Chapter 8, Problem 1P

A plane wave in air with an electric field amplitude of 20 V/m is incident normally upon the surface of a lossless, nonmagnetic medium with ϵ_r = 25. Determine the following:

1. (a) The reflection and transmission coefficients.
2. (b) The standing-wave ratio in the air medium.
3. (c) The average power densities of the incident, reflected, and transmitted waves.

To determine

The reflection coefficient $\left(\Gamma \right)$ and transmission coefficient $\left(\text{τ}\right)$ for the given condition.

Answer to Problem 1P

The reflection coefficient $\left(\Gamma \right)$ is $-0.67$ and transmission coefficient $\left(\text{τ}\right)$ is $0.33$.

Explanation of Solution

Given data:

The electric field amplitude $\left(E_{\text{0}}^{\text{i}}\right)$ of the wave is $20\text{V}/\text{m}$.

The permittivity $\left({\epsilon }_{\text{r}}\right)$ of the lossless medium is $25$.

Calculation:

The reflection coefficient $\left(\Gamma \right)$ for normal incidence is given by,

$\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}$ (1)

Here,

$\Gamma$ is the reflection coefficient.

${\eta }_1$ is intrinsic impedance of medium 1.

${\eta }_2$ is intrinsic impedance of medium 2.

Write formula to find intrinsic impedance.

${\eta }_{\text{i}}=\sqrt{\frac{{\mu }_{\text{i}}}{{\epsilon }_{\text{i}}}}$ (2)

Here,

${\eta }_{\text{i}}$ is intrinsic impedance of $i^{th}$ medium.

${\mu }_{\text{i}}$ is the permeability of the $i^{th}$ medium.

${\epsilon }_{\text{i}}$ is the permittivity of the $i^{th}$ medium.

So, find intrinsic impedance of the medium $1$ which is air.

${\eta }_1={\eta }_0$

Hence,

${\eta }_0=\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}$ (3)

Here,

${\eta }_0$ is the intrinsic impedance of air.

${\mu }_0$ is the permeability of air $\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)$.

${\epsilon }_0$ is the permittivity of air $\left(8.85\times {10}^{-12}\text{F}/\text{m}\right)$.

Substitute $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_0$ and $8.85\times {10}^{-12}\text{F}/\text{m}$ for ${\epsilon }_0$ in equation (3).

$\begin{array}{c}{\eta }_1=\sqrt{\frac{4\pi \times {10}^{-7}\text{H}/\text{m}}{8.85\times {10}^{-12}\text{F}/\text{m}}}\\ =120\pi \Omega \end{array}$

Now, find intrinsic impedance of medium 2.

${\eta }_2=\frac{{\eta }_0}{\sqrt{{\epsilon }_{\text{r}}}}$ (4)

Here,

${\epsilon }_{\text{r}}$ is the permittivity of the loseless medium.

Substitute $25$ for ${\epsilon }_{\text{r}}$ and $120\pi \Omega$ for ${\eta }_0$ in equation (4).

$\begin{array}{c}{\eta }_2=\frac{120\pi }{\sqrt{25}}\\ =24\pi \Omega \end{array}$

Substitute $120\pi \Omega$ for ${\eta }_1$ and $24\pi \Omega$ for ${\eta }_2$ in the equation (1).

$\begin{array}{c}\Gamma =\frac{24\pi \Omega -120\pi \Omega }{24\pi \Omega +120\pi \Omega }\\ =\frac{-96}{144}\\ =-0.67\end{array}$

The transmission coefficient $\left(\text{τ}\right)$ is given by,

$\text{τ}=1+\Gamma$ (5)

Here,

$\text{τ}$ is the transmission coefficient.

Substitute $-0.67$ for $\Gamma$ in equation (5).

$\begin{array}{c}\text{τ}=1+\left(-0.67\right)\\ =0.33\end{array}$

Conclusion:

Therefore, the reflection coefficient $\left(\Gamma \right)$ is $-0.67$ and transmission coefficient $\left(\text{τ}\right)$ is $0.33$.

To determine

The standing-wave ratio $\left(S\right)$ in the air medium.

Answer to Problem 1P

The value of standing-wave ratio $\left(S\right)$ in the air medium is $5$.

Explanation of Solution

The calculated value of reflection coefficient $\left(\Gamma \right)$ is $-0.67$.

Calculation:

The standing-wave ratio is given by,

$S=\frac{1+\left|\Gamma \right|}{1-\left|\Gamma \right|}$ (6)

Here,

$S$ is the standing-wave ratio.

Substitute $-0.67$ for $\Gamma$ in the equation (6).

