Question

Question:According to Snell's law, when light passes from an optically dense medium into a less dense one the propagation vector bends away from the normal (Fig. 9.28). In particular, if the light is incident at the critical angle

${\mathit{\theta }}_{\mathbf{c}}\mathbf{=}{\mathbf{sin}}^{\mathbf{-}}\left(\raisebox{1ex}{$n_2$}\!\left/ \!\raisebox{-1ex}{$n_1$}\right.\right)$

Then , and the transmitted ray just grazes the surface. If exceeds , there is no refracted ray at all, only a reflected one (this is the phenomenon of total internal reflection, on which light pipes and fiber optics are based). But the fields are not zero in medium ; what we get is a so-called evanescent wave, which is rapidly attenuated and transports no energy into medium 2.26

Figure 9.28

A quick way to construct the evanescent wave is simply to quote the results of Sect. 9.3.3, with and

${\mathbf{k}}_{\mathbf{T}}={\mathbf{k}}_{\mathbf{T}}\left(\mathbf{sin}{\theta }_{\mathbf{T}}\hat{\mathbf{x}}+\mathbf{cos}{\theta }_{\mathbf{T}}\hat{\mathbf{z}}\right)$

the only change is that

$\mathbf{sin}{\theta }_{\mathbf{T}}=\frac{{\mathbf{n}}_{\mathbf{1}}}{{\mathbf{n}}_{\mathbf{2}}}\mathbf{sin}{\theta }_I$

is now greater than, and

$\mathbf{cos}{\theta }_{\mathbf{T}}=\sqrt{\mathbf{1}-{\mathbf{sin}}^{\mathbf{2}}{\theta }_{\mathbf{T}}}$

is imaginary. (Obviously, can no longer be interpreted as an angle!)

(a) Show that

${\overrightarrow{\mathbf{E}}}_{\mathbf{T}}\left(r,t\right)\mathbf{=}{\overrightarrow{\mathbf{E}}}_{{\mathbf{0}}_{\mathbf{T}}}{\mathbf{e}}^{\mathbf{-}\mathbf{kz}}{\mathbf{e}}^{\mathbf{I}\left(\mathrm{kx}-\omega t\right)}$

Where

$\mathbf{k}\mathbf{\equiv }\frac{\mathbf{\omega }}{\mathbf{c}}\sqrt{{\left(n_1{\mathrm{sin\theta }}_1\right)}^{\mathbf{2}}\mathbf{-}{\mathbf{n}}_{\mathbf{2}}^{\mathbf{2}}}$

This is a wave propagating in the direction (parallel to the interface!), and attenuated in the direction.

(b) Noting that (Eq. 9.108) is now imaginary, use Eq. 9.109 to calculate theirreflection coefficient for polarization parallel to the plane of incidence. [Notice that you get reflection, which is better than at a conducting surface (see, for example, Prob. 9.22).]

(c) Do the same for polarization perpendicular to the plane of incidence (use the results of Prob. 9.17).

(d) In the case of polarization perpendicular to the plane of incidence, show that the (real) evanescent fields are

$$\begin{gathered} \mathbf{E}\left(\mathbf{r}\mathbf{,}\mathbf{t}\right)={\mathbf{E}}_{\mathbf{0}}{\mathbf{e}}^{-\mathbf{kz}}\mathbf{cos}\left(\mathbf{kx}-\omega \mathbf{t}\right)\hat{\mathbf{y}} \\ \mathbf{B}\left(\mathbf{r}\mathbf{,}\mathbf{t}\right)=\frac{{\mathbf{E}}_{\mathbf{0}}}{\omega }{\mathbf{e}}^{-\mathbf{kz}}\left[\mathbf{ksin}\left(\mathbf{kx}-\omega \mathbf{t}\right)\hat{\mathbf{x}}+\mathbf{kcos}\left(\mathbf{kx}-\omega \mathbf{t}\right)\hat{\mathbf{z}}\right] \end{gathered}$$

(e) Check that the fields in (d) satisfy all of Maxwell's equations (Eq. 9.67).

(f) For the fields in (d), construct the Poynting vector, and show that, on average, no energy is transmitted in the z direction.

Short answer

(a) The value ofelectric field component of a wave is${\overrightarrow{E}}_T\left(r,t\right)={\overrightarrow{E}}_{0_T}{e}^{-kz}{e}^{i\left(kx-\omega t\right)}$.

(b) The value of reflection coefficientfor polarization parallel to the plane of incidence is R =1.

(c) The value of reflection coefficient is R = 1 .

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

(e)

(i) The value of fields in Maxwell’s equations is$\nabla ·\text{E}=\text{0}$ .

(ii) The value of fields in Maxwell’s equations is $\nabla ·\text{B}=0$.

(iii) The value of fields in Maxwell’s equations is $\nabla \times \text{E}=-\frac{\partial \text{B}}{\partial t}$.

(iv) The value of fields in Maxwell’s equations is $\nabla \times \text{B}=\mu \epsilon \frac{\partial \text{E}}{\partial t}$.

