Chapter 10, Problem 76P

(a)

To determine

Find reflected and transmitted angles.

Answer to Problem 76P

The reflected and transmitted angles are $\underset{\_}{19.47°}$ and $\underset{\_}{90°}$, respectively.

Explanation of Solution

Calculation:

Consider the expression to find the incidence angle.

$\cos \left({\theta }_i\right)=a_k\cdot a_n$

Rearrange the expression for incidence angle.

${\theta }_i={\cos }^{-1}\left(a_k\cdot a_n\right)$        (1)

Consider the expression to find $a_k$.

$a_k=\frac{k}{\left|k\right|}$        (2)

Here,

$k$ is the propagation vector.

From the given magnetic field, the propagation vector is $k{a}_x+k\sqrt{8}{a}_z$.

Substitute $k{a}_x+k\sqrt{8}{a}_z$ for $k$ in Equation (2).

$\begin{array}{c}{a}_k=\frac{k{a}_x+k\sqrt{8}{a}_z}{\sqrt{k^2+{\left(k\sqrt{8}\right)}^2}}\\ =\left(\frac{1}{3}\right)\left(a_x+\sqrt{8}{a}_z\right)\end{array}$

From the given data, $a_n=a_z$.

Substitute $\left(\frac{1}{3}\right)\left(a_x+\sqrt{8}{a}_z\right)$ for $a_k$ and $a_z$ for $a_n$ in Equation (1).

$\begin{array}{c}{\theta }_i={\cos }^{-1}\left[\left(\frac{1}{3}\right)\left(a_x+\sqrt{8}{a}_z\right)\cdot a_z\right]\\ ={\cos }^{-1}\left(\frac{\sqrt{8}}{3}\right)\\ \cong 19.47°\end{array}$

The reflection angle is equal to incidence angle. Therefore, ${\theta }_r={\theta }_i$.

${\theta }_r=19.47°$

Consider the expression to find the angle of transmission ${\theta }_t$.

$\frac{\sin \left({\theta }_t\right)}{\sin \left({\theta }_i\right)}=\sqrt{\frac{{\mu }_1{\epsilon }_1}{{\mu }_2{\epsilon }_2}}$

Rewrite the expression for the given data.

$\begin{array}{c}\frac{\sin \left({\theta }_t\right)}{\sin \left({\theta }_i\right)}=\sqrt{\frac{{\mu }_{\text{o}}\left(9{\epsilon }_{\text{o}}\right)}{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}}\left\{\begin{array}{l}\because {\mu }_1={\mu }_{\text{o}}\\ \because {\epsilon }_1=9{\epsilon }_{\text{o}}\\ \because {\mu }_2={\mu }_{\text{o}}\text{ for air}\\ \because {\epsilon }_2={\epsilon }_{\text{o}}\text{ for air}\end{array}\right\}\\ =3\end{array}$

Rearrange the expression to find the angle of transmission ${\theta }_t$.

${\theta }_t={\sin }^{-1}\left[3\sin \left({\theta }_i\right)\right]$

Substitute $19.47°$ for ${\theta }_i$.

$\begin{array}{c}{\theta }_t={\sin }^{-1}\left[3\sin \left(19.47°\right)\right]\\ \cong 90°\end{array}$

Conclusion:

Thus, the reflected and transmitted angles are $\underset{\_}{19.47°}$ and $\underset{\_}{90°}$, respectively.

(b)

To determine

Find the value of k.

Answer to Problem 76P

The value of k is $\underset{\_}{3.333}$.

Explanation of Solution

Calculation:

The magnitude of propagation vector is equal to the value of phase constant. Write the expression for phase constant of the medium 1 (dielectric medium).

${\beta }_1=\omega \sqrt{{\mu }_1{\epsilon }_1}$

Rewrite the expression for given data.

${\beta }_1=\omega \sqrt{{\mu }_{\text{o}}\left(9{\epsilon }_{\text{o}}\right)}$

Substitute ${10}^9\text{rad}/\text{s}$ for $\omega$, $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_{\text{o}}$, and $\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}$ ${\epsilon }_{\text{o}}$.

$\begin{array}{c}{\beta }_1=\left({10}^9\text{rad}/\text{s}\right)\sqrt{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(9\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}\\ =10\text{rad}/\text{m}\end{array}$

Find the magnitude of the propagation vector.

