Chapter 12, Problem 12.9P

To determine

(a)

The percentage of the energy incident on the boundary is reflected.

Answer to Problem 12.9P

The percentage of the energy incident on the boundary reflected is $=6.25\times {10}^{-2}.$

Explanation of Solution

Given:

Two regions with region $1$ : $z<0$ and region $2$ : $z>0$ are perfect dielectrics.

Radian frequency of uniform plane wave $=3\times {10}^{10}\text{rad/s}\text{.}$

Wavelengths in two regions

   ${\text{λ}}_1=5\text{cm},{\text{λ}}_2=3\text{cm}$

Concept Used:

Use below formula to calculate the value of reflection coefficient which gives the value of the percentage of the energy incident on the boundary is reflected.

   $\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}=\frac{{\eta }_0\sqrt{1/{\epsilon }_{r2}^{\text{'}}}-{\eta }_0\sqrt{1/{\epsilon }_{r1}^{\text{'}}}}{{\eta }_0\sqrt{1/{\epsilon }_{r2}^{\text{'}}}+{\eta }_0\sqrt{1/{\epsilon }_{r1}^{\text{'}}}}=\frac{\sqrt{{\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}}-1}{\sqrt{{\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}}+1}$

Calculation:

We first use below formula:

   ${\epsilon }_{r1}^{\text{'}}={\left(\frac{2\pi c}{{\text{λ}}_1\omega }\right)}^2$ and ${\epsilon }_{r2}^{\text{'}}={\left(\frac{2\pi c}{{\text{λ}}_2\omega }\right)}^2$

So, we have ${\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}={\left({\text{λ}}_2/{\text{λ}}_1\right)}^2$

And with the $\mu ={\mu }_0$ in both the regions, we will calculate:

   $\begin{array}{c}\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}\\ =\frac{{\eta }_0\sqrt{1/{\epsilon }_{r2}^{\text{'}}}-{\eta }_0\sqrt{1/{\epsilon }_{r1}^{\text{'}}}}{{\eta }_0\sqrt{1/{\epsilon }_{r2}^{\text{'}}}+{\eta }_0\sqrt{1/{\epsilon }_{r1}^{\text{'}}}}\\ =\frac{\sqrt{{\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}}-1}{\sqrt{{\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}}+1}\\ =\frac{\left({\text{λ}}_2/{\text{λ}}_1\right)-1}{\left({\text{λ}}_2/{\text{λ}}_1\right)+1}\\ =\frac{{\text{λ}}_2-{\text{λ}}_1}{{\text{λ}}_2{\text{+λ}}_1}\\ =\frac{3-5}{3+5}\\ =\frac{-2}{8}\\ =-\frac{1}{4}\end{array}$

The fraction of the incident energy that is reflected is then ${\left|\Gamma \right|}^2=1/16=6.25\times {10}^{-2}.$

To determine

(b)

The percentage of the energy incident on the boundary is transmitted.

Answer to Problem 12.9P

The percentage of the energy incident on the boundary transmitted is $=\mathrm{0.938.}$

Explanation of Solution

Given:

Two regions with region $1$ : $z<0$ and region $2$ : $z>0$ are perfect dielectrics.

Radian frequency of uniform plane wave $=3\times {10}^{10}\text{rad/s}\text{.}$

Wavelengths in two regions

   ${\text{λ}}_1=5\text{cm},{\text{λ}}_2=3\text{cm}$

Concept Used:

Use below formula to calculate the value of reflection coefficient which gives the value of the percentage of the energy incident on the boundary is reflected.

   $\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}=\frac{{\eta }_0\sqrt{1/{\epsilon }_{r2}^{\text{'}}}-{\eta }_0\sqrt{1/{\epsilon }_{r1}^{\text{'}}}}{{\eta }_0\sqrt{1/{\epsilon }_{r2}^{\text{'}}}+{\eta }_0\sqrt{1/{\epsilon }_{r1}^{\text{'}}}}=\frac{\sqrt{{\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}}-1}{\sqrt{{\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}}+1}$

Then calculatethe percentage of the energy incident on the boundary transmitted by using formula $1-{\left|\Gamma \right|}^2.$

Calculation:

We first use below formula.

