... | ... | @@ -125,7 +125,7 @@ The light transmitted into metal will have complex wave number: |
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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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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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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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\epsilon_1\vec{E_1^{}}-\epsilon_2\vec{E_2^{}} = \sigma_f = 0
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```
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... | ... | @@ -138,7 +138,7 @@ Inside the metal, from the continuity equation for free charge one can solve the |
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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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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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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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... | ... | @@ -146,7 +146,7 @@ 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 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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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 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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## 4. References and further readings
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