Question 7.3: Any incident wave on the interface of two dielectrics may be...
Any incident wave on the interface of two dielectrics may be written in the form \boldsymbol{\textbf{E}} =\underline{E}_{//} \boldsymbol{{e}}_{//}+\underline{E} _\bot\boldsymbol{e}_\bot where \boldsymbol{e}_{//} and \boldsymbol{e}_⊥ are unit vectors that are perpendicular to the direction of propagation. \boldsymbol{e}_{//} is in the plane of incidence and \boldsymbol{e}_⊥ is perpendicular to this plane. a) Write in this form the expression of a plane wave that is incident at an angle θ and polarized linearly, polarized circularly and nonpolarized. b) Consider the case of a non-polarized incident wave. Calculate the energy reflection and transmission factors. It will be convenient to set n = n_2/n_1 . Determine the degree of polarization perpendicularly to the incidence plane for the reflected wave and for the transmitted wave. What are their values in the case of the incidence at the Brewster angle and at θ = 30° on the surface of the water.
a) The wave may be written in the form \underline{\boldsymbol{E}} =\underline{E}_{//} \boldsymbol{e}_{//}+\underline{E} _\bot \boldsymbol{e}_\bot where the components are complex, of the form \underline{E}_{//}=A_{//} e^{i(ωt-\boldsymbol{k.r})} and \underline{E} _\bot=A_\bot e^{i(ωt-\boldsymbol{k.r}+\phi )} . In the case of a linearly polarized wave, \phi =0 or π. In the case of a circularly polarized wave, we have A_{//}=A_\bot and \phi =\pm \pi /2 . In the case of a non-polarized wave, we may assume that A_{//}=A_\bot and \phi vary at random. The wave vector k lies in the plane of incidence and it makes with the normal to the interface an angle equal to the angle of incidence θ.
b) In the case of a non-polarized wave that is incident at an angle θ, the amplitudes of the incident, reflected and transmitted waves may be written as
\underline{\boldsymbol{E} }=\underline{E} _{//}\boldsymbol{e}_{//}+\underline{E} _\bot \boldsymbol{e}_\bot , \ \ \ \ \ \boldsymbol{E}^\prime=R_{//}\underline{E} _{//}\boldsymbol{e}_{//}+ R_\bot \underline{E} _\bot \boldsymbol{e}_\bot , \ \ \ \ \ \boldsymbol{ E}^{\prime \prime}=7_{//}\underline{E} _{//}\boldsymbol{e}_{//}+7_\bot \underline{E} _\bot \boldsymbol{e}_\botwith \underline{E} _{//}=Ae^{i(ωt-\boldsymbol{k.r})} \ \text {and} \ \underline{E} _\bot =Ae^{i(ωt-\boldsymbol{k.r}+\phi )} . The powers that are received, reflected and transmitted by the unit area of the interface are respectively
\lt S_Z\gt =7 \cos\theta =(1/\mu _ov_1)\lt (\boldsymbol{ReE})^2\gt \cos\theta =(n_1/\mu _oc)E_m^2 \cos \theta , \\ \lt S^\prime_Z\gt =7 ^\prime \cos\theta^\prime =(1/\mu _ov_1)\lt (\boldsymbol{ReE^\prime})^2\gt \cos\theta^\prime =(n_1/2 \mu _oc)E_m^2 (\boldsymbol{R}_{//}^2+\boldsymbol{R}_\bot ^2 ) \cos \theta , \\ \lt S^{\prime \prime}_Z\gt =7 ^{\prime \prime} \cos\theta ^{\prime \prime}=(1/\mu _ov_2)\lt (\boldsymbol{ReE^{\prime \prime}})^2\gt \cos\theta ^{\prime \prime}=(n_2/2\mu _oc)E_m^2(\boldsymbol{7}_{//}^2+\boldsymbol{7}_\bot ^2) \cos \theta ^{\prime \prime}.Thus, the energy reflection and transmission factors are
f_R=\frac{7 ^\prime \cos\theta ^\prime}{7 \cos \theta } =\frac{1}{2} (\boldsymbol{R}_{//}^2+\boldsymbol{R}_\bot ^2)= \tan^2(\theta -\theta ^{\prime \prime})\frac{1+ \tan^2\theta \tan^2\theta ^{\prime \prime} }{(\tan\theta +\tan\theta ^{\prime \prime})^2} , \\ f_T=\frac{7 ^{\prime \prime}\cos \theta ^{\prime \prime} }{7 \cos \theta } =\frac{n_2 \cos\theta ^{\prime \prime}}{2n_1 \cos\theta } (\boldsymbol{7}_{//}^2+\boldsymbol{7}_\bot ^2)=\frac{2 \tan\theta \tan\theta ^{\prime \prime}}{(\tan\theta +\tan\theta ^{\prime \prime})^2} \left[2+\tan^2(\theta -\theta ^{\prime \prime})\right] .They verify the conservation of energy equation f_T+f_R=1 .
