Chapter 12, Problem 12.18P

To determine

(a)

The reflective phase shift at the front surface of glass for the glass thickness $=\frac{\lambda }{2}.$

Answer to Problem 12.18P

The reflective phase shift at the front surface of glass is $0°.$

Explanation of Solution

Given:

Refractive index of glass $n=1.45$

Glass thickness $=\frac{\lambda }{2}$

Concept Used:

Calculate reflection coefficient $\Gamma =\frac{{\eta }_{in}-{\eta }_0}{{\eta }_{in}+{\eta }_0}.$

Use the value of input impedance as ${\eta }_{in}=j{\eta }_2\tan \beta l.$

Then, calculate phase by using the formula $\varphi ={\tan }^{-1}\left(\frac{{\Gamma }_i}{{\Gamma }_r}\right).$

Calculation:

Now there are three regions where region first is air, region two is the glass and the region three is perfectly conductor.

With the third region as the perfect conductor ${\eta }_3=0$.

Now, the input impedance for the structure as,

   ${\eta }_{in}=j{\eta }_2\tan \beta l$.

Then the reflection coefficient will be as below:

   $\begin{array}{l}\Gamma =\frac{{\eta }_{in}-{\eta }_0}{{\eta }_{in}+{\eta }_0}=\frac{j{\eta }_2\tan \beta l-{\eta }_0}{j{\eta }_2\tan \beta l+{\eta }_0}=\frac{j{\eta }_2\tan \beta l-{\eta }_0}{j{\eta }_2\tan \beta l+{\eta }_0}\times \frac{j{\eta }_2\tan \beta l-{\eta }_0}{j{\eta }_2\tan \beta l-{\eta }_0}\\ \Gamma =\frac{{({\eta }_2)}^2{\tan }^2\beta l-{({\eta }_0)}^2+j2{\eta }_0{\eta }_2\tan \beta l}{{({\eta }_2)}^2{\tan }^2\beta l+{({\eta }_0)}^2}={\Gamma }_r+j{\Gamma }_i\end{array}$

The reflective phase can be calculated as below:

We have ${\eta }_2=\left(\frac{{\eta }_0}{1.45}\right)$

   $\begin{array}{l}\varphi ={\tan }^{-1}\left(\frac{{\Gamma }_i}{{\Gamma }_r}\right)={\tan }^{-1}\left(\frac{2{\eta }_0{\eta }_2\tan \beta l}{{({\eta }_2)}^2{\tan }^2\beta l-{({\eta }_0)}^2}\right)={\tan }^{-1}\left[\frac{2\times {\eta }_0\left(\frac{{\eta }_0}{1.45}\right)\tan \beta l}{{\left(\frac{{\eta }_0}{1.45}\right)}^2{\tan }^2\beta l-{({\eta }_0)}^2}\right]\\ \varphi ={\tan }^{-1}\left[\frac{\left(2.90\right)\tan \beta l}{\tan \beta l-2.10}\right]\end{array}$

For $l=\frac{\lambda }{2}$ we will have $\beta l=\pi$

And so, we have, $\varphi \left(\frac{\lambda }{2}\right)=0$

To determine

(b)

The reflective phase shift at the front surface of glass for the glass thickness $l=\frac{\lambda }{4}.$

Answer to Problem 12.18P

The reflective phase shift at the front surface of glass is $71°.$

Explanation of Solution

Given:

Refractive index of glass $n=1.45$

glass thickness $l=\frac{\lambda }{4}$

Concept Used:

Calculate reflection coefficient $\Gamma =\frac{{\eta }_{in}-{\eta }_0}{{\eta }_{in}+{\eta }_0}.$

Use the value of input impedance as ${\eta }_{in}=j{\eta }_2\tan \beta l.$

Then, calculate phase by using the formula $\varphi ={\tan }^{-1}\left(\frac{{\Gamma }_i}{{\Gamma }_r}\right).$

Calculation:

Now there are three regions where region first is air, region two is the glass and the region three is perfectly conductor.

With the third region as the perfect conductor ${\eta }_3=0$.

Now the input impedance for the structure as,

   ${\eta }_{in}=j{\eta }_2\tan \beta l$.

