Question 9.17: Analyze the case of polarization perpendicular to the plane ...

Boundary conditions: \begin{cases}\text { (i) } \epsilon_{1} E_{1}^{\perp}=\epsilon_{2} E_{2}^{\perp}, & \text { (iii) } E _{1}^{\|}= E _{2}^{\|} \\ \text {(ii) } B_{1}^{\perp}=B_{2}^{\perp}, & \text { (iv) } \frac{1}{\mu_{1}} B _{1}^{\|}=\frac{1}{\mu_{2}} B _{2}^{\|}\end{cases}

Law of refraction: \frac{\sin \theta_{2}}{\sin \theta_{1}}=\frac{v_{2}}{v_{1}} .\left[\text { Note: } k _{I} \cdot r -\omega t= k _{R} \cdot r -\omega t= k _{T} \cdot r -\omega t, \text { at } z=0\right. , so we can drop all exponential factors in applying the boundary conditions.]

Boundary condition (i): 0 = 0 (trivial). Boundary condition \text { (iii): } \tilde{E}_{0_{I}}+\tilde{E}_{0_{R}}=\tilde{E}_{0_{T}}.

Boundary condition \text { (ii): } \frac{1}{v_{1}} \tilde{E}_{0_{I}} \sin \theta_{1}+\frac{1}{v_{1}} \tilde{E}_{0_{R}} \sin \theta_{1}=\frac{1}{v_{2}} \tilde{E}_{0_{T}} \sin \theta_{2} \Rightarrow \tilde{E}_{0_{I}}+\tilde{E}_{0_{R}}=\left(\frac{v_{1} \sin \theta_{2}}{v_{2} \sin \theta_{1}}\right) \tilde{E}_{0_{T}}.

But the term in parentheses is 1, by the law of refraction, so this is the same as (iii)

Boundary condition  ( iv ): \frac{1}{\mu_{1}}\left[\frac{1}{v_{1}} \tilde{E}_{0_{I}}\left(-\cos \theta_{1}\right)+\frac{1}{v_{1}} \tilde{E}_{0_{R}} \cos \theta_{1}\right]=\frac{1}{\mu_{2} v_{2}} \tilde{E}_{0_{T}}\left(-\cos \theta_{2}\right) \Rightarrow \tilde{E}_{0_{I}}-\tilde{E}_{0_{R}}=\left(\frac{\mu_{1} v_{1} \cos \theta_{2}}{\mu_{2} v_{2} \cos \theta_{1}}\right) \tilde{E}_{0_{T}} . Let  \alpha \equiv \frac{\cos \theta_{2}}{\cos \theta_{1}} ; \beta \equiv \frac{\mu_{1} v_{1}}{\mu_{2} v_{2}} . Then \tilde{E}_{0_{I}}-\tilde{E}_{0_{R}}=\alpha \beta \tilde{E}_{0_{T}}.

Solving for  \tilde{E}_{0_{R}} \text { and } \tilde{E}_{0_{T}}: 2 \tilde{E}_{0_{I}}=(1+\alpha \beta) \tilde{E}_{0_{T}} \Rightarrow \tilde{E}_{0_{T}}=\left(\frac{2}{1+\alpha \beta}\right) \tilde{E}_{0_{I}};

\tilde{E}_{0_{R}}=\tilde{E}_{0_{T}}-\tilde{E}_{0_{I}}=\left(\frac{2}{1+\alpha \beta}-\frac{1+\alpha \beta}{1+\alpha \beta}\right) \tilde{E}_{0_{I}} \Rightarrow \tilde{E}_{0_{R}}=\left(\frac{1-\alpha \beta}{1+\alpha \beta}\right) \tilde{E}_{0_{I}}.

Since \alpha \text { and } \beta are positive, it follows that 2 /(1+\alpha \beta) is positive, and hence the transmitted wave is in phase with the incident wave, and the (real) amplitudes are related by E_{0_{T}}=\left(\frac{2}{1+\alpha \beta}\right) E_{0_{I}} The reflected wave is in phase if \alpha \beta<1 \text { and } 180^{\circ} \text { out of phase if } \alpha \beta>1 ; the (real) amplitudes are related by E_{0_{R}}=\left|\frac{1-\alpha \beta}{1+\alpha \beta}\right| E_{0_{I}} .

These are the Fresnel equations for polarization perpendicular to the plane of incidence.

