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Reflection photoelasticity method
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## 1.Overview
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## 2.Theory [1]
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### 2.1. Reflection of polarized light on a insulator-conductor surface
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In a conductor, the solution of Maxwell equation gives electromagnetic waves with complex wave numbers. Without lost of generality we consider a specific solution correspond to frequency $`\omega`$.
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```math
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\vec{E}(z,t) = \tilde{E}e^{i(\tilde{k}z-\omega t)}\hat{x}
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```
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where $`\tilde{k} = k+i\kappa `$ and
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```math
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k=\omega\sqrt{\frac{\epsilon \mu}{2}}[\sqrt{1+(\frac{\sigma}{\epsilon \omega})^2}+1]^{1/2};~
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\kappa =\omega\sqrt{\frac{\epsilon \mu}{2}}[\sqrt{1+(\frac{\sigma}{\epsilon \omega})^2}-1]^{1/2}
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```
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Note when $`\sigma = 0 `$ the wave number becomes real.
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where $`\sigma`$ is the conductance of the conductor which connects the electric field and the free current density
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```math
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\vec{J}_f=\sigma \vec{E}
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```
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From the continuity equation for free charge one can solve the free charge density $`\rho_f(t) = e^{-(\sigma/\epsilon)t}\rho_f(0)`$, which means for a good conductor ($`\tau\equiv \epsilon/\sigma`$ is much larger than any interested time scale), $`\rho_f(t)=0`$ effectively, which leads to 4 boundary conditions at the interfaces:
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```math
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\epsilon_1\vec{E_1^{}}-\epsilon_2\vec{E_2^{}} = \sigma_f = 0
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```
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```math
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\vec{B_1^{}}-\vec{B_2^{}} = 0
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```
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```math
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\vec{E_1^{||}}-\vec{E_2^{||}}=0
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```
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```math
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\frac{1}{\mu_1}\vec{B_1}^{||}-\frac{1}{\mu_2}\vec{B_2}^{||}=\vec{K_f}\times \hat{n} = 0
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```
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The incident light contains
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```math
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\vec{E_I}(z,t) = \tilde{E_I}e^{i(k_1z-\omega t)}\hat{x}
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```
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```math
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\vec{B_I}(z,t) = \frac{k_1}{\omega}\tilde{E_I}e^{i(k_1z-\omega t)}\hat{y}
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```
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And the reflective light in air is
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```math
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\vec{E_R}(z,t) = \tilde{E_R}e^{i(-k_1z-\omega t)}\hat{x}
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```
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```math
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\vec{B_R}(z,t) = -\frac{k_1}{\omega}\tilde{E_R}e^{i(-k_1z-\omega t)}\hat{y}
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```
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Also, the transmitted light into the conductor is
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```math
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\vec{E_T}(z,t) = \tilde{E_T}e^{i(\tilde{k_2}z-\omega t)}\hat{x}
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```
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```math
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\vec{B_T}(z,t) = \frac{\tilde{k_2}}{\omega}\tilde{E_T}e^{i(\tilde{k_2}z-\omega t)}\hat{y}
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```
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Use the above expressions to solve the boundary conditions gives
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```math
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\tilde{E_R}=(\frac{1-\tilde{\beta}}{1+\tilde{\beta}})\tilde{E_I},\tilde{E_T}=(\frac{2}{1+\tilde{\beta}})\tilde{E_I}
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```
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where $`\tilde{\beta}=\tilde{k_2}\mu_1v_1/\mu_2\omega`$. For an ideal conductor, $`\sigma=\infty`$, which results in $`|\tilde{k_2}|=\infty`$, and therefore $`|\tilde{\beta}|=\infty`$. So
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```math
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\tilde{E_R}=-\tilde{E_I}, \tilde{E_T}=0
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```
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This gives the form of the reflection light:
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```math
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\vec{E_R}(z,t) = \tilde{E_I}e^{i(-k_1z-\omega t+\pi)}\hat{x}
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```
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```math
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\vec{B_R}(z,t) = -\frac{k_1}{\omega}\tilde{E_I}e^{i(-k_1z-\omega t+\pi)}\hat{y}
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```
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which defers from the incident light by a phase factor $\pi$ besides the change of the propagation direction.