$\begin{array}{c}\left|k\right|=\sqrt{k^2+{\left(k\sqrt{8}\right)}^2}\\ =3k\end{array}$

As ${\beta }_1=\left|k\right|$, write the expression.

$\begin{array}{c}{\beta }_1=\left|k\right|\\ 10\text{rad}/\text{m}=3k\end{array}$

Simplify the expression for k.

$\begin{array}{c}k=\frac{10}{3}\\ =3.333\end{array}$

Conclusion:

Thus, the value of k is $\underset{\_}{3.333}$.

(c)

To determine

Find the wavelength in the dielectric and in air.

Answer to Problem 76P

The wavelength in the dielectric and in air are $\underset{\_}{0.6283\text{m}}$ and $\underset{\_}{1.885\text{m}}$, respectively.

Explanation of Solution

Calculation:

Consider the expression to find the wavelength.

$\lambda =\frac{2\pi }{\beta }$        (1)

Modify the expression for wavelength in the dielectric medium.

${\lambda }_1=\frac{2\pi }{{\beta }_1}$

Substitute $10\text{rad}/\text{m}$ for ${\beta }_1$.

$\begin{array}{c}{\lambda }_1=\frac{2\pi }{10\text{rad}/\text{m}}\\ \cong 0.6283\text{m}\end{array}$

Write the expression for phase constant of the medium 2 (air).

${\beta }_2=\omega \sqrt{{\mu }_2{\epsilon }_2}$

Rewrite the expression for given data.

${\beta }_2=\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}$

Substitute ${10}^9\text{rad}/\text{s}$ for $\omega$, $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_{\text{o}}$, and $\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}$ ${\epsilon }_{\text{o}}$.

$\begin{array}{c}{\beta }_2=\left({10}^9\text{rad}/\text{s}\right)\sqrt{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}\\ =\frac{10}{3}\text{rad}/\text{m}\end{array}$

Modify Equation (1) for wavelength in the air (medium 2).

${\lambda }_2=\frac{2\pi }{{\beta }_2}$

Substitute $\frac{10}{3}\text{rad}/\text{m}$ for ${\beta }_2$.

$\begin{array}{c}{\lambda }_2=\frac{2\pi }{\frac{10}{3}\text{rad}/\text{m}}\\ \cong 1.885\text{m}\end{array}$

Conclusion:

Thus, the wavelength in the dielectric and in air are $\underset{\_}{0.6283\text{m}}$ and $\underset{\_}{1.885\text{m}}$, respectively.

(d)

To determine

Find the incident electric field.

Answer to Problem 76P

The incident electric field is $\underset{\_}{\left(23.6954a_x-8.3776a_z\right)\cos \left({10}^{9}t-kx-k\sqrt{8}z\right)\text{V}/\text{m}}$.

Explanation of Solution

Calculation:

Consider the expression to find the incident electric field for the given data.

$E_i={\eta }_{1}H\times a_k$        (2)

Find the intrinsic impedance of medium 1.

${\eta }_1=\sqrt{\frac{{\mu }_1}{{\epsilon }_1}}$

Rewrite the expression for given data.

${\eta }_1=\sqrt{\frac{{\mu }_{\text{o}}}{9{\epsilon }_{\text{o}}}}$

Substitute  $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_{\text{o}}$ and $\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}$ ${\epsilon }_{\text{o}}$.

$\begin{array}{c}{\eta }_1=\sqrt{\frac{4\pi \times {10}^{-7}\text{H}/\text{m}}{9\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}}\\ =40\pi \Omega \end{array}$

Substitute $40\pi \Omega$ for ${\eta }_1$, $0.2\cos \left({10}^{9}t-kx-k\sqrt{8}z\right)a_y\text{A}/\text{m}$ for $H$, and $\left(\frac{1}{3}\right)\left(a_x+\sqrt{8}{a}_z\right)$ for $a_k$.

