[Wien] birefringence

[Wien2k User]

Thank you very much

Le mer. 5 déc. 2018 à 06:36, Gavin Abo <gsabo at crimson.ua.edu> a écrit :

> I believe:
>
> delta_n=n_zz-n_xx
>
> However, you may check it for yourself:
>
> Birefringence := delta_n=n_e-n_o [1,2]
>
> Orthorhombic ɛ_xx and ɛ_yy perpendicular and ɛ_zz parallel [3] a bit
> similar to tetragonal that has ɛ_xx = ɛ_yy and ɛ_zz tensor (Table 1.1) [4]
>
> n_x = sqrt(ɛ_xx/ɛ_o) and n_z = sqrt(ɛ_zz/ɛ_o) [5,6]
>
> Section 1.4 of [4]:
>
> n_e = n_z (or n_zz) and n_o = n_x (or n_xx)
> [1] https://en.wikipedia.org/wiki/Birefringence#Uniaxial_materials
> [2]
> http://shodhganga.inflibnet.ac.in/bitstream/10603/4722/13/13_chapter%203.pdf
> (Equation 3.24)
> [3] https://pubs.acs.org/doi/abs/10.1021/jp410786w
> [4]
> http://www.physics.ucc.ie/fpetersweb/FrankWeb/courses/PY4118/Notes/Symmetry%20of%20Tensor%202.pdf
> [5] https://www.as-photonics.com/book_page/book-preview.pdf (Equations
> 2.13 and 2.15)
> [6]
> https://nanoed.tul.cz/pluginfile.php/1598/mod_resource/content/1/Optick%C3%A9%20vlastnosti%20krystal%C5%AF.pdf
> (slide 5)
>
>
> On 12/4/2018 5:54 PM, Wien2k User wrote:
>
> Dear Wien2k users;
>
> for a tetragonal material, the birefringence is given by :
> delta_n=n_xx-n_zz
> or
> delta_n=n_zz-n_xx
>
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