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Consider an Evanescent field are created when an oscillating electric and magnetic field concentrates its energy close to the source rather than propagating like an electromagnetic field would.

Consider an electric field depict for evanescent field is $\stackrel{鈫�}{{\mathrm{E}}_{\mathrm{T}}}\left(r.t\right)=\stackrel{鈫�}{{\mathrm{E}}_{\mathrm{T}}}{e}^{I(kx-wt)}$.

Consider the Maxwell鈥檚 equations are,

1. The Gauss law for no charge region,

$鈭�路\text{E}=0$
2. The Gauss law for magnetic field,

$鈭�路\text{B}=0$

3. The Maxwell law of induction is expressed as,

$鈭�脳\text{E}=-\frac{1}{c}\frac{鈭�\text{B}}{鈭�t}$

4. The Modified Ampere鈥檚 circuital law,

$鈭�脳\text{B}={渭}_0\left(\text{J}+{蔚}_0\frac{鈭�\text{E}}{鈭�t}\right)$

Consider the Fresnel equation for perpendicular polarization.

$$\begin{gathered} {\stackrel{鈫�}{E}}_{0T}=\frac{2}{1+伪尾}{\stackrel{鈫�}{E}}_{0I} \\ {\stackrel{鈫�}{E}}_{\text{0R}}=\frac{1-伪尾}{1+伪尾}{\stackrel{鈫�}{E}}_{0I} \end{gathered}$$

Here, ${\stackrel{鈫�}{E}}_{0R}$is magnitude of reflected electric wave,${\stackrel{鈫�}{E}}_{0T}$ is magnitude of transmitted electric wave, $伪$ is purely imaginary, data-custom-editor="chemistry" $\mathrm{尾}$ is real and ${\stackrel{鈫�}{E}}_{01}$ is magnitude of incident electricwave.

Write the formula ofelectric field component of a wave.

${\stackrel{\mathbf{鈫�}}{\mathbf{E}}}_{\mathbf{T}}\left(\mathbf{r}\mathbf{,}\mathbf{t}\right)={\stackrel{\mathbf{鈫�}}{\mathbf{E}}}_{\mathbf{0}\mathbf{T}}{\mathbf{e}}^{I\left({\mathbf{k}}_{\mathbf{T}}路\mathbf{r}路蝇\mathbf{t}\right)}$ 鈥︹�� (1)

Here,$\stackrel{鈫�}{E_{0T}}$ is magnitude of transmitted electric wave, $K_T$isevanescent wave, is radius and is frequency and is time.

Write the formula ofreflection coefficientfor polarization parallel to the plane of incidence.

$\mathbf{R}={\left|\frac{{\stackrel{\mathbf{鈫�}}{\mathbf{E}}}_{\mathbf{0}\mathbf{R}}}{{\stackrel{\mathbf{鈫�}}{\mathbf{E}}}_{\mathbf{0}\mathbf{I}}}\right|}^{\mathbf{2}}$ 鈥︹�� (2)

Here,${\mathbf{E}}_{\mathbf{0}\mathbf{R}}$ is magnitude of reflected electric wave, is magnitude of incident electricwave.
Write the formula ofreflection coefficient.

${\mathbf{E}}_{\mathbf{0}\mathbf{R}}=\left|\frac{\mathbf{1}-伪尾}{\mathbf{1}+伪尾}\right|{\mathbf{E}}_{\mathbf{0}I}$ 鈥︹�� (3)

Here, is purely imaginary, is real and is magnitude of incident electricwave.

Write the formula of (real) evanescent fields.

$\stackrel{\mathbf{鈫�}}{\mathbf{E}}\left(\mathbf{r}\mathbf{,}\mathbf{t}\right)={\stackrel{\mathbf{鈫�}}{\mathbf{E}}}_{{\mathbf{0}}_{\mathbf{T}}}{\mathbf{e}}^{I\left({\mathbf{K}}_{\mathbf{T}}路\mathbf{r}-蝇\mathbf{t}\right)}$ 鈥︹�� (4)

Here, is magnitude of transmitted electric wave, isevanescent wave, is radius and is frequency and is time.