$\begin{array}{c}{E}_i=\left(40\pi \Omega \right)\left[0.2\cos \left({10}^{9}t-kx-k\sqrt{8}z\right)a_y\text{A}/\text{m}\right]\times \left[\left(\frac{1}{3}\right)\left(a_x+\sqrt{8}{a}_z\right)\right]\\ =\left[\begin{array}{l}\left(40\pi \Omega \right)\left(0.2\right)\cos \left({10}^{9}t-kx-k\sqrt{8}z\right)\left(\frac{1}{3}\right)\left(a_y\times a_x\right)+\\ \left(40\pi \Omega \right)\left(0.2\right)\cos \left({10}^{9}t-kx-k\sqrt{8}z\right)\left(\frac{\sqrt{8}}{3}\right)\left(a_y\times a_z\right)\end{array}\right]\text{V}/\text{m}\\ \cong \left[\begin{array}{l}8.3776\cos \left({10}^{9}t-kx-k\sqrt{8}z\right)\left(-a_z\right)\\ +23.6954\cos \left({10}^{9}t-kx-k\sqrt{8}z\right)a_x\end{array}\right]\text{V}/\text{m}\left\{\begin{array}{l}\because a_y\times a_x=-a_z\\ \because a_y\times a_z=a_x\end{array}\right\}\\ \cong \left(23.6954a_x-8.3776a_z\right)\cos \left({10}^{9}t-kx-k\sqrt{8}z\right)\text{V}/\text{m}\end{array}$

Conclusion:

Thus, the incident electric field is $\underset{\_}{\left(23.6954a_x-8.3776a_z\right)\cos \left({10}^{9}t-kx-k\sqrt{8}z\right)\text{V}/\text{m}}$.

(e)

To determine

Find the transmitted and reflected electric fields.

Answer to Problem 76P

The transmitted and reflected electric fields are $\underset{\_}{1357\cos \left({10}^{9}t-3.333x\right)a_z\text{V}/\text{m}}$ and $\underset{\_}{\left(213.3a_x+75.4a_z\right)\cos \left({10}^{9}t-kx+k\sqrt{8}z\right)\text{V}/\text{m}}$, respectively.

Explanation of Solution

Calculation:

Consider the expression for transmitted electric field.

$E_t=-E_{to}\left[\cos \left({\theta }_t\right)a_x-\sin \left({\theta }_t\right)a_z\right]\cos \left[{10}^{9}t-{\beta }_{2}x\sin \left({\theta }_t\right)+{\beta }_{2}z\cos \left({\theta }_t\right)\right]$        (3)

Consider the polarization as parallel polarization.

Consider the expression of transmission coefficient in the parallel polarization of oblique incidence.

${\tau }_{\parallel }=\frac{2{\eta }_2\cos \left({\theta }_i\right)}{{\eta }_2\cos \left({\theta }_t\right)+{\eta }_1\cos \left({\theta }_i\right)}$

Rewrite the expression for the given data.

${\tau }_{\parallel }=\frac{2{\eta }_{\text{o}}\cos \left({\theta }_i\right)}{{\eta }_{\text{o}}\cos \left({\theta }_t\right)+{\eta }_1\cos \left({\theta }_i\right)}\left\{\because {\eta }_2={\eta }_{\text{o}}\right\}$

As ${\tau }_{\parallel }=\frac{E_{to}}{E_{io}}$, rewrite the expression.

$\frac{E_{to}}{E_{io}}=\frac{2{\eta }_{\text{o}}\cos \left({\theta }_i\right)}{{\eta }_{\text{o}}\cos \left({\theta }_t\right)+{\eta }_1\cos \left({\theta }_i\right)}$

Rearrange the expression for $E_{to}$.

$E_{to}=\left[\frac{2{\eta }_{\text{o}}\cos \left({\theta }_i\right)}{{\eta }_{\text{o}}\cos \left({\theta }_t\right)+{\eta }_1\cos \left({\theta }_i\right)}\right]E_{io}$

Substitute the corresponding parameters and find the value of $E_{to}$.

$\begin{array}{c}{E}_{to}=\left[\frac{2\left(120\pi \Omega \right)\cos \left(19.47°\right)}{\left(120\pi \Omega \right)\cos \left(90°\right)+\left(40\pi \Omega \right)\cos \left(19.47°\right)}\right]\left[\left(24\pi \right)\left(3\right)\text{V}/\text{m}\right]\\ \cong 1357\text{V}/\text{m}\end{array}$

Substitute $1357\text{V}/\text{m}$ for $E_{to}$, $90°$ for ${\theta }_t$, $3.333\text{rad}/\text{m}$ for ${\beta }_2$ in Equation (3).

$\begin{array}{c}{E}_t=-1357\left[\cos \left(90°\right)a_x-\sin \left(90°\right)a_z\right]\cos \left[{10}^{9}t-3.333x\sin \left(90°\right)+3.333z\cos \left(90°\right)\right]\text{V}/\text{m}\\ =1357\cos \left({10}^{9}t-3.333x\right)a_z\text{V}/\text{m}\end{array}$

Consider the expression for reflected electric field for the given data.

