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![reflective_circular](uploads/38015e452e9522ec66daf0d3c200a909/reflective_circular.png)
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Picture (b) in the figure above explains the observation in (a). Consider the electric field component of a monochromatic light. After passing the linear polarizer, in region (1), $`E_{I_1} = \sqrt{I_0} * e^{i(kz-\omega t)}\hat{y} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}(\hat{f}+\hat{s}) `$, where $`\hat{f}`$ and $`\hat{s}`$ are the directions of the fast and slow axis of the quarter-wave plate. After passing the quarter-wave plate the linear polarized light becomes a right hand circular polarized light in region (2). $`E_{I_2}=\frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}(-i\hat{f}+\hat{s})`$. After reflection at the mirror, which uses a layer of metal (typically silver) to reflect light, the electrodynamic boundary condition requires $`E_{I_3} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}[-i(-\hat{f})+(-\hat{s})]`$, which makes the reflected light left hand circular polarized. (See section 4) Passing the quarter-wave plate again introduce another $`\pi/2`$ phase lead to the $`\hat{f}`$ component, resulting in $`E_{I_4} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}[(-i)(-i)(-\hat{f})+(-\hat{s})] = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}\hat{x} `$, which is perpendicular to the linear polarizer. So no reflection light can go through the original polarizer again, creating a dark field.
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Picture (b) in the figure above explains the observation in (a). Consider the electric field component of a monochromatic light. After passing the linear polarizer, in region (1),
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```math
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E_{I_1} = \sqrt{I_0} * e^{i(kz-\omega t)}\hat{y} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}(\hat{f}+\hat{s})
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```
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, where $`\hat{f}`$ and $`\hat{s}`$ are the directions of the fast and slow axis of the quarter-wave plate. After passing the quarter-wave plate the linear polarized light becomes a right hand circular polarized light 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})
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```
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After reflection at the mirror, which uses a layer of metal (typically silver) to reflect light, the electrodynamic boundary condition requires
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```math
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E_{I_3} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}[-i(-\hat{f})+(-\hat{s})]
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```
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, which makes the reflected light left hand circular polarized. (See section 4) Passing the quarter-wave plate again introduce another $`\pi/2`$ phase lead to the $`\hat{f}`$ component, resulting in
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```math
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E_{I_4} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}[(-i)(-i)(-\hat{f})+(-\hat{s})] = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}\hat{x}
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```
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, which is perpendicular to the linear polarizer. So no reflection light can go through the original polarizer again, creating a dark field.
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### 2.2. Photoelasticity under the reflection polariscope
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![reflective_circular2](uploads/28f1e3f8a9384cb6ffc5a21e1d82fdc0/reflective_circular2.png)
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The above figure shows the idea of the photoelastic measurement for a specimen under the reflection polariscope, which requires the light to go through the specimen twice with different kind 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 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) $`E_{I_2} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}(-i\hat{f}+\hat{s}) = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}e^{i\phi}(-i\hat{m_1}+\hat{m_2})`$. 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): $`E_{I_3} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(kz-\omega t)}e^{i\phi}(-ie^{-i\Delta}\hat{m_1}+\hat{m_2})`$. After reflection, the electrodynamic boundary condition requires (see section 4) $`E_{I_4} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}(+ie^{-i\Delta}\hat{m_1}-\hat{m_2})`$ in region (4). Passing the specimen again gives another $`\Delta`$ phase lead to the $`\hat{m_1}`$ light component, which makes $`E_{I_5} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}(+ie^{-2i\Delta}\hat{m_1}-\hat{m_2}) =`$ $` \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}[(ie^{-2i\Delta}cos\phi+sin\phi)\hat{f}+(ie^{-2i\Delta}sin\phi-cos\phi)\hat{s}] `$ in region (5). Passing the quarter-wave plate again gives a $`-i`$ factor to the $`\hat{f}`$ component: $`E_{I_6} = \frac{1}{\sqrt{2}}\sqrt{I_0}e^{i(-kz-\omega t)}e^{i\phi}[-i(ie^{-2i\Delta}cos\phi+sin\phi)\hat{f}+(ie^{-2i\Delta}sin\phi-cos\phi)\hat{s}] `$. Finally, the light intensity that can be observed by the observer behind the linear polarizer is
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The above figure shows the idea of the photoelastic measurement for a specimen under the reflection polariscope, which requires the light to go through the specimen twice with different kind 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 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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```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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```
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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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```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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```
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After reflection, the electrodynamic boundary condition requires (see section 4)
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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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```
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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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```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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```
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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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```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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```
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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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... | ... | @@ -153,7 +194,12 @@ Note the incident and reflection light have reverse direction of rotation for th |
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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]
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[1]: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education
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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).
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[4]: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education
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