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![reflective_circular2](uploads/28f1e3f8a9384cb6ffc5a21e1d82fdc0/reflective_circular2.png)
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As shown in the above figure, under reflective polariscope, the circular polarized light goes through the photoelastic specimen twice with different chirality of polarization. Suppose the height (the size along $`z`$ direction) of the specimen is $`h`$, the pattern observed by the observer will be equivalent to the pattern under a transmission polariscope for a specimen under same condition but with $`h/2`$ height. This can be shown as following: suppose the principle direction for the stress tensor of the specimen point under consideration is $`\hat{m_1}`$ and $`\hat{m_2}`$ (corresponding to principle stress $`\sigma_1`$ and $`\sigma_2`$ respectively). Denote $`\phi = \alpha - \pi/4`$. Then in region (2)
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As shown in the above figure, under reflective polariscope, the circular polarized light goes through the photoelastic specimen twice with different chirality of polarization. Suppose the height (the size along $`z`$ direction) of the specimen is $`h`$, the pattern observed here will be the same as the pattern formed by a $`h/2`$ height specimen using transmissive polariscope. This can be shown as following: suppose the principle direction for the stress tensor of the specimen point under consideration is $`\hat{m_1}`$ and $`\hat{m_2}`$ (corresponding to principle stress $`\sigma_1`$ and $`\sigma_2`$ respectively). Denote $`\phi = \alpha - \pi/4`$. Then in region (2)
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```math
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E_{I_2} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}(-i\hat{f}+\hat{s}) = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}e^{i\phi}(-i\hat{m_1}+\hat{m_2})
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E_{2} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}(-i\hat{f}+\hat{s}) = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}e^{i\phi}(-i\hat{m_1}+\hat{m_2})
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```
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Passing the specimen results in a $`\Delta = 2\phi(\sigma_1-\sigma_2)/f_{\sigma}`$ phase lead to the $`\hat{m_1}`$ component of light. So in region (3):
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where $`I_0`$ is a constant as is called the background light intensity, $`k`$ is the wave number and $`\omega`$ is the frequency of the light, $`\hat{f}`$ and $`\hat{s}`$ is the fast and slow principle direction of the quarter-wave plate. Passing the specimen results in a $`\Delta = 2\phi(\sigma_1-\sigma_2)/f_{\sigma}`$ phase lead [5] to the $`\hat{m_1}`$ component of light. So in region (3):
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```math
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E_{I_3} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}e^{i\phi}(-ie^{-i\Delta}\hat{m_1}+\hat{m_2})
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E_{3} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}e^{i\phi}(-ie^{-i\Delta}\hat{m_1}+\hat{m_2})
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```
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After reflection, the electrodynamic boundary condition requires (see section 4)
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After reflection, the electrodynamic boundary condition requires a $`\pi`$ phase shift for both components of the light (see section 2.2.). So the reflected light in region (4) is:
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```math
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E_{I_4} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}(+ie^{-i\Delta}\hat{m_1}-\hat{m_2})
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E_{4} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}(+ie^{-i\Delta}\hat{m_1}-\hat{m_2})
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```
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in region (4). Passing the specimen again gives another $`\Delta`$ phase lead to the $`\hat{m_1}`$ light component, which makes
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Passing the specimen again gives another $`\Delta`$ phase lead to the $`\hat{m_1}`$ light component. Therefore, in region (5):
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```math
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E_{I_5} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}(+ie^{-2i\Delta}\hat{m_1}-\hat{m_2}) = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}[(ie^{-2i\Delta}cos\phi+sin\phi)\hat{f}+(ie^{-2i\Delta}sin\phi-cos\phi)\hat{s}]
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E_{5} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}(+ie^{-2i\Delta}\hat{m_1}-\hat{m_2}) = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}[(ie^{-2i\Delta}cos\phi+sin\phi)\hat{f}+(ie^{-2i\Delta}sin\phi-cos\phi)\hat{s}]
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```
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in region (5). Passing the quarter-wave plate again gives a $`-i`$ factor to the $`\hat{f}`$ component:
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Passing the quarter-wave plate again gives a $`-i`$ factor to the $`\hat{f}`$ component:
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```math
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E_{I_6} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}[-i(ie^{-2i\Delta}cos\phi+sin\phi)\hat{f}+(ie^{-2i\Delta}sin\phi-cos\phi)\hat{s}]
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E_{6} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}[-i(ie^{-2i\Delta}cos\phi+sin\phi)\hat{f}+(ie^{-2i\Delta}sin\phi-cos\phi)\hat{s}]
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```
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. Finally, the light intensity that can be observed by the observer behind the linear polarizer is
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Finally, the light intensity that can be observed by the observer behind the linear polarizer is
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```math
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I = |E_{I_6}\cdot \hat{y}|^2 = I_0sin^2\Delta = I_0sin^2(\frac{2\pi(\sigma_1-\sigma_2)}{f_{\sigma}}h)
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I = |E_{6}\cdot \hat{y}|^2 = I_0sin^2\Delta = I_0sin^2(\frac{2\pi(\sigma_1-\sigma_2)}{f_{\sigma}}h)
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```
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Note for a transmission polariscope the corresponding expression is $`I_0sin^2\frac{\Delta}{2}`$. This different needs to be taken care of when solving the contact forces by nonlinear fitting -- the stress-optic relation used in transmission polariscope can not be used directly to fit the patterns recorded by the reflection polariscope. The factor of 2 must be included.