Using equations [7.36] and [7.39], we obtain the degree of polarization perpendicularly to the plane of incidence for the reflected and transmitted waves:
\boldsymbol{R}_{//}=\frac{\underline{E}^\prime_m }{\underline{E}_m } =\frac{\underline{B}^\prime _m}{\underline{B}_m } =\frac{n_1 \cos\theta ^{\prime \prime}-n_2 \cos \theta }{n_1 \cos\theta ^{\prime \prime}+n_2 \cos\theta } =\frac{\tan(\theta ^{\prime \prime}-\theta )}{\tan(\theta +\theta ^{\prime \prime})} , \\ \boldsymbol{7}_{//}=\frac{\underline{E}^{\prime \prime}_m }{\underline{E}_m } =\frac{v_2}{v_1} \frac{\underline{B}^{\prime \prime} _m}{\underline{B}_m }=\frac{2n_1 \cos\theta }{n_1 \cos\theta ^{\prime \prime}+n_2 \cos\theta} =\frac{2 \cos\theta \sin\theta ^{\prime \prime}}{\sin(\theta +\theta ^{\prime \prime}) \cos(\theta -\theta ^{\prime \prime})}, [7.36]
\boldsymbol{ R}_\bot =\frac{\underline{E}^\prime_m }{\underline{E}_m } =\frac{\underline{B}^\prime _m}{\underline{B}_m } =\frac{\tan\theta ^{\prime \prime}-\tan\theta }{\tan \theta ^{\prime \prime}+\tan\theta }=\frac{n_1 \cos\theta – n_2 \cos\theta ^{\prime \prime}}{n_1 \cos\theta +n_2 \cos\theta ^{\prime \prime }}=\frac{\sin(\theta ^{\prime \prime}-\theta )}{\sin(\theta ^{\prime \prime}+\theta )} , \\ \boldsymbol{7}_\bot=\frac{\underline{E}^{\prime \prime}_m }{\underline{E}_m } =\frac{v_2}{v_1} \frac{\underline{B}^{\prime \prime} _m}{\underline{B}_m }=\frac{2n_1 \cos\theta }{n_1 \cos\theta +n_2 \cos\theta^{\prime \prime}} =\frac{2 \cos\theta \sin\theta ^{\prime \prime}} {\sin(\theta ^{\prime \prime }+\theta ) } \ . [7.39]
\boldsymbol{ p ^\prime=\frac{\bf R_\bot ^2-\bf R_{//}^2}{\bf R_\bot ^2+\bf R_{//}^2}} =\frac{2n(1-n^2) \cos\theta ^{\prime \prime} \cos\theta (\cos^2\theta -\cos^2\theta ^{\prime \prime})}{(1-n^2)^2 \cos^2\theta ^{\prime \prime} \cos^2\theta +n^2(\cos^2\theta ^{\prime \prime }-\cos^2\theta )^2} ,\boldsymbol{ p ^{\prime \prime}=\frac{7_\bot ^2-7_{//}^2}{7_\bot ^2+ 7_{//}^2} }=\frac{(n^2-1) (\cos^2\theta -\cos^2\theta ^{\prime \prime})}{(n^2+1)(\cos^2\theta +\cos^2\theta ^{\prime \prime})+4n \cos\theta \cos\theta ^{\prime \prime}} .
In the case of the air-water interface (n_1=1 \ \ \text {and} \ \ n_2=1.33), the Brewster angle is such that \tan \theta _B=n . In the case of Brewster incidence we obtain
\boldsymbol{p}^\prime =1 \ \ \ \text {and} \ \ \ \boldsymbol{p}^{\prime \prime}=\frac{(n^2-1) \cos(2\theta _B)}{n^2+1+2n \sin(2\theta _B)} =-\frac{(n^2-1)^2}{n^4+1+6n^2} =-0.04 .In the case of a non-polarized wave and an incidence angle θ = 30°, we find \theta ^{\prime \prime}22^\circ .08 ; then, \boldsymbol{p}^\prime=0.44 \text{ and } \boldsymbol{p}^{\prime \prime}=-0.0097 . Thus, at the Brewster incidence, the reflected wave is totally polarized perpendicularly to the plane of incidence, while the transmitted wave is slightly polarized in the plane of incidence . At an angle of incidence of 30°, the reflected wave is partially polarized perpendicularly to the plane of incidence, while the transmitted wave is very slightly polarized in the plane of incidence. It is the polarization of the reflected wave, which has an important variation as a function of the angle of incidence. The polarization of the transmitted wave varies very little because of the small energy reflection factor.