error is calculated, we will be able to calculate expected error from the less accurate red light, and determine whether or not it would be usable.

The laser used must have a wavelength less than the maximum expected defect of the diamond. Because any defects will be extremely small (on the order of one micron thick), we need light with a wavelength of around 500 nm. We could use either a green or red laser. Once expected error is calculated, we will be able to calculate expected error from the less accurate red light, and determine whether or not it would be usable.

== Future Updates ==

== Future Updates ==