Write the formula of (real) evanescent fields.

$\stackrel{\mathbf{鈫�}}{\mathbf{B}}\left(\mathbf{r}\mathbf{,}\mathbf{t}\right)=\frac{\mathbf{1}}{{谓}_{\mathbf{2}}}{\stackrel{\mathbf{鈫�}}{\mathbf{E}}}_{{\mathbf{0}}_{\mathbf{T}}}{\mathbf{e}}^{I\left(\mathbf{kT}路\mathbf{r}-蝇\mathbf{t}\right)}\hat{\mathbf{y}}$ $\stackrel{\mathbf{鈫�}}{\mathbf{B}}\left(\mathbf{r}\mathbf{,}\mathbf{t}\right)=\frac{\mathbf{1}}{{谓}_{\mathbf{2}}}{\stackrel{\mathbf{鈫�}}{\mathbf{E}}}_{{\mathbf{0}}_{\mathbf{T}}}{\mathbf{e}}^{I\left(\mathbf{kT}路\mathbf{r}-蝇\mathbf{t}\right)}\hat{\mathbf{y}}$ 鈥︹�� (5)

Here,${\stackrel{\mathbf{鈫�}}{\mathbf{E}}}_{\mathbf{0}\mathbf{T}}$ is magnitude of transmitted electric wave, $k_T$isevanescent wave, r is radius and is frequency and is time.

Write the formula of Maxwell鈥檚 equationsfor no charge region.

$鈭�路\mathbf{E}$ 鈥︹�� (6)

Here, E is electric field.

Write the formula of Maxwell鈥檚 equations formagnetic field.

$鈭�路B$$鈭�路\mathbf{B}$ 鈥︹�� (7)

Here, B is magnetic field.

Write the formula of Maxwell鈥檚 equations forinduction.

$$\begin{gathered} \mathbf{鈭�}\mathbf{脳}\mathbf{E}\mathbf{=}\left|\begin{array}{ccc}\hat{x}& \hat{Y}& \hat{Z}\\ \frac{鈭�}{鈭�x}& \frac{鈭�}{鈭�y}& \frac{鈭�}{鈭�z}\\ 0& E_y& 0\end{array}\right| \\ \mathbf{=}\mathbf{-}\frac{\mathbf{鈭�}{\mathbf{E}}_{\mathbf{y}}}{\mathbf{鈭�}\mathbf{z}}\hat{\mathbf{x}}\mathbf{+}\frac{\mathbf{鈭�}{\mathbf{E}}_{\mathbf{y}}}{\mathbf{鈭�}\mathbf{x}}\hat{\mathbf{z}} \end{gathered}$$ 鈥︹�� (8)

Here, ${\mathbf{E}}_{\mathbf{y}}$is electric field on y-axis, $\hat{\mathbf{x}}$ wave propagating in the X direction and $\hat{Z}$ is attenuated in the direction.

Write the formula of Maxwell鈥檚 equations forAmpere鈥檚 circuital law.

$$\begin{gathered} \mathbf{鈭�}\mathbf{脳}\mathit{B}\mathbf{=}\mathbf{\left|}\begin{array}{ccc}\hat{\mathbf{x}}& \hat{\mathbf{Y}}& \hat{\mathbf{Z}}\\ \frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{x}}& \frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{y}}& \frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{z}}\\ {\mathbf{B}}_{\mathbf{x}}& 0& {\mathbf{B}}_z\end{array}\mathbf{\right|} \\ =\left(\frac{鈭�{\mathbf{B}}_{\mathbf{z}}}{鈭�\mathbf{z}}-\frac{鈭�{\mathbf{B}}_{\mathbf{z}}}{鈭�\mathbf{x}}\right)\hat{\mathbf{y}} \end{gathered}$$ 鈥︹�� (9)

Here, ${\mathbf{B}}_{\mathbf{x}}$is magnetic field on z-axis, $\hat{\mathbf{x}}$ wave propagating in the x direction, $\hat{y}$wave propagating in the y direction and $\hat{z}$is attenuated in the z direction.

