Question

In writing Eqs. 9.76 and 9.77, I tacitly assumed that the reflected and transmitted waves have the same polarization as the incident wave—along the x direction. Prove that this must be so. [Hint: Let the polarization vectors of the transmitted and reflected waves be

${\hat{\mathbf{n}}}_{\mathbf{T}}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}{\mathit{\theta }}_{\mathbf{T}}\hat{\mathbf{x}}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_{\mathbf{T}}\hat{\mathbf{y}}\mathbf{,}\mathbf{}{\hat{\mathbf{n}}}_{\mathbf{R}}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}{\mathbf{\theta }}_{\mathbf{R}}\hat{\mathbf{x}}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}{\mathbf{\theta }}_{\mathbf{R}}\hat{\mathbf{y}}$prove from the boundary conditions that ${\mathit{\theta }}_{\mathbf{T}}\mathbf{=}{\mathit{\theta }}_{\mathbf{R}}\mathbf{=}\mathbf{0}$.]

Short answer

It is proved that ${\theta }_R={\theta }_T=0$.

Step-by-step solution

Expression for the reflection and transmission at normal incidence:

Let the xy plane form a boundary between the two linear media. A plane wave of frequency traveling in the z-direction and polarized in the x-direction.

Write the expression for reflected wave.

$$\begin{gathered} {\stackrel{\mathbf{~}}{\mathbf{E}}}_{\mathbf{R}}\mathbf{(}\mathit{z}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}{\stackrel{\mathbf{~}}{\mathbf{E}}}_{\mathbf{0}\mathbf{R}}{\mathit{e}}^{\mathbf{i}\mathbf{(}{\mathbf{k}}_{\mathbf{1}}\mathbf{z}\mathbf{-}\mathbf{\omega }\mathbf{t}\mathbf{)}}\hat{\mathbf{x}} \\ {\stackrel{\mathbf{~}}{\mathbf{B}}}_{\mathbf{R}}\mathbf{(}\mathit{z}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{v}}_{\mathbf{1}}}{\stackrel{\mathbf{~}}{\mathbf{E}}}_{\mathbf{0}\mathbf{R}}{\mathbf{e}}^{\mathbf{i}\mathbf{(}{\mathbf{k}}_{\mathbf{1}}\mathbf{z}\mathbf{-}\mathbf{\omega }\mathbf{t}\mathbf{)}}\hat{\mathbf{y}} \end{gathered}$$

Write the expression for the transmitted wave.

localid="1657519446367" $$\begin{gathered} {\stackrel{\mathbf{~}}{\mathbf{E}}}_{\mathbf{T}}\mathbf{(}\mathit{z}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}{\stackrel{\mathbf{~}}{\mathbf{E}}}_{\mathbf{0}\mathbf{T}}{\mathbf{e}}^{\mathbf{i}\mathbf{(}{\mathbf{k}}_{\mathbf{2}}\mathbf{z}\mathbf{-}\mathbf{\omega t}\mathbf{)}}\hat{\mathbf{x}} \\ {\stackrel{\mathbf{~}}{\mathbf{B}}}_{\mathbf{T}}\mathbf{(}\mathit{z}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{v}}_{\mathbf{2}}}{\stackrel{\mathbf{~}}{\mathbf{E}}}_{\mathbf{0}\mathbf{T}}{\mathbf{e}}^{\mathbf{i}\mathbf{(}{\mathbf{k}}_{\mathbf{2}}\mathbf{z}\mathbf{-}\mathbf{\omega t}\mathbf{)}}\hat{\mathbf{y}} \end{gathered}$$

Prove θT=θR=0:

Using a boundary condition,

$$\begin{gathered} E_1^{\text{'}\text{'}}=E_2^{\text{'}\text{'}} \\ {\tilde{E}}_{01}+{\tilde{E}}_{0R}={\tilde{E}}_{0T}...........\left(1\right) \end{gathered}$$

Again use boundary condition,

$$\begin{gathered} \frac{1}{{\mu }_1}{B}_1=\frac{1}{{\mu }_2}{B}_2 \\ {\tilde{E}}_{01}-{\tilde{E}}_{0R}=\beta {\tilde{E}}_{0T}.............\left(2\right) \end{gathered}$$

Hence, equation (1) is replaced as:

${\tilde{E}}_{o1}\hat{x}+{\tilde{E}}_{0R}{\hat{n}}_R={\tilde{E}}_{oT}{\hat{n}}_T$ ............(3)

Similarly, equation (2) is replaced as:

${\tilde{E}}_{01}\hat{y}-{\tilde{E}}_{0R}(\hat{z}\times {\hat{n}}_R)=\beta {\tilde{E}}_{0T}(\hat{z}\times {\hat{n}}_T)........\left(4\right)$

Substitute the known values in equation (3).

${\tilde{E}}_{01}\hat{x}+{\tilde{E}}_{0R}(\cos {\theta }_R\hat{x}+\sin {\theta }_R\hat{y})={\tilde{E}}_{0T}(\cos {\theta }_T\hat{x}+\sin {\theta }_T\hat{y})$ ….. (5)

Substitute the known values in equation (4).

$$\begin{gathered} {\tilde{E}}_{01}\hat{y}-{\tilde{E}}_{0R}(\hat{z}\times \cos {\theta }_R\hat{x}+\sin {\theta }_R\hat{y})=\beta {\tilde{E}}_{0T}(\hat{z}\times \cos {\theta }_T\hat{x}+\sin {\theta }_T\hat{y}) \\ {\tilde{E}}_{01}\hat{y}-{\tilde{E}}_{0R}(\cos {\theta }_R\hat{y}-\sin {\theta }_R\hat{x})=\beta {\tilde{E}}_{0T}(\cos {\theta }_T\hat{y}+\sin {\theta }_T\hat{x}) \end{gathered}$$ ….. (6)

Write the x component from equation (6).

${\tilde{E}}_{0R}\sin {\theta }_R=-{\tilde{E}}_{0T}\sin {\theta }_T$

Write the y-component from equation (5).

${\tilde{E}}_{0R}\sin {\theta }_R={\tilde{E}}_{0T}\sin {\theta }_T$

Hence, the above two equation can be satisfied only when.

Then, it is proved that,

${\theta }_R={\theta }_T=0$

Therefore, it is proved that ${\theta }_R={\theta }_T=0$.