Normal incidence on a Perfect Di-electric

Normal incidence on a Perfect Dielectric:-

Till now we focused on the propagation of a uniform plane wave in an unbounded medium either free space (or) dielectric.

we now consider a monochromatic uniform plane wave that travels through one medium and then enters another medium of infinite extent.

at this stage we assume that the interface between the two media is normal to the direction of propagation of the incoming wave.

The wave that is propagating in the first medium is called incident wave. Assume the direction of propagation of the incoming wave along positive z-direction.

The interface (or) the boundary is a plane z=0 in this case.

if direction of propagation was along +ve x-axis plane would be x=0 plane.
if direction of propagation was along +ve y-axis plane would be y=0 plane.

The wave reflected back into the same medium is called reflected wave and the wave that is propagating into the second medium is the transmitted wave.

incident and reflected waves are in opposite directions to each other.

Incident waves:-

\overrightarrow{E}_{i}=E_{io}\ e^{-\gamma _{1}z}\ \overrightarrow{a}_{x} .

\overrightarrow{H}_{i}=H_{io}\ e^{-\gamma _{1}z}\ \overrightarrow{a}_{y} .

\overrightarrow{H}_{i}=\frac{E_{io}}{\eta _{1}}\ e^{-\gamma _{1}z}\ \overrightarrow{a}_{y}.    the d.o.p of  H is  \overrightarrow{a}_{H}=\overrightarrow{a}_{k}X\overrightarrow{a}_{E}\Rightarrow \overrightarrow{a}_{z}X\overrightarrow{a}_{x}\Rightarrow \overrightarrow{a}_{y}.

Reflected waves:-

\overrightarrow{E}_{r}=E_{ro}\ e^{\gamma _{1}z}\ \overrightarrow{a}_{x} .

\overrightarrow{H}_{r}=-H_{ro}\ e^{\gamma _{1}z}\ \overrightarrow{a}_{y} .

\overrightarrow{H}_{r}=-\frac{E_{ro}}{\eta _{1}}\ e^{\gamma _{1}z}\ \overrightarrow{a}_{y}.  the d.o.p of  H is  \overrightarrow{a}_{H}=\overrightarrow{a}_{k}X\overrightarrow{a}_{E}\Rightarrow -\overrightarrow{a}_{z}X\overrightarrow{a}_{x}\Rightarrow -\overrightarrow{a}_{y}.

Transmitted waves:-

\overrightarrow{E}_{t}=E_{to}\ e^{-\gamma _{2}z}\ \overrightarrow{a}_{x} .

\overrightarrow{H}_{t}=H_{to}\ e^{\gamma _{2}z}\ \overrightarrow{a}_{y} .

\overrightarrow{H}_{t}=\frac{E_{to}}{\eta _{2}}\ e^{-\gamma _{2}z}\ \overrightarrow{a}_{y}.    the d.o.p of  H is  \overrightarrow{a}_{H}=\overrightarrow{a}_{k}X\overrightarrow{a}_{E}\Rightarrow \overrightarrow{a}_{z}X\overrightarrow{a}_{x}\Rightarrow \overrightarrow{a}_{y}.

Now the Transmission and reflection coefficients are defined as follows

reflection \ coefficient\ \rho (or)\ r= \frac{reflected\ electric\ field \ strength}{incident\ electric\ field \ strength} .

\rho _{E} = \frac{E_{ro}}{E_{io}} .

similarly the transmission coefficient is

transmission \ coefficient\ \tau (or)\ T= \frac{transmitted\ electric\ field \ strength}{incident\ electric\ field \ strength}.

\tau _{E} = \frac{E_{to}}{E_{io}} .

Derivation of coefficients:-

By using the boundary conditions,

the tangential components of E are continuous

i.e, \begin{vmatrix} E_{tan1} \end{vmatrix}=\begin{vmatrix} E_{tan2} \end{vmatrix} .

E_{io}+E_{ro}=E_{to} .

E_{io}+\rho \ E_{io}=\tau \ E_{io}.

1+\rho =\tau ------EQN(1).

the tangential components of H are discontinuous

i.e, \begin{vmatrix} H_{tan1} \end{vmatrix}-\begin{vmatrix} H_{tan2} \end{vmatrix} = J_{s} .  let us assume J_{s}=0 at the Boundary.

\begin{vmatrix} H_{tan1} \end{vmatrix}=\begin{vmatrix} H_{tan2} \end{vmatrix} .

H_{io}+H_{ro}=H_{to} .

\frac{E_{io}}{\eta _{1}}-\frac{E_{ro}}{\eta _{1}}=\frac{E_{to}}{\eta _{2}}\Rightarrow \frac{E_{io}}{\eta _{1}}-\rho \ \frac{E_{io}}{\eta _{1}}=\tau \ \frac{E_{io}}{\eta _{2}}.

\frac{1}{\eta _{1}}- \frac{\rho}{\eta _{1}}= \frac{\tau}{\eta _{2}}.

1-\rho =\frac{\eta _{1}}{\eta _{2}}\ \tau -------EQN(2).

by solving the above two equations the transmission and reflection coefficients  using electric field strength are

\tau _{E}=\frac{2\eta _{2}}{\eta _{2}+\eta _{1}}     and  \rho _{E}=\frac{\eta _{2}-\eta _{1}}{\eta _{2}+\eta _{1}} .

similarly  the transmission and reflection coefficients  using magnetic field strength are

\tau _{H}=\frac{\eta _{1}-\eta _{2}}{\eta _{2}+\eta _{1}}    and  \rho _{H}=\frac{2\eta _{1}}{\eta _{2}+\eta _{1}} .

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Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.