#54Fundamentals--Fraunhofer diffraction-I 常識集_夫朗和斐繞射_上

Our goal in this upcoming series is to introduce the concept and principles of optical coherence tomography (OCT). However, this requires some basic knowledge of optics. So we are going to talk about Fourier optics and Fraunhofer diffraction in these episodes, and we will use the result of diffraction to explain the Gaussian beam.

            本系列的目標是要和大家講解光學相干斷層掃描(Optical coherence tomography, OCT)的原理，但要讓大家理解需要建立一些基本的光學知識。這幾集必須和大家講解一下傅立葉光學和夫朗和斐繞射，之後我們會利用夫朗和斐繞射來介紹高斯光束。如果讀者不知道傅立葉轉換，請先看本站之前的介紹：

#‎23常識集_傅立葉轉換：https://goo.gl/ZChYZ0

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The reference for #54~#56 is Fundamentals of Photonics by Bahaa E. A. Saleh, Malvin Carl Teich 的 Chapter 4 Fourier Optics.
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How does an image look like after diffraction? 

影像經過繞射後長甚麼樣子？

What does it look like as the light comes to our eyes? According to the Huygens principle, every point on a wavefront is a source of wavelet, and the image at some distance d is the sum of diffraction of all wavelets. Let's see how does it look like to gain some insight about this issue. The following figure measures 3m x 3m:
           當光從物體傳到我們眼睛時， 他看起來會是怎樣呢？國高中學過海更士原理，物體上的每個點可以看成一個點波源，發出一個球面子波，而在距離d的某處的影像就是這些球面子波彼此干涉的結果。我們先實際看一下一個圖像傳播的情形可能比較有感覺。我們來看看這張圖，假設圖的長寬為3m x 3m：

As the light travels 1nm, the image now looks like:
如果我們讓影像往前走1nm，會看到的圖像是：

It does not change at all and it makes sense. What if the light travels 1cm?
看起來幾乎沒變，這挺合理的。但如果我們讓光往前走1cm呢？

It start blurring now. What if  the light travels 10 m?
可以看到開始糊掉了。那如果讓光走10m呢？

It is now basically unrecognizable. Why? Since each point of the original figure is now a new source of wavelet, the direction of transportation is not limited to the direction of the observer, and the new image comes from the sum of interference of all electric fields from each wavelet. If we stand far enough away, the light intensity is governed by the Fourier transform of the light source. The above figures are generated by MATLAB code below and please feel free to play with it.
基本上是面目全非了。為什麼呢？因為每個點都是一個新的點波源，他們傳播的方向並不局限於直行的方向，因此在距離d的一點的光強度是由光源每一點的光抵達該處之後電場疊加的結果來決定的。當觀測距離夠遠(符合夫朗和斐條件)的時候，看到的光強度分布是由原光源的傅立葉轉換決定的。上面這些圖的程式碼在文章最後面有附，有興趣者可以玩玩。

Spatial frequency 空間頻率

 We have previously discussed about the Fourier transform of time series signal. However, today we will deal with the concept of spatial frequency and its Fourier transform. From the basic electromagnetism, a wave propagating in a 3D space could be expressed as exp(iwt - ik‧r). If we only consider the spatial part, or the term exp(- ik‧r). k is called the wave vector, and its direction is the direction of wave propagation. As parallel light source pass through a transparent, each point on the transparent becomes a new source of wavelet. However, we could also interpret it in terms of various parallel light source with different direction, intensity and phases. As shown in the figure:
         在之前我們已經看過時間序列的訊號與他的傅立葉轉換是甚麼情形，但這裡我們要來討論空間頻率的概念和傅立葉轉換。如果各位學過基本的電磁學的話會知道，一個在空間中傳遞的波可以用 exp(iwt - ik‧r) 來表示，在這裡我們先只看空間的部分，也就是exp(- ik‧r)。其中向量k被稱為波向量(wave vector)，他的方向是波傳遞的方向。當平行光源通過一片有圖樣的幻燈片時，每一點會變成新的點波源，往四面八方傳送。但我們也可以想像成是這道平行光源被分成好多道不同方向的平行光，只是每個方向的平行光強度各異，且有不同的相位，如下圖所示：

Waves propagate toward different directions have different  k‧r = kx*x+ky*y+kz*z. However, it would be easier to interpret its direction if we express kx, ky in terms of angle. So we draw another figure:
不同方向的波，k‧r不相同，把內積展開來變成kx*x+ky*y+kz*z。但直接看kx, ky我們很難想像他到底是往哪個方向前進，所以我們希望能把kx, ky和角度扯上關連。所以我們畫出下圖：

Under para-axial approximation, we could write:
在近軸近似(para-axial approximation)的前提下，我們可以寫出：

Define spatial frequency νx=kx/2π, νy=ky/2π. Since λ = 2π/k, the above formula could now be written
如果我們定義空間頻率νx=kx/2π, νy=ky/2π，因為λ = 2π/k，所以上式可以改寫成：

,

That is to say, we could conceptually express the light passing through the transparent in terms of something like this:
也就是說現在我們在概念上可以把通過幻燈片的光寫成像這樣的形式：

Which is simply writing it in terms of the sum of parallel light sources from different directions.
希望到目前為止讀者還可以接受，其實就是把影像寫成來自不同方向的光的總和而已。

The Fourier transform of image 影像的傅立葉轉換

We have just written
剛剛我們寫出了

However, it is troublesome to consider the magnitude of A(νx, νy). If the light propagates in vacuum space, the magnitude is expected to be conserved. So let's see at z=0:
但很令人困擾的是，振幅A(νx, νy)到底是多少???我們預期說，這個振幅應該不會在真空中傳遞時改變，所以如果我們看在z = 0的地方：

And this shall be familiar to you. To make it true, A(νx, νy) must be the Fourier transform of f(x,y). So in fact we could write:
這算式看起來應該很熟悉，要讓上式為真，A(νx, νy)必須是f(x, y)的傅立葉轉換，所以我們把上面概念的算式寫成連續的形式就是：

And the image we will see at some distance z will be:
而在距透明片z的地方我們看到的影像就會是：

It should be noted that since kz = (k^2-kx^2-ky^2)^1/2, kz actually depends on νx, νy!
 這裡要小心的是因為kz = (k^2-kx^2-ky^2)^1/2，所以kz其實是νx, νy的函數，積分時不可以直接把他提出去喔。

In the next episode, we will integrate to calculate U(x,y,z). Stay tuned!
 這集就先到這邊~~下集我們要來算上面U(x,y,z)的積分式囉。

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MATLAB code for the above analysis：
%% load in figure
I0 = imread('biophysics.png');
I0=rgb2gray(I0);
I0=255-I0;
I0_double = double(I0)/255;
figure, imshow(I0);
[xR, yR] = size(I0);
%% far field
lambda = 500E-9; %unit: m
k = 2*pi/lambda;
d = 1E-9;
deltaX = 0.01; deltaY = 0.01;
[RangeX, RangeY] = meshgrid(-xR/2:1:xR/2);
RangeX_real = RangeX*deltaX; RangeY_real = RangeY*deltaY;
D_matrix = sqrt(RangeX_real.^2+RangeY_real.^2+d^2);
spherc_wave = (1./D_matrix).*exp(-1i*k*D_matrix);

I2 = conv2(spherc_wave, I0_double);
I2_central = I2(301:1:900, 301:1:900);
F2 = I2_central.*conj(I2_central);
%F2 = log(F2+1);
I2 = mat2gray(F2);
figure, imshow(I2);