Chapter 10, Problem 52P

(a)

To determine

Prove that the reflection and transmission coefficients are $R={\left(\frac{n_1-n_2}{n_1+n_2}\right)}^2$ and $T=\frac{4n_1{n}_2}{{\left(n_1+n_2\right)}^2}$.

Explanation of Solution

Calculation:

Consider the expression for reflection coefficient for average power.

$R=\frac{P_{r,\text{ave}}}{P_{i,\text{ave}}}$        (1)

Here,

$P_{r,\text{ave}}$ is the reflected average power and

$P_{i,\text{ave}}$ is the incident average power.

Consider the expression for reflected average power.

$P_{r,\text{ave}}=\frac{E_{ro}^2}{2{\eta }_1}$        (2)

Here,

$E_{ro}$ is the magnitude of reflected electric field at $z=0$ and

${\eta }_1$ is the intrinsic impedance of the medium-1 (dielectric medium).

Consider the expression for incident average power.

$P_{i,\text{ave}}=\frac{E_{io}^2}{2{\eta }_1}$        (3)

Here,

$E_{io}$ is the magnitude of incident electric field at $z=0$.

From Equations (2) and (3), substitute $\left(\frac{E_{ro}^2}{2{\eta }_1}\right)$ for $P_{r,\text{ave}}$ and $\left(\frac{E_{io}^2}{2{\eta }_1}\right)$ for $P_{i,\text{ave}}$ in Equation (1).

$\begin{array}{c}R=\frac{\left(\frac{E_{ro}^2}{2{\eta }_1}\right)}{\left(\frac{E_{io}^2}{2{\eta }_1}\right)}\\ ={\left(\frac{E_{ro}}{E_{io}}\right)}^2\end{array}$

$R={\Gamma }^2\left\{\because \Gamma =\frac{E_{ro}}{E_{io}}\right\}$        (4)

Here,

$\Gamma$ is the reflection coefficient.

Write the expression for reflection coefficient.

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

Here,

${\eta }_2$ is the intrinsic impedance of medium-2 (dielectric medium).

Substitute $\left(\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}\right)$ for $\Gamma$ in Equation (4).

$R={\left(\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}\right)}^2$        (5)

Consider the expression for intrinsic impedance of the medium-1 and medium-2 (dielectric mediums).

$\begin{array}{l}{\eta }_1=\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_1}}\\ {\eta }_2=\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_2}}\end{array}$

Substitute $\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_1}}$ for ${\eta }_1$ and $\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_2}}$ for ${\eta }_2$ in Equation (5).

$R={\left(\frac{\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_2}}-\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_1}}}{\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_2}}+\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_1}}}\right)}^2$

$R={\left(\frac{\sqrt{{\mu }_{\text{o}}{\epsilon }_1}-\sqrt{{\mu }_{\text{o}}{\epsilon }_2}}{\sqrt{{\mu }_{\text{o}}{\epsilon }_1}+\sqrt{{\mu }_{\text{o}}{\epsilon }_2}}\right)}^2$        (6)

Consider the expression for refractive indices of the dielectric media.

$\begin{array}{l}{n}_1=c\sqrt{{\mu }_{\text{o}}{\epsilon }_1}\\ n_2=c\sqrt{{\mu }_{\text{o}}{\epsilon }_2}\end{array}$

Rearrange the expressions.

$\begin{array}{l}\sqrt{{\mu }_{\text{o}}{\epsilon }_1}=\frac{n_1}{c}\\ \sqrt{{\mu }_{\text{o}}{\epsilon }_2}=\frac{n_2}{c}\end{array}$

Substitute $\left(\frac{n_1}{c}\right)$ for $\sqrt{{\mu }_{\text{o}}{\epsilon }_1}$ and $\left(\frac{n_2}{c}\right)$ for $\sqrt{{\mu }_{\text{o}}{\epsilon }_2}$ in Equation (6).

$\begin{array}{c}R={\left[\frac{\left(\frac{n_1}{c}\right)-\left(\frac{n_2}{c}\right)}{\left(\frac{n_1}{c}\right)+\left(\frac{n_2}{c}\right)}\right]}^2\\ ={\left(\frac{n_1-n_2}{n_1+n_2}\right)}^2\end{array}$

Consider the expression for transmission coefficient for average power.

$T=\frac{P_{t,\text{ave}}}{P_{i,\text{ave}}}$        (7)

Here,

$P_{t,\text{ave}}$ is the transmitted average power.

Consider the expression for transmitted average power.

$P_{t,\text{ave}}=\frac{E_{to}^2}{2{\eta }_2}$        (8)

Here,

$E_{to}$ is the magnitude of transmitted electric field at $z=0$.

