... | ... | @@ -60,97 +60,97 @@ In this section, the theoretical background of the reflective polariscope is rev |
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![reflective_circular2](uploads/28f1e3f8a9384cb6ffc5a21e1d82fdc0/reflective_circular2.png)
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As shown in the above figure, under a reflective polariscope, the circularly 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 a transmissive polariscope. This can be shown as follows: suppose the linear polarizer creates a polarized light along $\hat{y}$ direction. In region (1) denote the light vector as:
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```math
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As shown in the above figure, under a reflective polariscope, the circularly 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 linear polarizer creates a polarized light along $\hat{y}$ direction. In region (1) denote the light vector as:
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$$
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E_1 = \sqrt{I_0}e^{i(kz-\omega t)}\hat{y}
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```
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where $`I_0=|E_1|^2`$ is a constant called the background light intensity, $`k`$ is the wave number and $`\omega`$ is the frequency of the light. Suppose the principle direction for the stress tensor of the specimen point under consideration is $`\hat{m_1}`$ and $`\hat{m_2}`$ (corresponding to principle stress $`\sigma_1`$ and $`\sigma_2`$ respectively). Denote $`\phi = \alpha - \pi/4`$. Then, in the region (2)
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```math
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$$
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where \(I_0=|E_1|^2\) is a constant called the background light intensity, \(k\) is the wave number and \(\omega\) is the frequency of the light. 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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$$
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E_{2} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}(-i\hat{f}+\hat{s}) = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}e^{i\phi}(-i\hat{m_1}+\hat{m_2})
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```
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where $`\hat{f}`$ and $`\hat{s}`$ are the fast and slow principle directions of the quarter-wave plate. Passing the specimen results in a $`\Delta = 2\pi(\sigma_1-\sigma_2)/f_{\sigma}`$ phase lead [5] to the $`\hat{m_1}`$ component of light. So, in the region (3):
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```math
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$$
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where \(\hat{f}\) and \(\hat{s}\) are the fast and slow principle directions of the quarter-wave plate. Passing the specimen results in a \(\Delta = 2\pi(\sigma_1-\sigma_2)/f_{\sigma}\) phase lead [5] to the \(\hat{m_1}\) component of light. So in region (3):
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$$
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E_{3} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}e^{i\phi}(-ie^{-i\Delta}\hat{m_1}+\hat{m_2})
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```
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After reflection, the electrodynamic boundary condition requires a $`\pi`$ phase shift for both components of the light (see section 2.2.). So, the reflected light in the region (4) is:
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```math
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$$
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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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$$
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E_{4} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}(+ie^{-i\Delta}\hat{m_1}-\hat{m_2})
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```
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Passing the specimen again gives another $`\Delta`$ phase leading to the $`\hat{m_1}`$ light component. Therefore, in the region (5):
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```math
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$$
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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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$$
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E_{5} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}(+ie^{-2i\Delta}\hat{m_1}-\hat{m_2})
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```
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```math
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$$
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$$
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= \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}[(ie^{-2i\Delta}cos\phi+sin\phi)\hat{f}+(ie^{-2i\Delta}sin\phi-cos\phi)\hat{s}]
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```
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Passing the quarter-wave plate again gives a $`-i`$ factor to the $`\hat{f}`$ component:
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```math
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$$
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Passing the quarter-wave plate again gives a \(-i\) factor to the \(\hat{f}\) component:
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$$
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E_{6} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}[-i(ie^{-2i\Delta}cos\phi+sin\phi)\hat{f}+(ie^{-2i\Delta}sin\phi-cos\phi)\hat{s}]
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```
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$$
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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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$$
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I = |E_{6}\cdot \hat{y}|^2 = I_0sin^2\Delta = I_0sin^2(\frac{2\pi(\sigma_1-\sigma_2)}{f_{\sigma}}h)
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```
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Note for a transmission polariscope the corresponding expression is $`I_0sin^2\frac{\Delta}{2}`$. This difference needs to be taken care of when solving the contact forces by nonlinear fitting -- the stress-optic relation used in the transmission polariscope can not be used directly to fit the patterns recorded by the reflection polariscope. Also note when $`\Delta=0`$, i.e., for a stress-free specimen or no specimen, $`I=0`$. This is why metal looks black under a circular polarizer with a wave plate side towards the metal.
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$$
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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. Also note when \(\Delta=0\), i.e., for a stress-free specimen or no specimen, \(I=0\). This is why metal looks black under a circular polarizer with wave plate side towards the metal.
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### 2.2. Chirality change of circularly polarized light by reflection
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### 2.2. Chirality change of circular polarized light by reflection
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A mirror is usually implemented by a smooth layer of metal. Inside a metal (or any conductor), the solution of the Maxwell equation gives electromagnetic waves with complex wave numbers. Without loss of generality, we consider a monochromatic solution with frequency $`\omega`$.
