### #59 Optical coherence tomography-I 光學相干斷層掃描_上

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The reference of this series is:

Drexler, W. & Fujimoto, J. G. (2008). Optical Coherence Tomography - Technology and Applications. 1st ed. NY: Springer.

We will focus on the basic principles of OCT and demonstrate its feasibility by computer simulation. We will not talk about the technical details including the resolution problem, the interaction between tissues and light, and the effect of scattering. We hope everyone could have some inspiration from the design of OCT.

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## What OCT could do? 先看看OCT可以幹嘛吧

Before our start we shall see what OCT could do. It could resolve the reflected lights from different histological layers and is also one of the optical biopsy. The most well-established application of OCT is the examination of retina in ophthalmology.

Let's see how retina looks like under OCT:

Figure source:Englander, Miriam; Xu, David; Kaiser, Peter K. (2014). Ophthalmology.1st ed. Fig. 130-2.

Most of us don't know what a retina should look like, so let's see a standard histology slide:

Figure source: Ross, M. H., Pawlina, W. (2011). Histology - a Text and Atlas. 6th ed. Fig.24-10.
There are 10 layers in retina, and we could examine the section by OCT without cutting a small piece of the retina of patients.

## The setting of OCT OCT的基本儀器架設

Let's first see the experimental setting of OCT. It is actually the Michelson interferometer familiar to most of us. However, the light source is "low-coherent" instead of high-coherent. And we will know why soon.

Figure source:Drexler, W. & Fujimoto, J. G. (2008). Optical Coherence Tomography - Technology and Applications. 1st ed. Fig. 2-3.
The light from light source with electric field Ei is separated into 2 light by a 50/50 beam splitter. One in the reference light path while the other hit upon our sample. There are many layers in our sample, and light reflect between different layers due to the difference in refractive index. Assume the reflectivity at depth zs to be rs(zs). The reflected lights hit the beam splitter again and we only detect the ones hit our detector. We should noted that the light pass through the beam splitter twice and the magnitude of electric field is therefore half the original.

## The basic setting of our model 模型的基本設定

After some brief explanation about the experimental setting, we could now define some variables for our further analysis. As the light first arrive at the beam splitter, its electric field Ei could be assumed as:

s(k) represent the spectrum of light source. As the light return back to the beam splitter, assume the length of reference light path  z_R and the reflectivity of reference light path r_R, the electric field of light goes reference light path becomes:
s(k)表示光源的光譜分布，我們後面會再看看他的影響。 當兩道光束被分開又再回到分光鏡時，假設參考光徑單趟路程z_R，路上的反射鏡反射率為r_R，則走參考光徑的光現在變成了E_R，我們知道E_R應該可以寫成：
$E_R&space;=&space;\frac{E_i}{\sqrt{2}}r_Re^{i2kz_R}$
What about the sample light path? Because our sample have different reflectivity at different depth, its functionality could be written as r_s(z_s). For simplification, it is assumed to be the sum of many delta functions:

$r_S(z_S)&space;=&space;\sum_{n}&space;r_{S_n}\delta(z_S-z_{S_n})$
Because the differences in refractive index across layers are usually small, r_sn << 1, the reflected light could therefore be approximated as:

So the current of detector could then be written as:

In which rho means the responsibility of detector. If we set  $S(k)=\left&space;\langle&space;|s(k)|^2&space;\right&space;\rangle$, $R_R&space;=&space;|r_R|^2,&space;R_{S_n}=|r_{S_n}|^2$, then the above equation could be expanded as:
rho是儀器的反應情形。如果我們令 $S(k)=\left&space;\langle&space;|s(k)|^2&space;\right&space;\rangle$$R_R&space;=&space;|r_R|^2,&space;R_{S_n}=|r_{S_n}|^2$,則上式可以展開成下面的樣子：
There are 3 terms in the detector current. What do they mean? How could they help us delineate the histological layers of our sample? Stay tuned and we will talk about it later.