(f) The value of Poynting vector is .

Step-by-step solution

Write the given data from the question.

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 $\overrightarrow{{\mathrm{E}}_{\mathrm{T}}}\left(r.t\right)=\overrightarrow{{\mathrm{E}}_{\mathrm{T}}}{e}^{I(kx-wt)}$.

Consider the Maxwell’s equations are,

1. The Gauss law for no charge region,

$\nabla ·\text{E}=0$
2. The Gauss law for magnetic field,

$\nabla ·\text{B}=0$

3. The Maxwell law of induction is expressed as,

$\nabla \times \text{E}=-\frac{1}{c}\frac{\partial \text{B}}{\partial t}$

4. The Modified Ampere’s circuital law,

$\nabla \times \text{B}={\mu }_0\left(\text{J}+{\epsilon }_0\frac{\partial \text{E}}{\partial t}\right)$

Consider the Fresnel equation for perpendicular polarization.

$$\begin{gathered} {\overrightarrow{E}}_{0T}=\frac{2}{1+\alpha \beta }{\overrightarrow{E}}_{0I} \\ {\overrightarrow{E}}_{\text{0R}}=\frac{1-\alpha \beta }{1+\alpha \beta }{\overrightarrow{E}}_{0I} \end{gathered}$$

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

Determine the formula of electric field component of a wave, reflection coefficient for polarization parallel to the plane of incidence, reflection coefficient, (real) evanescent fields, fields in Maxwell’s equations and Poynting vector.

Write the formula ofelectric field component of a wave.

${\overrightarrow{\mathbf{E}}}_{\mathbf{T}}\left(\mathbf{r}\mathbf{,}\mathbf{t}\right)={\overrightarrow{\mathbf{E}}}_{\mathbf{0}\mathbf{T}}{\mathbf{e}}^{I\left({\mathbf{k}}_{\mathbf{T}}·\mathbf{r}·\omega \mathbf{t}\right)}$ …… (1)

Here,$\overrightarrow{E_{0T}}$ 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.

$\mathbf{R}={\left|\frac{{\overrightarrow{\mathbf{E}}}_{\mathbf{0}\mathbf{R}}}{{\overrightarrow{\mathbf{E}}}_{\mathbf{0}\mathbf{I}}}\right|}^{\mathbf{2}}$ …… (2)

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

${\mathbf{E}}_{\mathbf{0}\mathbf{R}}=\left|\frac{\mathbf{1}-\alpha \beta }{\mathbf{1}+\alpha \beta }\right|{\mathbf{E}}_{\mathbf{0}I}$ …… (3)

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

Write the formula of (real) evanescent fields.

$\overrightarrow{\mathbf{E}}\left(\mathbf{r}\mathbf{,}\mathbf{t}\right)={\overrightarrow{\mathbf{E}}}_{{\mathbf{0}}_{\mathbf{T}}}{\mathbf{e}}^{I\left({\mathbf{K}}_{\mathbf{T}}·\mathbf{r}-\omega \mathbf{t}\right)}$ …… (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.

$\overrightarrow{\mathbf{B}}\left(\mathbf{r}\mathbf{,}\mathbf{t}\right)=\frac{\mathbf{1}}{{\nu }_{\mathbf{2}}}{\overrightarrow{\mathbf{E}}}_{{\mathbf{0}}_{\mathbf{T}}}{\mathbf{e}}^{I\left(\mathbf{kT}·\mathbf{r}-\omega \mathbf{t}\right)}\hat{\mathbf{y}}$ $\overrightarrow{\mathbf{B}}\left(\mathbf{r}\mathbf{,}\mathbf{t}\right)=\frac{\mathbf{1}}{{\nu }_{\mathbf{2}}}{\overrightarrow{\mathbf{E}}}_{{\mathbf{0}}_{\mathbf{T}}}{\mathbf{e}}^{I\left(\mathbf{kT}·\mathbf{r}-\omega \mathbf{t}\right)}\hat{\mathbf{y}}$ …… (5)

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

Write the formula of Maxwell’s equationsfor no charge region.

$\nabla ·\mathbf{E}$ …… (6)

Here, E is electric field.

Write the formula of Maxwell’s equations formagnetic field.

$\nabla ·B$$\nabla ·\mathbf{B}$ …… (7)

Here, B is magnetic field.

