Target Diamond Structural Analysis

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The Target Diamond

Section in progress

Probing the Diamond's Structure

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.

Ideal Thickness Calculation

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.

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, $\omega$ 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

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

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

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

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

Because all three waves are reflections of the same original wave, they all have the same amplitude and frequency. However, the processes of reflection and transmission will modify the amplitudes 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 = 100, so we are only concerned with that of the half-silvered mirror. The recorded amplitude will be equal to $C_{0}A_{laser}$. Because all reflected 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

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

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.

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

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

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

$R={\frac {n_{2}-n_{1}}{n_{2}+n_{1}}}^{2}$

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 $C_{1}$:

$C_{1}=0.17189$

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

$C_{2}=0.117876$

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

$C_{3}=0.003483$

$C_{4}=0.000103$

$C_{5}=0.000003$

Using just these five waves, we can recalculate our shape term.

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

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.

Future Updates

• Add diagrams to this page
• Compensate for interference and internal reflection