$\begin{array}{c}S=\frac{1+\left|-0.67\right|}{1-\left|-0.67\right|}\\ =\frac{1.67}{0.33}\\ =5\end{array}$

Conclusion:

Therefore, the value of standing-wave ratio $\left(\text{S}\right)$ in the air medium is $5$.

To determine

The average power density of the incident $\left(S_{\text{av}}^{\text{i}}\right)$, reflected $\left(S_{\text{av}}^{\text{r}}\right)$ and transmitted $\left(S_{\text{av}}^{\text{t}}\right)$ wave.

Answer to Problem 1P

The value of average power density of the incident wave $\left(S_{\text{av}}^{\text{i}}\right)$ is $0.53\text{W}/{\text{m}}^{\text{2}}$, of reflected wave $\left(S_{\text{av}}^{\text{r}}\right)$ is $0.24\text{W}/{\text{m}}^{\text{2}}$ and of transmitted wave $\left(S_{\text{av}}^{\text{t}}\right)$ is $0.28\text{W}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given data:

The electric field amplitude $\left(E_{\text{0}}^{\text{i}}\right)$ of the wave is $20\text{V}/\text{m}$.

The calculated value of intrinsic impedance of air $\left({\eta }_0\right)$ is $120\pi \Omega$.

The calculated value of reflection coefficient $\left(\Gamma \right)$ is $-0.67$.

Calculation:

The average power density of the incident $\left(S_{\text{av}}^{\text{i}}\right)$ is given by,

$\left(S_{\text{av}}^{\text{i}}\right)=\frac{{\left|E_{\text{0}}^{\text{i}}\right|}^2}{2{\eta }_0}$ (7)

Here,

$E_{\text{0}}^{\text{i}}$ is the electric field amplitude of the wave.

$S_{\text{av}}^{\text{i}}$ is the average power density of the incident wave.

Substitute $20\text{V}/\text{m}$ for $E_{\text{0}}^{\text{i}}$ and $120\pi \Omega$ for ${\eta }_0$ in equation (7).

$\begin{array}{c}\left(S_{\text{av}}^{\text{i}}\right)=\frac{{\left|20\text{V}/\text{m}\right|}^2}{2\times 120\pi \Omega }\\ =\frac{400}{2\times 120\pi }\text{W}/{\text{m}}^{\text{2}}\\ =0.53\text{W}/{\text{m}}^{\text{2}}\end{array}$

The average power density of the reflected $\left(S_{\text{av}}^{\text{r}}\right)$ is given by,

$\left(S_{\text{av}}^{\text{r}}\right)={\left|\Gamma \right|}^2\left(S_{\text{av}}^{\text{i}}\right)$ (8)

Substitute $-0.67$ for $\Gamma$ and $0.53\text{W}/{\text{m}}^{\text{2}}$ $S_{\text{av}}^{\text{i}}$ for  in equation (8).

$\begin{array}{c}\left(S_{\text{av}}^{\text{r}}\right)={\left|-0.67\right|}^2\times 0.53\text{W}/{\text{m}}^{\text{2}}\\ =0.24\text{W}/{\text{m}}^{\text{2}}\end{array}$

The average power density of the transmitted $\left(S_{\text{av}}^{\text{t}}\right)$ is given by,

$\left(S_{\text{av}}^{\text{r}}\right)={\left|\text{τ}\right|}^2\frac{{\left|E_{\text{0}}^{\text{i}}\right|}^2}{2{\eta }_2}$ (9)

Substitute $0.33$ for $\text{τ}$, $24\pi \Omega$ for ${\eta }_2$ and $20\text{V}/\text{m}$ for $E_{\text{0}}^{\text{i}}$ in equation (9).

$\begin{array}{c}\left(S_{\text{av}}^{\text{r}}\right)={\left|0.33\right|}^2\times \frac{{\left|20\text{V}/\text{m}\right|}^2}{2\times 24\pi \Omega }\\ ={\left|0.33\right|}^2\times \frac{400}{2\times 24\pi }\text{W}/{\text{m}}^{\text{2}}\\ =0.28\text{W}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the value of average power density of the incident wave $\left(S_{\text{av}}^{\text{i}}\right)$ is $0.53\text{W}/{\text{m}}^{\text{2}}$, of reflected wave $\left(S_{\text{av}}^{\text{r}}\right)$ is $0.24\text{W}/{\text{m}}^{\text{2}}$ and of transmitted wave $\left(S_{\text{av}}^{\text{t}}\right)$ is $0.28\text{W}/{\text{m}}^{\text{2}}$.