matlab中光波叠加的可视化

我写了一些快速代码来可视化空间中两个不同振幅的波的叠加，点源几何。这在khanacademy CS平台上起作用。但我无法在matlab中再现精确的现象。我得到的只是一个嘈杂的图像。这与随机数生成中的差异有关吗？我不知道random（0,1）（在JS中）和rand（在matlab中）有多不同

下面是matlab代码

象平面上x，y点的波叠加函数

function S = Super(refamp,objamp,x,y,a,lambda)
    r1 = sqrt(a*a+x*x+y*y); %a is in z-axis
    S = refamp+(objamp*cos(2*pi*r1/(lambda/(10^6))));

测试脚本

close all;
clear all;
clc;

a=10;       %distance from source to image plane

width = 1024;
height =1024;

im = zeros(width); % the image
x=1;
y=1;
A0 = 3; % amplitude of reference wave
A1 = 1; % amplitude of object wave  A0>>A1: A0/A1>=3
lambda = 632; % wavelength in nanometers
% generate the superposition in space width*height at a along z-axis
for y=1:height
    for x=1:width
        s = Super(A0,A1,x-(width/2),y-(height/2),a, lambda);
        r=rand;
        if(r<(s/(A0+A1)))
            im(x,y) = 1;
    end
end

%display the image
figure
imshow(im,[])
title('test image')

全部关闭；
清除所有；
clc；
a=10；%源到像面距离
宽度=1024；
高度=1024；
im=零（宽度）；%形象
x=1；
y=1；
A0=3；%参考波振幅
A1=1；%目标波振幅A0>>A1:A0/A1>=3
λ=632；%波长单位：纳米
%沿z轴在a处生成空间宽度*高度的叠加
y=1时：高度
对于x=1：宽度
s=超级（A0，A1，x-（宽度/2），y-（高度/2），a，λ）；
r=兰特；
如果（r主要的问题是你的天平关闭了，所以你看不到干涉模式。如果你仔细考虑每件东西有多大/多远，它就会正确，你就能看到干涉模式

第二个问题是，你的代码将真正受益于矢量化。我在下面展示了这一点——这样做可以大大加快执行速度

function Interference
a=1000 * 10^-9;       #% distance from source to image plane
width = 10000 * 10^-9;
height= 10000 * 10^-9;
size = 700;
A0 = 3; %# amplitude of reference wave
A1 = 1; %# amplitude of object wave  A0>>A1: A0/A1>=3
lambda = 632 * 10^-9; #% wavelength in nanometers

x=linspace(0,width,size); #% vector from 0 to width
y=linspace(0,height,size); #% vector from 0 to height
[X,Y]=meshgrid(x,y); #% matrices of x and y values at each position

s=Super(A0, A1, X-(width/2), Y-(height/2), a, lambda); #% size-by-size (700x700) 
r=rand(size); #% 700x700 matrix of random values on [0 1]

im = zeros(size);
im(r<(s/(A0+A1))) = 1; %# do this all at once instead of pixel-by-pixel

#% display the image
figure
imshow(im,[])
title('test image')
end #% end of function Interference


#% Super is now vectorized, so you can give it a matrix of values for x and y
function S = Super(refamp,objamp,x,y,a,lambda)
    r1 = sqrt(a.*a+x.*x+y.*y); #% dot notation: multiply element-wise
    S = refamp+(objamp*cos(2*pi*r1/(lambda)));
end #% end of function Super

功能干扰
a=1000*10^-9；#%从源到图像平面的距离
宽度=10000*10^-9；
高度=10000*10^-9；
尺寸=700；
A0=3；%#参考波振幅
A1=1；%#目标波振幅A0>>A1:A0/A1>=3
λ=632*10^-9；#%波长（纳米）
x=linspace（0，宽度，大小）；#%0到宽度的向量
y=linspace（0，高度，大小）；#%从0到高度的向量
[X，Y]=每个位置X和Y值的网格（X，Y）#%矩阵
s=Super（A0，A1，X-（宽度/2），Y-（高度/2），a，λ）；#尺寸百分比（700x700）
r=rand（size）；#%700x700[0 1]上随机值矩阵
im=零（大小）；
im（r+1：我正在写这一切，而你走在我前面：）最近我经常遇到这种情况。非常感谢。我仍在思考这一点…特别是高度和宽度：Dbtw，考虑到我试图模拟的物理系统，缩放仍处于关闭状态。像素的宽度等值应以几微米为单位，因此我发布了正确的工作代码
function i = Interference(width, height, sizeh,sizev,z)
% parameters explained
% width: is the horizontal pixel pitch in microns
% height: is the vertical pixel pitch in microns
% size is the width=height of the CCD in number of pixels
% z is distance from source to image plane
A0 = 3; %# amplitude of reference wave
A1 = 1; %# amplitude of object wave  A0>>A1: A0/A1>=3
lambda = 635 * 10^-9; % wavelength in nanometers

%the linspace was wrong
x=linspace(0,width*sizeh,sizeh); % vector from 0 to width of size 'size'
y=linspace(0,height*sizev,sizev); % vector from 0 to height of size 'size'
[X,Y]=meshgrid(x,y); % matrices of x and y values at each position

s=Super(A0, A1, X-((width*sizeh)/2), Y-((height*sizev)/2), z, lambda); % size-by-size (1024x1024) 
r=rand(size); % 1024x1024 matrix of random values on [0 1]
%i=s;
im = zeros(size);
im(r<(s/(A0+A1))) = 1; %# do this all at once instead of pixel-by-pixel
i=im;
end % end of function Interference


% Super is now vectorized, so you can give it a matrix of values for x and y
function S = Super(refamp,objamp,x,y,a,lambda)
    r1 = sqrt(a.*a+x.*x+y.*y); % dot notation: multiply element-wise
    S = refamp+(objamp*cos(2*pi*r1/(lambda)));
end % end of function Super

width = 2.8 * 10^-6;
height= 2.8 * 10^-6; %pixel size
%  sizeh = 16; %image size in pixels
%  sizev = 16;
 sizeh = 1600; %image size in pixels
 sizev = 1200;
  int_z = 100*10^-3; % z dist in m
%  xes1 = 100;
 %xes2 = ;
 int_im = Interference(width,height,sizeh, sizev,int_z);
 int_im = int_im/max(max(int_im)); % normalize
 int_im = (int_im-0.5)*2; % enhance visualy

% display the image
figure
imshow(int_im,[])