Chapter 10, Problem 73P

(a)

To determine

Find incidence angle.

Answer to Problem 73P

The incidence angle is $\underset{\_}{36.87°}$.

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 electric field, the propagation vector is $4a_y+3a_z$.

Substitute $4a_y+3a_z$ for $k$ in Equation (2).

$\begin{array}{c}{a}_k=\frac{4a_y+3a_z}{\sqrt{4^2+3^2}}\\ =\frac{4a_y+3a_z}{5}\end{array}$

From the given data, $a_n=a_y$.

Substitute $\left(\frac{4a_y+3a_z}{5}\right)$ for $a_k$ and $a_y$ for $a_n$ in Equation (1).

$\begin{array}{c}{\theta }_i={\cos }^{-1}\left[\left(\frac{4a_y+3a_z}{5}\right)\cdot a_y\right]\\ ={\cos }^{-1}\left(\frac{4}{5}\right)\\ \cong 36.87°\end{array}$

Conclusion:

Thus, the incidence angle is $\underset{\_}{36.87°}$.

(b)

To determine

Find the time-average power in air.

Answer to Problem 73P

The time-average power in air is $\underset{\_}{106.1a_y+79.58a_z\text{mW}/{\text{m}}^2}$.

Explanation of Solution

Calculation:

Consider the expression to find the time-average power in air.

$P_{\text{ave}}=\frac{E_{\text{o}}^2}{2{\eta }_{\text{o}}}{a}_k$        (3)

Here,

$E_{\text{o}}$ is the magnitude of electric field.

${\eta }_{\text{o}}$ is the intrinsic impedance in the air, which is $120\pi \Omega$.

Find $E_{\text{o}}$ from the given data.

$\begin{array}{c}{E}_{\text{o}}=\sqrt{8^2+{\left(-6\right)}^2}\text{V}/\text{m}\\ =10\text{V}/\text{m}\end{array}$

Substitute $10\text{V}/\text{m}$ for $E_{\text{o}}$, $120\pi \Omega$ for ${\eta }_{\text{o}}$, and $\left(\frac{4a_y+3a_z}{5}\right)$ for $a_k$ in Equation (3).

$\begin{array}{c}{P}_{\text{ave}}=\frac{{\left(10\text{V}/\text{m}\right)}^2}{2\left(120\pi \Omega \right)}\left(\frac{4a_y+3a_z}{5}\right)\\ \cong 106.1a_y+79.58a_z\text{mW}/{\text{m}}^2\end{array}$

Conclusion:

Thus, the time-average power in air is $\underset{\_}{106.1a_y+79.58a_z\text{mW}/{\text{m}}^2}$.

(c)

To determine

Find the reflected and transmitted electric fields.

Answer to Problem 73P

The reflected and transmitted electric fields are $\underset{\_}{-\left(1.518a_y+2.02a_z\right)\sin \left(\omega t+4y-3z\right)\text{V}/\text{m}}$ and $\underset{\_}{\left(1.879a_y-5.968a_z\right)\sin \left(\omega t-9.539y-3z\right)\text{V}/\text{m}}$, respectively.

Explanation of Solution

Calculation:

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

$E_r=E_{ro}\left[\sin \left({\theta }_r\right)a_y+\cos \left({\theta }_r\right)a_z\right]\sin \left(\omega t-k_r\cdot r\right)$        (4)

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

${\theta }_r=36.87°$

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

${\Gamma }_{\parallel }=\frac{{\eta }_2\cos \left({\theta }_t\right)-{\eta }_1\cos \left({\theta }_i\right)}{{\eta }_2\cos \left({\theta }_t\right)+{\eta }_1\cos \left({\theta }_i\right)}$        (5)

Here,

${\eta }_1$ is the intrinsic impedance of the medium 1 (air),

${\eta }_2$ is the intrinsic impedance for medium 2 (dielectric half-space),

${\theta }_t$ is the angle of transmission, and

${\theta }_i$ is the angle of incidence.

Rewrite Equation (5) for the given data.