From Equations (3) and (8), substitute $\left(\frac{E_{to}^2}{2{\eta }_2}\right)$ for $P_{t,\text{ave}}$ and $\left(\frac{E_{io}^2}{2{\eta }_1}\right)$ for $P_{i,\text{ave}}$ in Equation (7).

$\begin{array}{c}T=\frac{\left(\frac{E_{to}^2}{2{\eta }_2}\right)}{\left(\frac{E_{io}^2}{2{\eta }_1}\right)}\\ =\left(\frac{{\eta }_1}{{\eta }_2}\right){\left(\frac{E_{to}}{E_{io}}\right)}^2\\ =\left(\frac{{\eta }_1}{{\eta }_2}\right){\left(\tau \right)}^2\left\{\because \tau =\frac{E_{to}}{E_{io}}\right\}\\ =\left(\frac{{\eta }_1}{{\eta }_2}\right){\left(1+\Gamma \right)}^2\left\{\because \tau =1+\Gamma \right\}\end{array}$

Substitute $\left(\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}\right)$ for $\Gamma$.

$T=\left(\frac{{\eta }_1}{{\eta }_2}\right){\left[1+\left(\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}\right)\right]}^2$

Substitute $\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_1}}$ for ${\eta }_1$ and $\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_2}}$ for ${\eta }_2$.

$\begin{array}{c}T=\left(\frac{\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_1}}}{\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_2}}}\right){\left[1+\left(\frac{\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_2}}-\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_1}}}{\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_2}}+\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_1}}}\right)\right]}^2\\ =\left(\frac{\sqrt{{\mu }_{\text{o}}{\epsilon }_2}}{\sqrt{{\mu }_{\text{o}}{\epsilon }_1}}\right){\left(1+\frac{\sqrt{{\mu }_{\text{o}}{\epsilon }_1}-\sqrt{{\mu }_{\text{o}}{\epsilon }_2}}{\sqrt{{\mu }_{\text{o}}{\epsilon }_1}+\sqrt{{\mu }_{\text{o}}{\epsilon }_2}}\right)}^2\end{array}$

Substitute $\left(\frac{n_1}{c}\right)$ for $\sqrt{{\mu }_{\text{o}}{\epsilon }_1}$ and $\left(\frac{n_2}{c}\right)$ for $\sqrt{{\mu }_{\text{o}}{\epsilon }_2}$.

$\begin{array}{c}T=\left[\frac{\left(\frac{n_2}{c}\right)}{\left(\frac{n_1}{c}\right)}\right]{\left[1+\frac{\left(\frac{n_1}{c}\right)-\left(\frac{n_2}{c}\right)}{\left(\frac{n_1}{c}\right)+\left(\frac{n_2}{c}\right)}\right]}^2\\ =\left(\frac{n_2}{n_1}\right){\left(1+\frac{n_1-n_2}{n_1+n_2}\right)}^2\\ =\frac{4n_1{n}_2}{{\left(n_1+n_2\right)}^2}\end{array}$

Conclusion:

Thus, the expressions of reflection and transmission coefficients are shown.

(b)

To determine

Find the ratio $\left(\frac{n_1}{n_2}\right)$ for the given data.

Answer to Problem 52P

The ratio $\left(\frac{n_1}{n_2}\right)$ is $\underset{\_}{5.828\text{ or }0.1716}$.

Explanation of Solution

Calculation:

Write the expression when the reflected and transmitted waves have the same average power.

$P_{r,\text{ave}}=P_{t,\text{ave}}$

From Equations (1) and (7), substitute $\left(R{P}_{i,\text{ave}}\right)$ for $P_{r,\text{ave}}$ and $\left(T{P}_{i,\text{ave}}\right)$ for $P_{t,\text{ave}}$.

$\begin{array}{l}\left(R{P}_{i,\text{ave}}\right)=\left(T{P}_{i,\text{ave}}\right)\\ R=T\end{array}$

From Part (a), substitute ${\left(\frac{n_1-n_2}{n_1+n_2}\right)}^2$ for R and $\frac{4n_1{n}_2}{{\left(n_1+n_2\right)}^2}$ for T.

$\begin{array}{l}{\left(\frac{n_1-n_2}{n_1+n_2}\right)}^2=\frac{4n_1{n}_2}{{\left(n_1+n_2\right)}^2}\\ {\left(n_1-n_2\right)}^2=4n_1{n}_2\\ n_1^2-6n_1{n}_2+n_2^2=0\\ {\left(\frac{n_1}{n_2}\right)}^2-6\left(\frac{n_1}{n_2}\right)+1=0\end{array}$

Simplify the quadratic expression.

$\frac{n_1}{n_2}=5.828\text{ or }0.1716$

Conclusion:

Thus, the ratio $\left(\frac{n_1}{n_2}\right)$ is $\underset{\_}{5.828\text{ or }0.1716}$.