### #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

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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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## 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:

It does not change at all and it makes sense. What if the light travels 1cm?

It start blurring now. What if  the light travels 10 m?

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.

## 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:

Under para-axial approximation, we could write:

Define spatial frequency νx=kx/2π, νy=ky/2π. Since λ = 2π/k, the above formula could now be written

That is to say, we could conceptually express the light passing through the transparent in terms of something like this:

Which is simply writing it in terms of the sum of parallel light sources from different directions.

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

We have just written

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:

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:

And the image we will see at some distance z will be:

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：
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);