Write the formula of Poynting vector.

$\mathbf{S}=\frac{\mathbf{1}}{{渭}_{\mathbf{2}}}\left(\mathbf{E}脳\mathbf{B}\right)$ 鈥︹�� (10)

Here, $渭$ is permeability, E is electric field and B is magnetic field.

Construct the graphical representation of evanescent wave is,

Figure 1

Determine the electric field component of a wave is,

Substitute$k_T\left(\sin {胃}_T\hat{x}+\cos {胃}_T\hat{z}\right)路\left(x\hat{x}+y\hat{y}+z\hat{z}\right)$ for $k_T路r$ into equation (1).

$$\begin{gathered} k_T\left(x\sin {胃}_T+z\cos {胃}_T\right)=x{k}_T\sin {胃}_T+iz{k}_T\sqrt{{\sin }^2{胃}_T-1} \\ =kx+ikz \end{gathered}$$

Here, substitute $k=\frac{蝇n_1}{c}\sin {胃}_1$and $k=\frac{蝇}{c}\sqrt{n_1^2{\sin }^2{胃}_1-n_2^2}$ into above equation.

${\stackrel{鈫�}{E}}_T\left(r,t\right)={\stackrel{鈫�}{E}}_{0T}{e}^{-kz}{e}^{i\left(kx-蝇t\right)}$

Therefore, the value of electric field component of a wave is${\stackrel{鈫�}{E}}_T\left(r,t\right)={\stackrel{鈫�}{E}}_{0T}{e}^{-kz}{e}^{i\left(kx-蝇t\right)}$ .

Determine the reflection coefficient is the ratio of incident wave to the reflected wave. It is mathematically expressed as,

Substitute $\left(伪-尾\right)$ for ${\stackrel{鈫�}{E}}_{0R}$ and $\left(伪+尾\right)$ for ${\stackrel{鈫�}{E}}_{01}$ into equation (2).

$$\begin{gathered} R={\left|\frac{伪-尾}{伪+尾}\right|}^2 \\ R=\left(\frac{ia-尾}{ia+尾}\right)\left(\frac{-ia-尾}{-ia+尾}\right) \\ =\frac{a^2+{尾}^2}{a^2+{尾}^2} \\ =1 \end{gathered}$$

Therefore, the reflection coefficient is unity means, 100% reflection is observed.

Determine theFresnel equation for perpendicular polarization.

$E_{0R}=\left|\frac{1-伪尾}{1+伪尾}\right|E_{0I}$

Determine the reflection coefficient.

$R={\left|\frac{1-伪尾}{1+伪尾}\right|}^2$

Substitute ia for into equation (3).

$$\begin{gathered} R={\left|\frac{1-ia尾}{1+ia尾}\right|}^2 \\ =\frac{\left(1-ia尾\right)\left(1+ia尾\right)}{\left(1+ia尾\right)\left(1-ia尾\right)} \\ =1 \end{gathered}$$

Therefore, the value of reflection coefficient is R =1 .

Determine thetransmitted wave of (real) evanescent electric fields.

Substitute $kx+ikz-蝇t$ for $k_T路r$ into equation (4).

$$\begin{gathered} \left(r,t\right)={\stackrel{鈫�}{E}}_{0T}{e}^{-kt}{e}^{I\left(kx-蝇t\right)}\hat{y} \\ ={\stackrel{鈫�}{E}}_0{e}^{-kz}\cos \left(kx-蝇t\right)\hat{y} \end{gathered}$$

Phase constant should be chosen that ${\stackrel{鈫�}{E}}_{0T}$ is real.

$\text{E}\left(\text{r,t}\right)=E_0{e}^{-kz}\cos \left(kx-蝇t\right)\hat{y}$

Therefore, the value of transmitted wave of (real) evanescent fields is .

