91影视

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.

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

Write the expression for the transmitted wave.

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

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,

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

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:

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

Substitute the known values in equation (3).

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

Substitute the known values in equation (4).

$$\begin{gathered} {\tilde{E}}_{01}\hat{y}-{\tilde{E}}_{0R}(\hat{z}脳\cos {胃}_R\hat{x}+\sin {胃}_R\hat{y})=尾{\tilde{E}}_{0T}(\hat{z}脳\cos {胃}_T\hat{x}+\sin {胃}_T\hat{y}) \\ {\tilde{E}}_{01}\hat{y}-{\tilde{E}}_{0R}(\cos {胃}_R\hat{y}-\sin {胃}_R\hat{x})=尾{\tilde{E}}_{0T}(\cos {胃}_T\hat{y}+\sin {胃}_T\hat{x}) \end{gathered}$$ 鈥�.. (6)

Write the x component from equation (6).

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

Write the y-component from equation (5).

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

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

Then, it is proved that,

${胃}_R={胃}_T=0$

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