${\Gamma }_{\parallel }=\frac{\frac{{\eta }_{\text{o}}}{2}\cos \left({\theta }_t\right)-{\eta }_{\text{o}}\cos \left({\theta }_i\right)}{\frac{{\eta }_{\text{o}}}{2}\cos \left({\theta }_t\right)+{\eta }_{\text{o}}\cos \left({\theta }_i\right)}$

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

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

Rearrange the expression for $E_{ro}$.

$E_{ro}=\left[\frac{\frac{{\eta }_{\text{o}}}{2}\cos \left({\theta }_t\right)-{\eta }_{\text{o}}\cos \left({\theta }_i\right)}{\frac{{\eta }_{\text{o}}}{2}\cos \left({\theta }_t\right)+{\eta }_{\text{o}}\cos \left({\theta }_i\right)}\right]E_{io}$        (6)

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.

$\begin{array}{c}\frac{\sin \left({\theta }_t\right)}{\sin \left({\theta }_i\right)}=\sqrt{\frac{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}{\left(2{\mu }_{\text{o}}\right)\left(2{\epsilon }_{\text{o}}\right)}}\left\{\begin{array}{l}\because {\mu }_1={\mu }_{\text{o}}\\ \because {\epsilon }_1={\epsilon }_{\text{o}}\\ \because {\mu }_2=2{\mu }_{\text{o}}\text{ for half-space dielectric}\\ \because {\epsilon }_2=2{\epsilon }_{\text{o}}\text{ for half-space dielectric}\end{array}\right\}\\ =0.5\end{array}$

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

${\theta }_t={\sin }^{-1}\left[0.5\sin \left({\theta }_i\right)\right]$

Substitute $36.87°$ for ${\theta }_i$.

$\begin{array}{c}{\theta }_t={\sin }^{-1}\left[0.5\sin \left(36.87°\right)\right]\\ =17.46°\end{array}$

Substitute $10\text{V}/\text{m}$ for $E_{io}$, $120\pi \Omega$ for ${\eta }_{\text{o}}$, $36.87°$ for ${\theta }_i$, and $17.46°$ for ${\theta }_t$ in Equation (6).

$\begin{array}{c}{E}_{ro}=\left\{\frac{\left(\frac{120\pi \Omega }{2}\right)\cos \left(17.46°\right)-\left(120\pi \Omega \right)\cos \left(36.87°\right)}{\left(\frac{120\pi \Omega }{2}\right)\cos \left(17.46°\right)+\left(120\pi \Omega \right)\cos \left(36.87°\right)}\right\}\left(10\text{V}/\text{m}\right)\\ \cong -2.53\text{V}/\text{m}\end{array}$

From the given Figure, find $k_r$.

$\begin{array}{c}{k}_r=-k_{ry}{a}_y+k_{rz}{a}_z\\ =-k_r\cos \left({\theta }_r\right)a_y+k_r\sin \left({\theta }_r\right)a_z\\ =-\left(5\right)\cos \left(36.87°\right)a_y+\left(5\right)\sin \left(36.87°\right)a_z\\ =-4a_y+3a_z\end{array}$

From the given Figure, find $k_r\cdot r$.

$\begin{array}{c}{k}_r\cdot r=\left(0a_x-4a_y+3a_z\right)\cdot \left(x{a}_x+y{a}_y+z{a}_z\right)\\ =-4y+3z\end{array}$

Substitute $-2.53\text{V}/\text{m}$ for $E_{ro}$, $36.87°$ for ${\theta }_r$, and $-4y+3z$ for $k_r\cdot r$ in Equation (4).

$\begin{array}{c}{E}_r=\left(-2.53\text{V}/\text{m}\right)\left[\sin \left(36.87°\right)a_y+\cos \left(36.87°\right)a_z\right]\sin \left[\omega t-\left(-4y+3z\right)\right]\\ \cong -\left(1.518a_y+2.02a_z\right)\sin \left(\omega t+4y-3z\right)\text{V}/\text{m}\end{array}$

Consider the expression for transmitted electric field.

$E_t=E_{to}\left[\sin \left({\theta }_t\right)a_y-\cos \left({\theta }_t\right)a_z\right]\sin \left(\omega t-k_t\cdot r\right)$        (7)

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.