Determine thetransmitted wave of (real) evanescent magnetic fields.

Substitute $kx+ikz-蝇t$ for, $\frac{ck}{蝇n_2}$ for $\sin {胃}_T$ and $i\frac{ck}{蝇n_2}$ for $\cos {胃}_T$ into equation (5).

$$\begin{gathered} \stackrel{鈫�}{B}\left(r,t\right)=\frac{1}{{谓}_2}{\stackrel{鈫�}{E}}_{0T}{e}^{-kz}{e}^{I\left(kx-蝇t\right)}\left(-i\frac{ck}{蝇n_2}\hat{x}+\frac{ck}{蝇n_2}\hat{z}\right) \\ =\frac{1}{{谓}_2}{E}_0{e}^{-kz}\frac{c}{蝇n_2}Re\left\{\left[\cos \left(kx-蝇t\right)+i\sin \left(kx-蝇t\right)\right]\left[-ik\hat{x}+k\hat{z}\right]\right\} \\ =\frac{1}{蝇}{E}_0{e}^{-kz}\left[k\sin \left(kx-蝇t\right)\hat{x}+k\cos \left(kx-蝇t\right)\hat{z}\right] \end{gathered}$$

Here, ${谓}_2=\frac{c}{n_2}$,

Therefore, the value of transmitted wave (real) evanescent electric and magnetic fields are $\text{E}\left(\text{r,t}\right)=E_0{e}^{-kz}\cos \left(kx-蝇t\right)\hat{y}$and $\text{B}\left(\text{r,t}\right)=\frac{E_0}{蝇}{e}^{-kz}\left[k\sin \left(kx-蝇t\right)\hat{x}+k\cos \left(kx-蝇t\right)\hat{z}\right]$.

Determine the Maxwell鈥檚 equationsfor no charge region.

Substitute $E_0{e}^{-kz}\cos \left(kx-蝇t\right)\hat{y}$ for E into equation (6).

$$\begin{gathered} 鈭�路\text{E}=\frac{鈭�}{鈭�y}\left[E_0{e}^{-kz}\cos \left(kx-蝇t\right)\right] \\ =0 \end{gathered}$$

Therefore, the value of fields in Maxwell鈥檚 equations is$鈭�路\text{E}=\text{0}$.

Determine the fields in Maxwell鈥檚 equations is $鈭�路B=\text{0}$.

Substitute $\frac{E_0}{蝇}{e}^{-kz}\left[k\sin \left(kx-蝇t\right)\hat{x}+k\cos \left(kx-蝇t\right)\hat{z}\right]$ for B into equation (7).

$$\begin{gathered} 鈭�路\text{B}=\frac{鈭�}{鈭�x}\left[\frac{E_0}{蝇}{e}^{-kz}k\sin \left(kx-蝇t\right)\right]+\frac{鈭�}{鈭�z}\left[\frac{E_0}{蝇}{e}^{-kz}k\cos \left(kx-蝇t\right)\right] \\ =\frac{E_0}{蝇}\left[e^{-kz}kK\cos \left(kx-蝇t\right)-k{e}^{kz}k\cos \left(kx-蝇t\right)\right] \\ =0 \end{gathered}$$

Therefore, the value of fields in Maxwell鈥檚 equations is$鈭�路\text{B}=0$ .

Determine the Maxwell鈥檚 equations forinduction.

Substitute $E_0{e}^{-kz}\cos \left(kx-蝇t\right)\hat{y}$ for into equation (8).

$$\begin{gathered} 鈭�脳\text{E}=-k{E}_0{e}^{-kz}\cos \left(kx-蝇t\right)\hat{x}-E_0{e}^{-kz}k\sin \left(kx-蝇t\right)\hat{z} \\ -\frac{鈭�\text{B}}{鈭�t}==-k{E}_0{e}^{-kz}\cos \left(kx-蝇t\right)\hat{x}-E_0{e}^{-kz}k\sin \left(kx-蝇t\right)\hat{z} \end{gathered}$$

Therefore, thevalue of fields in Maxwell鈥檚 equations is $鈭�脳\text{E}=-\frac{鈭�\text{B}}{鈭�t}$.

