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Derivation

If a plane wave is incident on a dielectric with an angle ThetasubI and wavevector ksubI then a transmitted ray will be created with a corresponding angle of transmittance as shown in Fig. 1. The transmitted wavevector is given by:

$\mathbf {k_{T}} =k_{T}sin(\theta _{T}){\hat {x}}+k_{T}cos(\theta _{T}){\hat {z}}$

If $n_{1}>n_{2}$, then $sin(\theta _{T})$ becomes greater than one since in the relation $sin(\theta _{T})={\frac {n_{1}}{n_{2}}}sin(\theta _{I})$ obtained from snell's law both ${\frac {n_{1}}{n_{2}}}$ and $sin(\theta _{I})$ are greater than one . As a result of this $cos(\theta _{T})$ becomes complex:

$cos(\theta _{T})={\sqrt {1-sin^{2}(\theta _{T})}}=i{\sqrt {sin^{2}(\theta _{T})-1}}$

The electric field of the transmitted plane wave is given by $\mathbf {E_{T}} =\mathbf {E_{0}} e^{i(\mathbf {k_{T}} \cdot \mathbf {r} -\omega t)}$ and so evaluating this further one obtains:

$\mathbf {E_{T}} =\mathbf {E_{0}} e^{i(\mathbf {k_{T}} \cdot \mathbf {r} -\omega t)}=\mathbf {E_{0}} e^{i(k_{T}sin(\theta _{T})+k_{T}cos(\theta _{T})-\omega t)}$

$\mathbf {E_{T}} =\mathbf {E_{0}} e^{i(xk_{T}sin(\theta _{T})+zk_{T}i{\sqrt {sin^{2}(\theta _{T})-1}}-\omega t)}$

$\mathbf {E_{T}} =\mathbf {E_{0}} e^{-\kappa z}e^{i(kx-\omega t)}$

where $\kappa ={\frac {\omega }{c}}{\sqrt {(n_{1}sin(\theta _{I})^{2}-n_{2}^{2}}}$ and $k={\frac {\omega n_{1}}{c}}sin(\theta _{I})$

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