Virtual lab basic experiment tutorial -8. Fourier transform (1)

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Preface

This is the tenth collective work of Dachuang team , It is a new chapter for another branch of physical optics – Various phenomena of Fourier optics that can reflect this spectral characteristic have been explored and tried .
（ There are still some problems in the holographic part of the current test , Welcome to try to discuss ）

One 、 What is Fourier transform

1、 Formula transformation

Time domain Fourier transform
Its Fourier transform is $F(j\omega )=\int f(t)e^{-j\omega t}dt$
The inverse Fourier transform is ：$f(t)=\int F(j\omega )e^{j\omega t}dw$
From the inverse Fourier transform ： Time domain signals can have different amplitudes , Linear superposition of sinusoidal functions with different frequency components .

Spatial Fourier transform
$$F(f_x,f_y)=\int \int f(x,y)e^{-2\pi j(f_x+f_y)}dxdy$$
Space frequency
$$f_x=\frac{cos\alpha }{\lambda }{f}_y=\frac{cos\beta }{\lambda }$$
When the incident wavelength is fixed , Spatial frequency is a quantity related to angle , That is, the spatial frequency is related to the incident direction of light , Different directions represent different frequency components . And Fourier transform , Will be $f(x,y)$ To the frequency domain , Observe its spatial frequency components . Through the spectrum diagram , We can observe each frequency component of the original function and its corresponding amplitude .
Inverse Fourier transform ：$$f(x,y)=\int \int F(x,y)e^{2\pi j(f_x+f_y)}d{f}_{x}d{f}_y$$
It can be seen from the inverse Fourier transform , Object function f(x,y) It can be seen as different amplitudes , Linear superposition of plane waves with different directions .

2、 Angular spectrum function （ spectrum ）

The spatial spectrum is called angular spectrum .
$$F(f_x,f_y)=\int \int f(x,y)e^{-2\pi j(f_x+f_y)}dxdy$$

３、 Fourier transform properties of lens

Examine the light field distribution on the rear focal plane .
$$g(x,y)=\frac{1}{i\lambda f}{e}^{i\frac{k}{2f}(1-\frac{d_1}{f})(x^2+y^2)}F(f_x,f_y)$$
here $f_x=\frac{x}{f\lambda },f_y=\frac{y}{f\lambda }$
The Fourier spectrum of the object function is obtained on the back focal plane of the lens . But the phase factor in front of the Fourier transform , This leads to the bending of the light field on the back focal plane .
Discuss several special situations ：
The object is located in the front focal plane of the lens ($d_1=f$)
$$g(x,y)=\frac{1}{i\lambda f}F(f_x,f_y)$$
The object is close to the front surface of the lens ($d_1=\mathrm{０}$)
$$g(x,y)=\frac{1}{i\lambda f}{e}^{i\frac{k}{2f}(x^2+y^2)}F(f_x,f_y)$$
The object is placed behind the lens
$$g(x,y)=\frac{fA}{\lambda d^2}{e}^{i\frac{k}{2d}(x^2+y^2)}$$

４、 Holography

To record the phase , Optical holography uses the principle of interference , Record the amplitude and phase information in the form of interference fringes , Then the diffraction effect is used to reproduce the three-dimensional image containing all the information of the object .
1. Record
$$\{\begin{array}{l}O(r,t)=A_o(r)e^{i({\psi }_O-\omega t)}\\ R(r,t)=A_R(r)e^{i({\psi }_R-\omega t)}\end{array}$$
Then the reference light and the object light interfere with each other , Its light intensity is
$$I(r,t)=｜O{｜}^2+｜R{｜}^2+2｜O｜｜R｜cos({\psi }_O-{\psi }_R)$$
The third term represents the interference effect between two waves , The fringe shape is determined by the phase difference , therefore , Density of interference fringes , Orientation and strength , Contrast reflects the phase distribution .
2. repeat
Let the amplitude transmittance of the holographic negative be proportional to the light intensity
$$\tau (x,y)={\tau }_0+\beta I(x,y)$$
When the original reference light is used for illumination reproduction , Its transmitted light field
$$A_{r}ec(r,t)=R(r,t)\tau (r)=R[{\tau }_0+\beta (|O{|}^2+|R{|}^2)]+\beta R{R}^{\ast }O+\beta R^2{O}^{\ast }$$
The first term in the formula is the same transmitted light field as the reproduced light wave , The second item is the virtual image +1 level , The third term is conjugate image -1 level .