To construct the graphs, note that \alpha \beta=\beta \frac{\sqrt{1-\sin ^{2} \theta / \beta^{2}}}{\cos \theta}=\frac{\sqrt{\beta^{2}-\sin ^{2} \theta}}{\cos \theta} , where θ is the angle of incidence, so, for \beta=1.5, \alpha \beta=\frac{\sqrt{2.25-\sin ^{2} \theta}}{\cos \theta} . [In the figure, the minus signs on the vertical axis should be decimal points.]

Is there a Brewster’s angle? Well, E_{0_{R}}=0 \text { would mean that } \alpha \beta=1, and hence that

\alpha=\frac{\sqrt{1-\left(v_{2} / v_{1}\right)^{2} \sin ^{2} \theta}}{\cos \theta}=\frac{1}{\beta}=\frac{\mu_{2} v_{2}}{\mu_{1} v_{1}} \text {, or } 1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \sin ^{2} \theta=\left(\frac{\mu_{2} v_{2}}{\mu_{1} v_{1}}\right)^{2} \cos ^{2} \theta , so

1=\left(\frac{v_{2}}{v_{1}}\right)^{2}\left[\sin ^{2} \theta+\left(\mu_{2} / \mu_{1}\right)^{2} \cos ^{2} \theta\right] . \text { Since } \mu_{1} \approx \mu_{2}, \text { this means } 1 \approx\left(v_{2} / v_{1}\right)^{2} , which is only true for optically
indistinguishable media, in which case there is of course no reflection—but that would be true at any angle, not just at a special “Brewster’s angle”. [If \mu_{2} were substantially different from \mu_{1} , and the relative velocities were just right, it would be possible to get a Brewster’s angle for this case, at

But the media would be very peculiar.]

By the same token, \delta_{R}  is either always 0, or always π, for a given interface—it does not switch over as you change θ, the way it does for polarization in the plane of incidence. In particular, if \beta=3 / 2, \text { then } \alpha \beta>1 , for

\alpha \beta=\frac{\sqrt{2.25-\sin ^{2} \theta}}{\cos \theta}>1 \text { if } 2.25-\sin ^{2} \theta>\cos ^{2} \theta, \text { or } 2.25>\sin ^{2} \theta+\cos ^{2} \theta=1.

In general, for \beta>1, \alpha \beta>1, \text { and hence } \delta_{R}=\pi \text {. For } \beta<1, \alpha \beta<1, \text { and } \delta_{R}=0 .

At normal incidence, α = 1, so Fresnel’s equations reduce to E_{0_{T}}=\left(\frac{2}{1+\beta}\right) E_{0_{I}} ; E_{0_{R}}=\left|\frac{1-\beta}{1+\beta}\right| E_{0_{I}} ,
consistent with Eq. 9.82. 

\tilde{E}_{0_{R}}=\left(\frac{1-\beta}{1+\beta}\right) \tilde{E}_{0_{I}}, \quad \tilde{E}_{0_{T}}=\left(\frac{2}{1+\beta}\right) \tilde{E}_{0_{I}}                     (9.82)

Reflection and Transmission coe!cients: R=\left(\frac{E_{0_{R}}}{E_{0_{I}}}\right)^{2}=\left(\frac{1-\alpha \beta}{1+\alpha \beta}\right)^{2} . Referring to Eq. 9.116,

T \equiv \frac{I_{T}}{I_{I}}=\frac{\epsilon_{2} v_{2}}{\epsilon_{1} v_{1}}\left(\frac{E_{0_{T}}}{E_{0_{I}}}\right)^{2} \frac{\cos \theta_{T}}{\cos \theta_{I}}=\alpha \beta\left(\frac{2}{\alpha+\beta}\right)^{2}                             (9.116)

T=\frac{\epsilon_{2} v_{2}}{\epsilon_{1} v_{1}} \alpha\left(\frac{E_{0_{T}}}{E_{0_{I}}}\right)^{2}=\alpha \beta\left(\frac{2}{1+\alpha \beta}\right)^{2}.

R+T=\frac{(1-\alpha \beta)^{2}+4 \alpha \beta}{(1+\alpha \beta)^{2}}=\frac{1-2 \alpha \beta+\alpha^{2} \beta^{2}+4 \alpha \beta}{(1+\alpha \beta)^{2}}=\frac{(1+\alpha \beta)^{2}}{(1+\alpha \beta)^{2}}=1.