Chapter 8: Problem 11
Consider total reflection at an interface between two nonmagnetic media, with relative refractive index \(n=c_{1} / c_{2}<1\). For angles of incidence \(\theta_{1}\) exceeding the critical angle of (8.6.42), Snell's law gives $$ \sin \theta_{2}=\frac{\sin \theta_{1}}{n}>1, $$ which implies that \(\theta_{2}\) is a complex angle with an imaginary cosine, $$ \cos \theta_{2}=\left(1-\sin ^{2} \theta_{2}\right)^{1 / 2}=j\left(\frac{\sin ^{2} \theta_{1}}{n^{2}}-1\right)^{1 / 2} $$ Substitute these relations in the case I reflection coefficient (8.6.28) to establish $$ R_{\mathbf{E} \perp}=e^{-i 2 \phi_{\perp}}, $$ where $$ \tan \phi_{\perp}=\frac{\left(\sin ^{2} \theta_{1}-n^{2}\right)^{1 / x}}{\cos \theta_{1}} $$ That is, the magnitude of the reflection coefficient is unity, but the phase of the reflected wave depends upon angle. Similarly show for case II from (8.6.36), that \(R_{\text {III }}=e^{-\text {jod with }}\) $$ \tan \phi \|=\frac{\left(\sin ^{2} \theta_{1}-n^{2}\right)^{1 / 2}}{n^{2} \cos \theta_{1}}=\frac{1}{n^{2}} \tan \phi_{\perp} . $$ Note that the two phase shifts are different, so that in general the state of polarization of an incident wave is altered.
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