Two 、VirtualLab Simulation

1. Fourier basis

We study Fourier transform .
First , Set up the light path diagram as shown in the figure , The plane wave reaches the lens through an aperture , The lens focal length is set to 100mm, Put the aperture in front of the lens 100mm It's about , The detector is placed behind the aperture and the lens 100mm That is, observe the distribution of light at the rear focal plane .


Raw Data Detector Digital detectors can directly detect the amplitude and phase of light （ choose camera detector Can also be ）,Angular Spectrum Visualizer The angular spectrum detector performs Fourier transform on light .

The final light path can also be built without identity operator, Set the detector to the rear focus of the lens .

Use Classic Field Tracing After operation , You can get such a simulation image .


The previous figure shows the complex amplitude distribution after passing through the aperture , The latter figure shows the complex amplitude distribution on the back focal plane after Fourier transform through the lens .

Click on the left image , Click on Manipulations The Fourier transform button under the menu , Perform manual Fourier transform , The following experimental results can be obtained .

You can see , The effect is consistent .
The following figure shows the simulation results of the angular spectrum detector , The angular spectrum detector itself is the function of Fourier transform , You can see , The effect is also consistent with the image after passing through the lens .
Next , We will just light the path of Identity Operator Delete the device , Connect the digital detector directly , Run .

You can see , Because the object function at the focal point behind the lens is equivalent to a Fourier transform , The complex amplitude obtained by the detector is still consistent with the Fourier change .

2. Holographic recording

The distance of the whole optical path is designed as follows .

First , Create a new empty file , Add a Gaussian wave to it .

Double click to set the wavelength of Gaussian wave 532nm And the girdle radius is 200um X 200um.

After the wave source is set , add to Ideal Components Ideal beam splitter in , And set its position .

Double click to set its position as Z=25mm.
Then add the ideal lens , The setting positions are 10mm and 45mm. The beam expanding system is composed of two lenses .

The focal length of the first lens is set to 5mm.

The focal length of the second lens is set to 50mm.

Add an ideal flat mirror , The position is 50mm.

The angle is set to around y Shaft rotation -30°.

Place sine grating , The position is set to 10mm.

Set its cycle to 45um.

Set the two aperture, The positions are 31.603mm and 90mm. Use these two apertures to simulate holographic dry plate .

The aperture above is set to 5mm×2mm.

The aperture below is set to 6mm×2mm.

Put two more Camera Dectector.

Connect the optical path after all components are placed , Note the beam splitter and fixed mirror The light path between is a red line , And mirror Between the blue line （ The reflected light in the software is represented by a red line , The transmitted light is represented by a blue line ）
We can go through 3D View to observe the construction of components , stay optical setup Click on the 3D identification .

You can see the three-dimensional position view of components .

adopt Ray tracing system analyzer You can see the light path diagram .


and classical field tracing Then you can see the pattern .

３. Holographic reconstruction

First, according to the Gaussian light in holographic recording and the same setting as the two lenses .

Then drag stored function

Double click to enter, and then click set Import the data recorded on the holographic dry plate in the holographic recording process （ Convert the results detected by the detector , stay manipulations Into the form of transmittance function ）.

After that, add Aperture, And double click to set its shape and size , And modify Relative Edge Width Parameter is 20％

Its position is set to x The shaft -1.22mm.

（ Same as in holographic recording , Because the level selector is a programmable element , Here we also omit it , Of course, this may cause some problems ）

Then place identity operator, And set its x The axial position is 1.22m, And around y Shaft rotation -30 degree .

Then place the detector , Just change the interpolation method to Accelerated Sinc, Other parameters can be defaulted .

final -1 The light path of level is built as shown in the figure , Be careful Aperture and identity operator stay x The positions in the axial direction are always opposite to each other .

about 0 Level and +1 Level light path construction only requires Aperture and identity operator stay x Change the position in the axis direction .


（ To be continued ）

summary

This article is written by members of Da Chuang team ： Tang Yiheng 、 Help Yang Yu 、 Huang Yinuo 、 Li Sitong 、 Ming Yue jointly completed .
This is the first exploration of Fourier transform . We do encounter various problems , I hope you can discuss and see how to solve it after you are interested in trying .