Target Diamond Structural Analysis

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The Target Diamond

Section in progress

Probing the Diamond's Structure

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 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.

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.

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

Light is a wave, and can be expressed as

$\Psi =A_{laser}\sin(\omega t+d)\,$

where $A_{laser}$ is the amplitude of the initial laser beam, $\omega$ 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

$\Psi _{Front}=\Psi _{1}=C_{1}A_{laser}\sin(\omega t+d_{1})\,$

$\Psi _{Back}=\Psi _{2}=C_{2}A_{laser}\sin(\omega t+d_{2})\,$

$\Psi _{Mirror}=\Psi _{0}=C_{0}A_{laser}\sin(\omega t)\,$

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

Ideal Amplitude Calculation

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.

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 $C_{0}A_{laser}$. 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

$A=A_{laser}C_{0}\,$

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 $A$ is not the same as $A_{laser}$.

Ideal Thickness Calculation

The first wave reflection.

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

$A_{12}^{2}=C_{1}A^{2}+C_{2}A^{2}+2C_{1}C_{2}A^{2}\cos(d_{2}-d_{1})$

$A_{12}^{2}/A^{2}=C_{1}+C_{2}+2C_{1}C_{2}\cos(d_{2}-d_{1})$

Because the wave reflecting off the back of the diamond travels through the diamond twice, the term $d_{2}-d_{1}$ is twice the thickness of the diamond, in radians. Because this measurement is in unhelpful units, we can multiply it by the wavelength $\lambda$ and divide by $2\pi$ for the thickness $\tau$ in meters.

${\frac {(d_{2}-d_{1})\lambda }{4\pi }}=\tau$

Therefore, we can rewrite the earlier equation in terms of $\tau$:

$A_{12}^{2}/A^{2}=C_{1}+C_{2}+2C_{1}C_{2}\cos({\frac {4\pi \tau }{\lambda }})$

Ideal Shape Calculation

Of course, thickness is not the only thing we need. After calculating $\tau$ 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

$A_{012}^{2}=C_{0}^{2}A^{2}+C_{1}^{2}A^{2}+C_{2}^{2}A^{2}+2C_{1}C_{2}A^{2}\cos(d_{2}-d_{1})+2C_{0}C_{1}A^{2}\cos(-d_{1})+2C_{0}C_{2}A^{2}\cos(-d_{2})$

Although this equation looks very complicated, we know that

$(d_{2}-d_{1})={\frac {4\pi \tau }{\lambda }}$

so

$d_{2}={\frac {4\pi \tau }{\lambda }}+d_{1}$

To further simplify, we need a "shape term". This term will represent the difference in the distance traveled by waves 0 and 1. If the phase-shift between 0 and 1 is constant, the diamond is flat; otherwise, it is deformed by some distance expressed by s.

$s={\frac {d_{1}\lambda }{4\pi }}$

We can rewrite this in terms of $d_{1}$:

$d_{1}={\frac {4\pi s}{\lambda }}$

Simplifying our initial equation, we find that

${\frac {A_{012}^{2}}{A^{2}}}=C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+2C_{1}C_{2}\cos({\frac {4\pi \tau }{\lambda }})+2C_{0}C_{1}\cos(-{\frac {4\pi s}{\lambda }})+2C_{0}C_{2}\cos(-{\frac {4\pi (\tau +s)}{\lambda }})$

Because the cosine is an even function, we can slightly simplify this further.

${\frac {A_{012}^{2}}{A^{2}}}=C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+2C_{1}C_{2}\cos({\frac {4\pi \tau }{\lambda }})+2C_{0}C_{1}\cos({\frac {4\pi s}{\lambda }})+2C_{0}C_{2}\cos({\frac {4\pi (\tau +s)}{\lambda }})$

This equation, athough complicated, is completely solvable, as the only unknown is the s-term.

Compensating for Internal Reflection

Internal reflection, where C is a coefficient to be multiplied by the amplitude. The angles have been exaggerated from zero degrees.

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.

The coefficient of reflectivity R can be calculated from

$R=\left({\frac {n_{2}-n_{1}}{n_{2}+n_{1}}}\right)^{2}$

where $n_{1}$ is the index of refraction for air and $n_{2}$ is the index of refraction for diamond. We can look these up and calculate R:

$R=0.17189\,$

We also need the coefficient of transmission T. However, because $R+T=1$, calculation is easy.