$E_r=E_{ro}\left[\cos \left({\theta }_r\right)a_x+\sin \left({\theta }_r\right)a_z\right]\cos \left({10}^{9}t-kx+k\sqrt{8}z\right)$        (4)

Consider the expression of reflection coefficient in the parallel polarization of oblique incidence.

${\Gamma }_{\parallel }=\frac{{\eta }_2\cos \left({\theta }_t\right)-{\eta }_1\cos \left({\theta }_i\right)}{{\eta }_2\cos \left({\theta }_t\right)+{\eta }_1\cos \left({\theta }_i\right)}$

Rewrite the expression for the given data.

${\Gamma }_{\parallel }=\frac{{\eta }_{\text{o}}\cos \left({\theta }_t\right)-{\eta }_1\cos \left({\theta }_i\right)}{{\eta }_{\text{o}}\cos \left({\theta }_t\right)+{\eta }_1\cos \left({\theta }_i\right)}$

As ${\Gamma }_{\parallel }=\frac{E_{ro}}{E_{io}}$, rewrite the expression.

$\frac{E_{ro}}{E_{io}}=\frac{{\eta }_{\text{o}}\cos \left({\theta }_t\right)-{\eta }_1\cos \left({\theta }_i\right)}{{\eta }_{\text{o}}\cos \left({\theta }_t\right)+{\eta }_1\cos \left({\theta }_i\right)}$

Rearrange the expression for $E_{ro}$.

$E_{ro}=\left[\frac{{\eta }_{\text{o}}\cos \left({\theta }_t\right)-{\eta }_1\cos \left({\theta }_i\right)}{{\eta }_{\text{o}}\cos \left({\theta }_t\right)+{\eta }_1\cos \left({\theta }_i\right)}\right]E_{io}$

Substitute the corresponding parameters and find the value of $E_{ro}$.

$\begin{array}{c}{E}_{ro}=\left[\frac{\left(120\pi \Omega \right)\cos \left(90°\right)-\left(40\pi \Omega \right)\cos \left(19.47°\right)}{\left(120\pi \Omega \right)\cos \left(90°\right)+\left(40\pi \Omega \right)\cos \left(19.47°\right)}\right]\left[-\left(24\pi \right)\left(3\right)\text{V}/\text{m}\right]\\ =226.1946\text{V}/\text{m}\end{array}$

Substitute $226.1946\text{V}/\text{m}$ for $E_{ro}$ and $19.47°$ for ${\theta }_r$ in Equation (4).

$\begin{array}{c}{E}_r=226.1946\left[\cos \left(19.47°\right)a_x+\sin \left(19.47°\right)a_z\right]\cos \left({10}^{9}t-kx+k\sqrt{8}z\right)\text{V}/\text{m}\\ \cong \left(213.3a_x+75.4a_z\right)\cos \left({10}^{9}t-kx+k\sqrt{8}z\right)\text{V}/\text{m}\end{array}$

Conclusion:

Thus, the transmitted and reflected electric fields are $\underset{\_}{1357\cos \left({10}^{9}t-3.333x\right)a_z\text{V}/\text{m}}$ and $\underset{\_}{\left(213.3a_x+75.4a_z\right)\cos \left({10}^{9}t-kx+k\sqrt{8}z\right)\text{V}/\text{m}}$, respectively.

(f)

To determine

Find the Brewster angle.

Answer to Problem 76P

The Brewster angle is $\underset{\_}{18.43°}$.

Explanation of Solution

Calculation:

Consider the expression to find the Brewster angle for the parallel-polarized wave.

$\tan {\theta }_{B_{\parallel }}=\sqrt{\frac{{\epsilon }_2}{{\epsilon }_1}}$

Rewrite the expression for the given data.

$\begin{array}{c}\tan {\theta }_{B_{\parallel }}=\sqrt{\frac{{\epsilon }_{\text{o}}}{9{\epsilon }_{\text{o}}}}\\ =\frac{1}{3}\end{array}$

Rearrange the expression for Brewster angle.

$\begin{array}{c}{\theta }_{B_{\parallel }}={\tan }^{-1}\left(\frac{1}{3}\right)\\ \cong 18.43°\end{array}$

Conclusion:

Thus, the Brewster angle is $\underset{\_}{18.43°}$.