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```math
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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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$$
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\vec{E}(z,t) = \tilde{E}e^{i(\tilde{k}z-\omega t)}\hat{x}
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```
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where the complex wave number $`\tilde{k} = k+i\kappa `$ and
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```math
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$$
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where the complex wave number \(\tilde{k} = k+i\kappa \) and
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$$
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k=\omega\sqrt{\frac{\epsilon \mu}{2}}[\sqrt{1+(\frac{\sigma}{\epsilon \omega})^2}+1]^{1/2};~
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\kappa =\omega\sqrt{\frac{\epsilon \mu}{2}}[\sqrt{1+(\frac{\sigma}{\epsilon \omega})^2}-1]^{1/2}
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```
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where $`\epsilon`$ is the electric permittivity, $`\mu`$ is the magnetic permeability, $`\sigma`$ is the conductance of the conductor which connects the electric field $`\vec{E}`$ and the free current density: $`\vec{J}`$
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```math
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$$
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where \(\epsilon\) is the electric permittivity, \(\mu\) is the magnetic permeability, \(\sigma\) is the conductance of the conductor which connects the electric field \(\vec{E}\) and the free current density: \(\vec{J}\)
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$$
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\vec{J}_f=\sigma \vec{E}
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```
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Note, when $`\sigma=0`$, $`\vec{E}(z,t)`$ is the solution of Maxwell equations in vacuum (and is a good approximation in air), whose form has already been used in section 2.1.. Using subscript 1 to denote quantities in air and subscript 2 to denote quantities in metal, the incident light, which is in air, writes:
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```math
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$$
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Note, when \(\sigma=0\), \(\vec{E}(z,t)\) is the solution of Maxwell equations in vacuum (and is a good approximation in air), whose form has already been used in section 2.1.. Using subscript 1 to denote quantities in air and subscript 2 to denote quantities in metal, the incident light, which is in air, writes:
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$$
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\vec{E_I}(z,t) = \tilde{E_I}e^{i(k_1z-\omega t)}\hat{x}
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```
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```math
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$$
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$$
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\vec{B_I}(z,t) = \frac{k_1}{\omega}\tilde{E_I}e^{i(k_1z-\omega t)}\hat{y}
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```
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$$
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Similarly, the reflected light, also in the air, writes:
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```math
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$$
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\vec{E_R}(z,t) = \tilde{E_R}e^{i(-k_1z-\omega t)}\hat{x}
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```
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```math
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$$
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$$
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\vec{B_R}(z,t) = -\frac{k_1}{\omega}\tilde{E_R}e^{i(-k_1z-\omega t)}\hat{y}
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```
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$$
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The light transmitted into metal will have complex wave numbers:
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```math
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$$
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\vec{E_T}(z,t) = \tilde{E_T}e^{i(\tilde{k_2}z-\omega t)}\hat{x}
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```
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```math
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$$
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$$
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\vec{B_T}(z,t) = \frac{\tilde{k_2}}{\omega}\tilde{E_T}e^{i(\tilde{k_2}z-\omega t)}\hat{y}
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```
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Inside the metal, from the continuity equation for free charge one can solve the free charge density $`\rho_f(t) = e^{-t/\tau}\rho_f(0)`$. For a metal that is a good conductor ($`\tau`$ is much larger than any interested time scale), $`\rho_f(t)=0`$ effectively. As a result, the boundary condition at the air-metal surface becomes:
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```math
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$$
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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:
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$$
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\epsilon_1\vec{E_1^{\bot}}-\epsilon_2\vec{E_2^{\bot}} = \sigma_f = 0
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```
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```math
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$$
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$$
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\vec{B_1^{\bot}}-\vec{B_2^{\bot}} = 0
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```
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```math
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$$
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$$
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\vec{E_1^{\parallel}}-\vec{E_2^{\parallel}}=0
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```
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```math
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$$
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$$
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\frac{1}{\mu_1}\vec{B_1^{\parallel}}-\frac{1}{\mu_2}\vec{B_2^{\parallel}}=\vec{K_f}\times \hat{n} = 0
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```
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In the above equations, $`\vec{K_f}`$ is the surface current of free charges, $`\vec{E_1}=\vec{E_I}+\vec{E_R}`$, $`\vec{E_2}=\vec{E_T}`$, $`\vec{B_1}=\vec{B_I}+\vec{B_R}`$, $`\vec{B_2}=\vec{B_T}`$. Solving the boundary conditions gives:
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```math
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$$
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In the above equations, \(\vec{K_f}\) is the surface current of free charges, \(\vec{E_1}=\vec{E_I}+\vec{E_R}\), \(\vec{E_2}=\vec{E_T}\), \(\vec{B_1}=\vec{B_I}+\vec{B_R}\), \(\vec{B_2}=\vec{B_T}\). Solving the boundary conditions gives:
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$$
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\tilde{E_R}=(\frac{1-\tilde{\beta}}{1+\tilde{\beta}})\tilde{E_I},\tilde{E_T}=(\frac{2}{1+\tilde{\beta}})\tilde{E_I}
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```
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where $`\tilde{\beta}=\tilde{k_2}\mu_1/k_1\mu_2`$. For an ideal conductor, $`\sigma=\infty`$, which results in $`|\tilde{k_2}|=\infty`$, and therefore $`|\tilde{\beta}|=\infty`$. So
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```math
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$$
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where \(\tilde{\beta}=\tilde{k_2}\mu_1/k_1\mu_2\). For an ideal conductor, \(\sigma=\infty\), which results in \(|\tilde{k_2}|=\infty\), and therefore \(|\tilde{\beta}|=\infty\). So
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$$
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\tilde{E_R}=-\tilde{E_I}, \tilde{E_T}=0
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```
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This means a $`\pi`$ phase shift is given by the reflection on an ideal mirror. Note the above calculation is for a linear polarized light. For a circularly polarized light, after reflection, both its $`\hat{x}`$ and $`\hat{y}`$ components (can be regarded as linear polarized light themselves) are shifted by a phase factor $`\pi`$. So the relative phase difference between the components of light remains unchanged. Therefore, the rotation direction of $`\vec{E_R}`$ in the xy plane remains the same as $`\vec{E_I}`$. However, the direction of the propagation is changed, so the chirality of the light must be reversed. This effect is shown in section 2.1. figure region (3) and region (4).
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$$
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This means a \(\pi\) 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 phase difference between the components of light 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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## 4. References and further readings
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... | ... | @@ -165,4 +165,4 @@ Web of Conference 140, 03049 (2017).](#https://www.epj-conferences.org/articles/ |
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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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