According to Snell’s law, when light passes from an optically dense medium into a less dense one \left(n_{1}>n_{2}\right) the propagation vector k bends away from the normal (Fig. 9.28). In particular, if the light is incident at the critical angle
\theta_{c} \equiv \sin ^{-1}\left(n_{2} / n_{1}\right) (9.200)
then \theta_{T}=90^{\circ}, and the transmitted ray just grazes the surface. If \theta_{I} \text { exceeds } \theta_{c}, there is no refracted ray at all, only a reflected one (this is the phenomenon of total internal reflection, on which light pipes and fiber optics are based). But the fields are not zero in medium 2; what we get is a so-called evanescent wave, which is rapidly attenuated and transports no energy into medium 2 .^{26}
A quick way to construct the evanescent wave is simply to quote the results of Sect. 9.3.3, with k_{T}=\omega n_{2} / c and
k _{T}=k_{T}\left(\sin \theta_{T} \hat{ x }+\cos \theta_{T} \hat{ z }\right) ;
the only change is that
\sin \theta_{T}=\frac{n_{1}}{n_{2}} \sin \theta_{I}is now greater than 1, and
\cos \theta_{T}=\sqrt{1-\sin ^{2} \theta_{T}}=i \sqrt{\sin ^{2} \theta_{T}-1}is imaginary. (Obviously, \theta_{T} can no longer be interpreted as an angle!)
(a) Show that
\tilde{ E }_{T}( r , t)=\tilde{ E }_{0_{T}} e^{-\kappa z} e^{i(k x-\omega t)} (9.201)
where
\kappa \equiv \frac{\omega}{c} \sqrt{\left(n_{1} \sin \theta_{I}\right)^{2}-n_{2}^{2}} \quad \text { and } \quad k \equiv \frac{\omega n_{1}}{c} \sin \theta_{I} (9.202)
This is a wave propagating in the x direction (parallel to the interface!), and attenuated in the z direction.
(b) Noting that α (Eq. 9.108) is now imaginary, use Eq. 9.109 to calculate the reflection coefficient for polarization parallel to the plane of incidence. [Notice that you get 100% reflection, which is better than at a conducting surface (see, for example, Prob. 9.22).]
\alpha \equiv \frac{\cos \theta_{T}}{\cos \theta_{I}} (9.108)
\tilde{E}_{0_{R}}=\left(\frac{\alpha-\beta}{\alpha+\beta}\right) \tilde{E}_{0_{I}}, \quad \tilde{E}_{0_{T}}=\left(\frac{2}{\alpha+\beta}\right) \tilde{E}_{0_{I}} (9.109)
(c) Do the same for polarization perpendicular to the plane of incidence (use the results of Prob. 9.17).
(d) In the case of polarization perpendicular to the plane of incidence, show that the (real) evanescent fields are
\left.\begin{array}{l} E ( r , t)=E_{0} e^{-\kappa z} \cos (k x-\omega t) \hat{ y } \\B ( r , t)=\frac{E_{0}}{\omega} e^{-\kappa z}[\kappa \sin (k x-\omega t) \hat{ x }+k \cos (k x-\omega t) \hat{ z }]\end{array}\right\} (9.203)
(e) Check that the fields in (d) satisfy all of Maxwell’s equations (Eq. 9.67).
\left. \begin{matrix} \text { (i) } \nabla \cdot E =0 \text {, } & \text { (iii) } \nabla \times E =-\frac{\partial B }{\partial t} \text {, } \\ \text { (ii) } \nabla \cdot B =0 \text {, } & \text { (iv) } \quad \nabla \times B =\mu \epsilon \frac{\partial E }{\partial t} \text {, } \end{matrix} \right\} (9.67)
(f) For the fields in (d), construct the Poynting vector, and show that, on average, no energy is transmitted in the z direction