$\begin{array}{c}{\tau }_{\parallel }=\frac{2\left(\frac{{\eta }_{\text{o}}}{2}\right)\cos \left({\theta }_i\right)}{\frac{{\eta }_{\text{o}}}{2}\cos \left({\theta }_t\right)+{\eta }_{\text{o}}\cos \left({\theta }_i\right)}\\ =\frac{2{\eta }_{\text{o}}\cos \left({\theta }_i\right)}{{\eta }_{\text{o}}\cos \left({\theta }_t\right)+2{\eta }_{\text{o}}\cos \left({\theta }_i\right)}\end{array}$

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

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

Rearrange the expression for $E_{to}$.

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

Substitute $10\text{V}/\text{m}$ for $E_{io}$, $120\pi \Omega$ for ${\eta }_{\text{o}}$, $36.87°$ for ${\theta }_i$, and $17.46°$ for ${\theta }_t$.

$\begin{array}{c}{E}_{to}=\left[\frac{2\left(120\pi \Omega \right)\cos \left(36.87°\right)}{\left(120\pi \Omega \right)\cos \left(17.46°\right)+2\left(120\pi \Omega \right)\cos \left(36.87°\right)}\right]\left(10\text{V}/\text{m}\right)\\ \cong 6.265\text{V}/\text{m}\end{array}$

From the given Figure, find $k_t$.

$k_t=k_{ty}{a}_y+k_{tz}{a}_z$

$k_t=k_t\cos \left({\theta }_t\right)a_y+k_t\sin \left({\theta }_t\right)a_z$        (8)

Find $k_t$.

$\begin{array}{c}{k}_t={\beta }_2\\ =\omega \sqrt{{\mu }_2{\epsilon }_2}\\ =\omega \sqrt{\left(2{\mu }_{\text{o}}\right)\left(2{\epsilon }_{\text{o}}\right)}\\ =2\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}\end{array}$

Find $k_i$.

$\begin{array}{c}{k}_i={\beta }_1\\ =\omega \sqrt{{\mu }_1{\epsilon }_1}\\ =\omega \sqrt{\left({\mu }_{\text{o}}\right)\left({\epsilon }_{\text{o}}\right)}\\ =\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}\end{array}$

Find the value of $k_t$.

$\begin{array}{c}\frac{k_t}{k_i}=\frac{2\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}}{\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}}\\ =2\end{array}$

Simplify the expression.

$\begin{array}{c}{k}_t=2k_i\\ =2{\beta }_1\\ =2\left(5\right)\left\{\because {\beta }_1=5\text{rad}/\text{m}\right\}\\ =10\end{array}$

Substitute 10 for $k_t$ and $17.46°$ for ${\theta }_t$ in Equation (8).

$\begin{array}{c}{k}_t=10\cos \left(17.46°\right)a_y+\left(10\right)\sin \left(17.46°\right)a_z\\ \cong 9.539a_y+3a_z\end{array}$

From the given Figure, find $k_t\cdot r$.

$\begin{array}{c}{k}_t\cdot r=\left(0a_x+9.539a_y+3a_z\right)\cdot \left(x{a}_x+y{a}_y+z{a}_z\right)\\ =9.539y+3z\end{array}$

Substitute $6.265\text{V}/\text{m}$ for $E_{to}$, $17.46°$ for ${\theta }_t$, and $9.539y+3z$ for $k_t\cdot r$ in Equation (7).

$\begin{array}{c}{E}_t=\left(6.265\text{V}/\text{m}\right)\left[\sin \left(17.46°\right)a_y-\cos \left(17.46°\right)a_z\right]\sin \left[\omega t-\left(9.539y+3z\right)\right]\\ \cong \left(1.879a_y-5.968a_z\right)\sin \left(\omega t-9.539y-3z\right)\text{V}/\text{m}\end{array}$

Conclusion:

Thus, the reflected and transmitted electric fields are $\underset{\_}{-\left(1.518a_y+2.02a_z\right)\sin \left(\omega t+4y-3z\right)\text{V}/\text{m}}$ and $\underset{\_}{\left(1.879a_y-5.968a_z\right)\sin \left(\omega t-9.539y-3z\right)\text{V}/\text{m}}$, respectively.