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Note for a transmission polariscope the corresponding expression is $`I_0sin^2\frac{\Delta}{2}`$. This different needs to be taken care of when solving the contact forces by nonlinear fitting -- the stress-optic relation used in transmission polariscope can not be used directly to fit the patterns recorded by the reflection polariscope.
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### 2.2. Chirality change of circular polarized light by reflection
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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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A mirror is usually implemented by a smooth layer of metal. Inside a metal (or any conductor), the solution of Maxwell equation gives electromagnetic waves with complex wave numbers. Without lost of generality, we consider a monochromatic solution with 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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where the complex wave number $`\tilde{k} = k+i\kappa `$ 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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where $`\sigma`$ is the conductance of the conductor which connects the electric field $`\vec{E}`$ and the free current density: $`\vec{J}`$
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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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Note, when $`\sigma=0`$, $\vec{E}(z,t)$ is the solution of Maxwell in vacuum (and is a good approximation in air), whose form has already been used in section 2.1.. Therefore, the incident light, which is in air, writes:
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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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\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_1^{}}-\vec{B_2^{}} = 0
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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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Similarly, the reflected light, also in air, writes:
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```math
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\vec{E_1^{||}}-\vec{E_2^{||}}=0
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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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\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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\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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The incident light contains
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The light transmitted into metal will have complex wave number:
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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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\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_I}(z,t) = \frac{k_1}{\omega}\tilde{E_I}e^{i(k_1z-\omega t)}\hat{y}
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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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And the reflective light in air is
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Inside the metal, from the continuity equation for free charge one can solve the free charge density $`\rho_f(t) = e^{-t/\tau}\rho_f(0)`$. For a metal that is a good conductor ($`\tau'$ is much larger than any interested time scale), $`\rho_f(t)=0`$ effectively. As a result, the boundary condition at the air-metal surface becomes (recall that quantities has subscript 1 means in air and subscript 2 means in the metal):
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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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\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_R}(z,t) = -\frac{k_1}{\omega}\tilde{E_R}e^{i(-k_1z-\omega t)}\hat{y}
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\vec{B_1^{}}-\vec{B_2^{}} = 0
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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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\vec{E_1^{||}}-\vec{E_2^{||}}=0
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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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\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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Use the above expressions to solve the boundary conditions gives
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In the above expression, $\vec{E_1}=\vec{E_I}+\vec{E_R}$, $\vec{E_2}=\vec{E_T}$, $\vec{B_1}=\vec{B_I}+\vec{B_R}$, $\vec{B_2}=\vec{B_T}$. The boundary condition solves to be:
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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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... | ... | @@ -159,26 +146,19 @@ where $`\tilde{\beta}=\tilde{k_2}\mu_1v_1/\mu_2\omega`$. For an ideal conductor, |
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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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This means a $`\phi`$ phase shift is given by the reflection on an ideal mirror. Note the above calculation is for a linear polarized light. For a circular polarized light, after reflection, both its $`\hat{x}`$ and $`\hat{y}`$ components (can be regarded as linear polarized light themselves) are shifted by a phase factor $`\pi`$. So the relative difference of the components remains unchanged. Therefore, the rotation direction of $`\vec{E_R}`$ in the xy plane remains the same as $`\vec{E_I}`$. However, the direction of the propagation is changed, so the chirality of the light must be reversed. This effect is shown in the section 2.1. figure region (3) and region (4).
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For a circular polarized light, after reflection, both its $`\hat{x}`$ and $`\hat{y}`$ components are shifted by a phase factor $`\pi`$. So the relative difference of the components remains unchanged. Therefore, the rotation direction of the $`\vec{E_R}`$ in the xy plane remains the same as $`\vec{E_I}`$. However, the direction of the propagation is changed, this results the change of chirality.
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## 4. References and further readings
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[1] [Puckett, J.G. and Daniels, K.E., 2013. Equilibrating temperaturelike variables in jammed granular subsystems. Physical Review Letters, 110(5), p.058001.](#https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.110.058001)
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## 4. References
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[1] Puckett, J.G. and Daniels, K.E., 2013. Equilibrating temperaturelike variables in jammed granular subsystems. Physical Review Letters, 110(5), p.058001.
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[2] [Daniels, K.E., Kollmer, J.E. and Puckett, J.G., 2017. Photoelastic force measurements in granular materials. Review of Scientific Instruments, 88(5), p.051808.](#https://aip.scitation.org/doi/10.1063/1.4983049)
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[2] Daniels, K.E., Kollmer, J.E. and Puckett, J.G., 2017. Photoelastic force measurements in granular materials. Review of Scientific Instruments, 88(5), p.051808.
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[3] [Y. Zhao, J. Barés, H. Zheng, and R. P. Behringer, EPJ
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Web of Conference 140, 03049 (2017).](#https://www.epj-conferences.org/articles/epjconf/abs/2017/09/epjconf162348/epjconf162348.html)
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[3] Y. Zhao, J. Barés, H. Zheng, and R. P. Behringer, EPJ
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Web of Conference 140, 03049 (2017).
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[4] Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education
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[4]: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education
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[5] [J.W. Phillips (Ed.), TAM 326—experimental stress analysis, University of Illinois, Champaign, Ill (2000), pp. 1-62](#http://www.ifsc.usp.br/~lavfis/images/BDApostilas/ApEfFotoelastico/photoelasticity.pdf)
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[<go back to home](https://git-xen.lmgc.univ-montp2.fr/PhotoElasticity/Main/wikis/home) |
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