$T=0.82811\,$

This tells us that about 83% of the laser will be transmitted through the diamond at each reflection. This tells us $C_{1}$:

$C_{1}=R=0.17189\,$

For $C_{2}$, we must take into account two transmissions and one reflection. The calculation is easy:

$C_{2}=RT^{2}=0.117876\,$

We can continue and calculate $C_{3}$, $C_{4}$, and $C_{5}$.

$C_{3}=R^{3}T^{2}=0.003483\,$

$C_{4}=R^{5}T^{2}=0.000103\,$

$C_{5}=R^{7}T^{2}=0.000003\,$

Generally, for $C_{n}$ where $n>2$:

$C_{n}=T^{2}R^{2(n-1)-1}\,$

Using just the five waves $C_{1}$ to $C_{5}$ to begin with, we can recalculate our shape term.

${\frac {A_{12345}^{2}}{A^{2}}}=C_{1}^{2}+C_{2}^{2}+C_{3}^{2}+C_{4}^{2}+C_{5}^{2}+2C_{1}C_{2}\cos(d_{2}-d_{1})$

$+2C_{1}C_{3}\cos(d_{3}-d_{1})+2C_{1}C_{4}\cos(d_{4}-d_{1})+2C_{1}C_{5}\cos(d_{5}-d_{1})\,$

$+2C_{2}C_{3}\cos(d_{3}-d_{2})+2C_{2}C_{4}\cos(d_{4}-d_{2})+2C_{2}C_{5}\cos(d_{5}-d_{2})\,$

$+2C_{3}C_{4}\cos(d_{4}-d_{3})+2C_{3}C_{5}\cos(d_{5}-d_{3})+2C_{4}C_{5}\cos(d_{5}-d_{4})\,$

This is an unnerving equation. However, we only have eleven C-terms that need to be calculated. We can begin with the sum of C-terms and compare it to the idealized version.

$C_{1}^{2}+C_{2}^{2}+C_{3}^{2}+C_{4}^{2}+C_{5}^{2}=0.043453\,$

$C_{1}^{2}+C_{2}^{2}=0.043441\,$

Because the difference between these terms is three orders of magnitude less than the measurement, we can count it as error and do not need to compensate for it.

Next, we need to calculate all of the inner products.

$2C_{1}C_{2}=0.0405\,$

$2C_{1}C_{3}=0.0012\,$

$2C_{1}C_{4}=0.0000\,$

$2C_{1}C_{5}=0.0000\,$

$2C_{2}C_{3}=0.0008\,$

$2C_{2}C_{4}=0.0000\,$

$2C_{2}C_{5}=0.0000\,$

$2C_{3}C_{4}=0.0000\,$

$2C_{3}C_{5}=0.0000\,$

$2C_{4}C_{5}=0.0000\,$

Once again, the terms become very small very quickly. Because even the largest internal-reflection induced term ($2C_{2}C_{3}$) 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 $\Psi _{2}$ 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.

Interference

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 $f(x,y,0,t)$ will look like at $f(x,y,D,t)$. To do this, we can integrate over the product of this equation times a propogator g.

$g(x_{i},t_{i},x_{f},t_{f})\,$

Because the propogator is actually in terms of the differences between the x and t values, we will write the difference between the x-vectors as Δx and the difference between the times as $\Delta t$.

The integral is therefore

$\int {dt_{i}}\int {f(x_{i},t_{i})g(\Delta \mathbf {x} ,\Delta t)dx_{i}dy_{i}}$

We know that

$f(x_{i},t_{i})\,$

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

${\frac {d^{2}y}{dt^{2}}}-c^{2}{\frac {d^{2}y}{dx^{2}}}=0$

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.

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

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.

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

Therefore,

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

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.

$\delta ^{4}(\Delta \mathbf {x} ,\Delta t)={\frac {1}{16\pi ^{4}}}\int {e^{-i\mathbf {k} \cdot \Delta \mathbf {x} }e^{i\omega \Delta t}d^{3}kd\omega }$

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.

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

This is a complicated equation, but it can be solved for G. Once G is calculated, we can apply an inverse Fourier transform and find g; we can then plug this into

$\int {dt_{i}}\int {f(x_{i},t_{i})g(\Delta x,\Delta t)dx_{i}dy_{i}}$

and calculate the effects of the spherical wavelets.

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.

Future Updates

• Compensate for other sources of error
• Calculate expected error