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### circular polarizer
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The incident light with circular polarization writes:
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```math
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\vec{E_I}(z,t) = \tilde{E_I}(e^{i(k_1z-\omega t)}\hat{x}+e^{i(k_1z-\omega t+\pi/2)}\hat{y})
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```
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```math
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\vec{B_I}(z,t) = -\frac{k_1}{\omega}\tilde{E_I}(e^{i(k_1z-\omega t)}\hat{y}+e^{i(k_1z-\omega t+\pi/2)}\hat{x})
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```
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Following same calculations as in the linear polarization case for components of the circular polarized light, we get the polarization for the reflective light:
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```math
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\vec{E_R}(z,t) = \tilde{E_I}(e^{i(-k_1z-\omega t+\pi)}\hat{x}+e^{i(-k_1z-\omega t+3\pi/2)}\hat{y})
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```
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```math
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\vec{B_R}(z,t) = -\frac{k_1}{\omega}\tilde{E_I}(e^{i(-k_1z-\omega t+\pi)}\hat{y}+e^{i(-k_1z-\omega t+3\pi/2)}\hat{x})
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```
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Note the incident and reflection light have reverse direction of rotation for their $`\vec{E}`$ vector and $`\vec{B}`$ vector. So we have proved that on a conductor the direction of circular polarization is switched after reflection on a conductor.
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### 2.2. Photoelasticity under reflective light
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Note: how to think of the reflected light go through the particle has different polarization with the incoming light? How does this change the photo-elastic response?
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The whole problem has reflection symmetry so the final result would not be fundamentally different. proof?
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## 3. Implementation of the reflective photoelasticimetry
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There are two ways to implement the reflective photoelasticimetry. One is by using particles with reflective bases. The other is by putting particles on mirror. Using particles with reflective bases can be achieved by paiting particles using.... The advantage is that it can be used on air tables ...
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### 3.1. Paint on particles
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Duke reflective setup: base as mirror, particle are mirrored
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### 3.2. Mirror base
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mirror base has the advatange that no additional process needs to be done on particles. So the transparent photo-elastic particles used in other experiments can be used on this setup. No more purchase/cast of particles particular for this geometry is needed. Also, the tilting of the particles will not be a problem. The disadvantage is that the reflection of the particle height makes the detection of the particle boundaries really hard. (fig) Also for the base-driven uniform shear system, the split between strips or rings will leave a dark line inside the photo-elastic response pattern, introducing errors in the stress estimations.
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Mirror also provides a higher reflection ratio R than painted particles (check?)
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### 3.3. Example implementations
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## 4. Common issues with the reflective photoelascimetry
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### 4.1. Non-uniform light distribution
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#### 4.1.1. Sensitivity to tilting
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When using the reflective photoelasticity. Small titling of the particles may cause a non-negligible change of the light intensity reflected by that particle. (fig) Since all the stress solver (either the qualitative $G^2$ method or the quantitative inverse solver method) depends on the intensity value, this issue will introduce errors for the stress estimations. However, this issue can be resolved by renormalize the light intensity by divide the polarized image to a pure light image. Fig. show a calibration result for particle with different tilting angle (light intensity).
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#### 4.1.2. Light source
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### 4.2. Signal to noise ratio: non perfect reflection: comparing to other kinds of imagings.
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Current technology using mirror effect powders to paint the base of the particle to create the reflection of light. However, the reflection ratio is much smaller than a real mirror. So the signal to noise ratio may be smaller than the experiments using the transparent particles. However, the
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## 5. References
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[1]: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education
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[<go back to home](https://git-xen.lmgc.univ-montp2.fr/PhotoElasticity/Main/wikis/home) |
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