Determine theMaxwell鈥檚 equations forinduction.

Substitute $\frac{E_0}{蝇}{e}^{-kz}\left[k\sin \left(kx-蝇t\right)\hat{x}+k\cos \left(kx-蝇t\right)\hat{z}\right]$for B into equation (9).

$$\begin{gathered} 鈭�脳\text{B}=\left[-\frac{E_0}{蝇}{k}^2{e}^{-kz}\sin \left(kx-蝇t\right)+-\frac{E_0}{蝇}{k}^2{e}^{kz}\sin \left(kx-蝇t\right)\right]\hat{y} \\ =\left(k^2-k^2\right)\frac{E_0}{蝇}{e}^{-kz}\sin \left(kx-蝇t\right)\hat{y} \end{gathered}$$

Substitute $\left(\frac{蝇n_1}{c}\sin {胃}_1\right)$ for K and $\left(\frac{蝇}{c}\sqrt{n_1^2{\sin }^2{胃}_1-n_2^2}\right)$ for K into above equation.

$$\begin{gathered} k^2-k^2={\left(\frac{蝇}{c}\right)}^2\left[n_1^2{\sin }^2{胃}_1-{\left(n_1\sin {胃}_1\right)}^2+{\left(n_2\right)}^2\right] \\ ={\left(\frac{n_2蝇}{c}\right)}^2 \\ ={蝇}^2{蔚}_2{渭}_2 \end{gathered}$$

Here, $\frac{n_2^2}{c^2}={蔚}_2{渭}_2$

Therefore,

$鈭�脳\text{B}={蔚}_2{渭}_2蝇E_0{e}^{-kz}\sin \left(kx-蝇t\right)\hat{y}$

Compute ${渭}_2{蔚}_2\frac{鈭�\text{E}}{鈭�t}$

${渭}_2{蔚}_2\frac{鈭�\text{E}}{鈭�t}={渭}_2{蔚}_2{E}_0{e}^{-kz}蝇\sin \left(kx-蝇t\right)\hat{y}$

Therefore, thevalue of fields in Maxwell鈥檚 equations is $鈭�脳\text{B}=渭蔚\frac{鈭�\text{E}}{鈭�t}$.

Hence, the fields in part (d) satisfy all the Maxwell鈥檚 equations.

Determine the Poynting vector.

Substitute $E_0{e}^{-kz}\cos \left(kx-蝇t\right)\hat{y}$ for E and $\frac{E_0}{蝇}{e}^{-kz}\left[k\sin \left(kx-蝇t\right)\hat{x}+k\cos \left(kx-蝇t\right)\hat{z}\right]$ for B into equation (10)

$$\begin{gathered} S=\frac{1}{{渭}_2}\frac{E_0^2}{蝇}{e}^{kz}\left|\begin{array}{ccc}\hat{x}& \hat{y}& \hat{z}\\ 0& \cos \left(kx-蝇t\right)& 0\\ k\sin \left(kx-蝇t\right)& 0& k\sin \left(kx-蝇t\right)\end{array}\right| \\ =\frac{E_0^2}{{渭}_2蝇}{e}^{-2kz}\left[k{\cos }^2\left(kx-蝇t\right)\hat{x}-k\sin \left(kx-蝇t\right)\cos \left(kx-蝇t\right)\hat{z}\right] \end{gathered}$$

Hence, Poynting vector is $\frac{E_0^2}{{渭}_2蝇}{e}^{-2kz}\left[k{\cos }^2\left(kx-蝇t\right)\hat{x}-k\sin \left(kx-蝇t\right)\cos \left(kx-蝇t\right)\hat{z}\right]$.

Average over a complete cycle.

Therefore, the value of Poynting vector is .

Hence, as a result, when the average is calculated, it is shown that energy transmission occurs along the -direction rather than the -direction.