   ${\epsilon }_{r1}^{\text{'}}={\left(\frac{2\pi c}{{\text{λ}}_1\omega }\right)}^2$ and ${\epsilon }_{r2}^{\text{'}}={\left(\frac{2\pi c}{{\text{λ}}_2\omega }\right)}^2$

So, we have ${\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}={\left({\text{λ}}_2/{\text{λ}}_1\right)}^2$

And with the $\mu ={\mu }_0$ in both the regions, we will calculate:

   $\begin{array}{l}\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}=\frac{{\eta }_0\sqrt{1/{\epsilon }_{r2}^{\text{'}}}-{\eta }_0\sqrt{1/{\epsilon }_{r1}^{\text{'}}}}{{\eta }_0\sqrt{1/{\epsilon }_{r2}^{\text{'}}}+{\eta }_0\sqrt{1/{\epsilon }_{r1}^{\text{'}}}}=\frac{\sqrt{{\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}}-1}{\sqrt{{\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}}+1}=\frac{\left({\text{λ}}_2/{\text{λ}}_1\right)-1}{\left({\text{λ}}_2/{\text{λ}}_1\right)+1}\\ =\frac{{\text{λ}}_2-{\text{λ}}_1}{{\text{λ}}_2{\text{+λ}}_1}=\frac{3-5}{3+5}=-\frac{1}{4}\end{array}$

The fraction of the transmitted energy that is reflected is then $1-{\left|\Gamma \right|}^2=1-6.25\times {10}^{-2}=\mathrm{0.938.}$

To determine

(c)

The standing wave ratio in region $1$.

Answer to Problem 12.9P

The standing wave ratio in region $1$ is $s=1.67$.

Explanation of Solution

Given:

Two regions with region $1$ : $z<0$ and region $2$ : $z>0$ are perfect dielectrics.

Radian frequency of uniform plane wave $=3\times {10}^{10}\text{rad/s}\text{.}$

Wavelengths in two regions

   ${\text{λ}}_1=5\text{cm},{\text{λ}}_2=3\text{cm}$

Concept Used:

Using the below formula to calculate the value of reflection coefficient which gives the value of the percentage of the energy incident on the boundary is reflected.

   $\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}=\frac{{\eta }_0\sqrt{1/{\epsilon }_{r2}^{\text{'}}}-{\eta }_0\sqrt{1/{\epsilon }_{r1}^{\text{'}}}}{{\eta }_0\sqrt{1/{\epsilon }_{r2}^{\text{'}}}+{\eta }_0\sqrt{1/{\epsilon }_{r1}^{\text{'}}}}=\frac{\sqrt{{\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}}-1}{\sqrt{{\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}}+1}$

Use $s=\frac{1+\left|\Gamma \right|}{1-\left|\Gamma \right|}$ to calculate the standing wave ratio.

Calculation:

We first use below formula.

   ${\epsilon }_{r1}^{\text{'}}={\left(\frac{2\pi c}{{\text{λ}}_1\omega }\right)}^2$ and ${\epsilon }_{r2}^{\text{'}}={\left(\frac{2\pi c}{{\text{λ}}_2\omega }\right)}^2$

So, we have ${\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}={\left({\text{λ}}_2/{\text{λ}}_1\right)}^2$

And with the $\mu ={\mu }_0$ in both the regions, we will calculate

   $\begin{array}{l}\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}=\frac{{\eta }_0\sqrt{1/{\epsilon }_{r2}^{\text{'}}}-{\eta }_0\sqrt{1/{\epsilon }_{r1}^{\text{'}}}}{{\eta }_0\sqrt{1/{\epsilon }_{r2}^{\text{'}}}+{\eta }_0\sqrt{1/{\epsilon }_{r1}^{\text{'}}}}=\frac{\sqrt{{\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}}-1}{\sqrt{{\epsilon }_{r1}^{\text{'}}/{\epsilon }_{r2}^{\text{'}}}+1}=\frac{\left({\text{λ}}_2/{\text{λ}}_1\right)-1}{\left({\text{λ}}_2/{\text{λ}}_1\right)+1}\\ =\frac{{\text{λ}}_2-{\text{λ}}_1}{{\text{λ}}_2{\text{+λ}}_1}=\frac{3-5}{3+5}=-\frac{1}{4}\end{array}$

The standing wave ratio is:

   $s=\frac{1+\left|\Gamma \right|}{1-\left|\Gamma \right|}=\frac{1+1/4}{1-1/4}=\frac{5}{3}=1.67$