Write the formula of Maxwell’s equations forinduction.

$$\begin{gathered} \mathbf{\nabla }\mathbf{\times }\mathbf{E}\mathbf{=}\left|\begin{array}{ccc}\hat{x}& \hat{Y}& \hat{Z}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 0& E_y& 0\end{array}\right| \\ \mathbf{=}\mathbf{-}\frac{\mathbf{\partial }{\mathbf{E}}_{\mathbf{y}}}{\mathbf{\partial }\mathbf{z}}\hat{\mathbf{x}}\mathbf{+}\frac{\mathbf{\partial }{\mathbf{E}}_{\mathbf{y}}}{\mathbf{\partial }\mathbf{x}}\hat{\mathbf{z}} \end{gathered}$$ …… (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’s equations forAmpere’s circuital law.

$$\begin{gathered} \mathbf{\nabla }\mathbf{\times }\mathit{B}\mathbf{=}\mathbf{\left|}\begin{array}{ccc}\hat{\mathbf{x}}& \hat{\mathbf{Y}}& \hat{\mathbf{Z}}\\ \frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}& \frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{y}}& \frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{z}}\\ {\mathbf{B}}_{\mathbf{x}}& 0& {\mathbf{B}}_z\end{array}\mathbf{\right|} \\ =\left(\frac{\partial {\mathbf{B}}_{\mathbf{z}}}{\partial \mathbf{z}}-\frac{\partial {\mathbf{B}}_{\mathbf{z}}}{\partial \mathbf{x}}\right)\hat{\mathbf{y}} \end{gathered}$$ …… (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.

$\mathbf{S}=\frac{\mathbf{1}}{{\mu }_{\mathbf{2}}}\left(\mathbf{E}\times \mathbf{B}\right)$ …… (10)

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

(a) Determine the value of electric field component of a wave

Construct the graphical representation of evanescent wave is,

Figure 1

Determine the electric field component of a wave is,

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

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

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

${\overrightarrow{E}}_T\left(r,t\right)={\overrightarrow{E}}_{0T}{e}^{-kz}{e}^{i\left(kx-\omega t\right)}$

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

(b) Determine the value of reflection coefficient for polarization parallel to the plane of incidence.

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

Substitute $\left(\alpha -\beta \right)$ for ${\overrightarrow{E}}_{0R}$ and $\left(\alpha +\beta \right)$ for ${\overrightarrow{E}}_{01}$ into equation (2).

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

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

(c) Determine the value of reflection coefficient.

Determine theFresnel equation for perpendicular polarization.

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

Determine the reflection coefficient.

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

Substitute ia for into equation (3).

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

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

(d) Determine the (real) evanescent fields.

Determine thetransmitted wave of (real) evanescent electric fields.

Substitute $kx+ikz-\omega t$ for $k_T·r$ into equation (4).

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

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

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

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

Determine thetransmitted wave of (real) evanescent magnetic fields.

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

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

Here, ${\nu }_2=\frac{c}{n_2}$,

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

(e) Determine the Maxwell’s equations.

Determine the Maxwell’s equationsfor no charge region.

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

$$\begin{gathered} \nabla ·\text{E}=\frac{\partial }{\partial y}\left[E_0{e}^{-kz}\cos \left(kx-\omega t\right)\right] \\ =0 \end{gathered}$$

Therefore, the value of fields in Maxwell’s equations is$\nabla ·\text{E}=\text{0}$.

Determine the fields in Maxwell’s equations is $\nabla ·B=\text{0}$.

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

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

Therefore, the value of fields in Maxwell’s equations is$\nabla ·\text{B}=0$ .

Determine the Maxwell’s equations forinduction.

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

$$\begin{gathered} \nabla \times \text{E}=-k{E}_0{e}^{-kz}\cos \left(kx-\omega t\right)\hat{x}-E_0{e}^{-kz}k\sin \left(kx-\omega t\right)\hat{z} \\ -\frac{\partial \text{B}}{\partial t}==-k{E}_0{e}^{-kz}\cos \left(kx-\omega t\right)\hat{x}-E_0{e}^{-kz}k\sin \left(kx-\omega t\right)\hat{z} \end{gathered}$$

Therefore, thevalue of fields in Maxwell’s equations is $\nabla \times \text{E}=-\frac{\partial \text{B}}{\partial t}$.

Determine theMaxwell’s equations forinduction.

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

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

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

$$\begin{gathered} k^2-k^2={\left(\frac{\omega }{c}\right)}^2\left[n_1^2{\sin }^2{\theta }_1-{\left(n_1\sin {\theta }_1\right)}^2+{\left(n_2\right)}^2\right] \\ ={\left(\frac{n_2\omega }{c}\right)}^2 \\ ={\omega }^2{\epsilon }_2{\mu }_2 \end{gathered}$$

Here, $\frac{n_2^2}{c^2}={\epsilon }_2{\mu }_2$

Therefore,

$\nabla \times \text{B}={\epsilon }_2{\mu }_2\omega E_0{e}^{-kz}\sin \left(kx-\omega t\right)\hat{y}$

Compute ${\mu }_2{\epsilon }_2\frac{\partial \text{E}}{\partial t}$

${\mu }_2{\epsilon }_2\frac{\partial \text{E}}{\partial t}={\mu }_2{\epsilon }_2{E}_0{e}^{-kz}\omega \sin \left(kx-\omega t\right)\hat{y}$

Therefore, thevalue of fields in Maxwell’s equations is $\nabla \times \text{B}=\mu \epsilon \frac{\partial \text{E}}{\partial t}$.

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

(f) Determine the Poynting vector.

Determine the Poynting vector.

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

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

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

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.