Then the reflection coefficient will be as below:

   $\begin{array}{l}\Gamma =\frac{{\eta }_{in}-{\eta }_0}{{\eta }_{in}+{\eta }_0}=\frac{j{\eta }_2\tan \beta l-{\eta }_0}{j{\eta }_2\tan \beta l+{\eta }_0}=\frac{j{\eta }_2\tan \beta l-{\eta }_0}{j{\eta }_2\tan \beta l+{\eta }_0}\times \frac{j{\eta }_2\tan \beta l-{\eta }_0}{j{\eta }_2\tan \beta l-{\eta }_0}\\ \Gamma =\frac{{({\eta }_2)}^2{\tan }^2\beta l-{({\eta }_0)}^2+j2{\eta }_0{\eta }_2\tan \beta l}{{({\eta }_2)}^2{\tan }^2\beta l+{({\eta }_0)}^2}={\Gamma }_r+j{\Gamma }_i\end{array}$

The reflective phase can be calculated as below: -

We have ${\eta }_2=\left(\frac{{\eta }_0}{1.45}\right)$

   $\begin{array}{l}\varphi ={\tan }^{-1}\left(\frac{{\Gamma }_i}{{\Gamma }_r}\right)={\tan }^{-1}\left(\frac{2{\eta }_0{\eta }_2\tan \beta l}{{({\eta }_2)}^2{\tan }^2\beta l-{({\eta }_0)}^2}\right)={\tan }^{-1}\left[\frac{2\times {\eta }_0\left(\frac{{\eta }_0}{1.45}\right)\tan \beta l}{{\left(\frac{{\eta }_0}{1.45}\right)}^2{\tan }^2\beta l-{({\eta }_0)}^2}\right]\\ \varphi ={\tan }^{-1}\left[\frac{\left(2.90\right)\tan \beta l}{\tan \beta l-2.10}\right]\end{array}$

For $l=\frac{\lambda }{4}$ we will have $\beta l=\frac{\pi }{2}$

And so, we have, $\varphi \left(\frac{\lambda }{4}\right)={\tan }^{-1}\left(2.90\right)=71°$

To determine

(c)

The reflective phase shift at the front surface of glass for the glass thickness $l=\frac{\lambda }{8}.$

Answer to Problem 12.18P

The reflective phase shift at the front surface of glass is $-69.2°\text{or}291°.$

Explanation of Solution

Given:

Refractive index of glass $n=1.45$

glass thickness $l=\frac{\lambda }{8}$

Concept Used:

Calculate reflection coefficient $\Gamma =\frac{{\eta }_{in}-{\eta }_0}{{\eta }_{in}+{\eta }_0}.$

Use the value of input impedance as ${\eta }_{in}=j{\eta }_2\tan \beta l.$

Then, calculate phase by using the formula $\varphi ={\tan }^{-1}\left(\frac{{\Gamma }_i}{{\Gamma }_r}\right).$

Calculation:

Now there are three regions where region first is air, region two is the glass and the region three is perfectly conductor.

With the third region as the perfect conductor ${\eta }_3=0$.

Now the input impedance for the structure as,

   ${\eta }_{in}=j{\eta }_2\tan \beta l$.

Then the reflection coefficient will be as below: -

   $\begin{array}{l}\Gamma =\frac{{\eta }_{in}-{\eta }_0}{{\eta }_{in}+{\eta }_0}=\frac{j{\eta }_2\tan \beta l-{\eta }_0}{j{\eta }_2\tan \beta l+{\eta }_0}=\frac{j{\eta }_2\tan \beta l-{\eta }_0}{j{\eta }_2\tan \beta l+{\eta }_0}\times \frac{j{\eta }_2\tan \beta l-{\eta }_0}{j{\eta }_2\tan \beta l-{\eta }_0}\\ \Gamma =\frac{{({\eta }_2)}^2{\tan }^2\beta l-{({\eta }_0)}^2+j2{\eta }_0{\eta }_2\tan \beta l}{{({\eta }_2)}^2{\tan }^2\beta l+{({\eta }_0)}^2}={\Gamma }_r+j{\Gamma }_i\end{array}$

The reflective phase can be calculated as below:

We have ${\eta }_2=\left(\frac{{\eta }_0}{1.45}\right)$

   $\begin{array}{l}\varphi ={\tan }^{-1}\left(\frac{{\Gamma }_i}{{\Gamma }_r}\right)={\tan }^{-1}\left(\frac{2{\eta }_0{\eta }_2\tan \beta l}{{({\eta }_2)}^2{\tan }^2\beta l-{({\eta }_0)}^2}\right)={\tan }^{-1}\left[\frac{2\times {\eta }_0\left(\frac{{\eta }_0}{1.45}\right)\tan \beta l}{{\left(\frac{{\eta }_0}{1.45}\right)}^2{\tan }^2\beta l-{({\eta }_0)}^2}\right]\\ \varphi ={\tan }^{-1}\left[\frac{\left(2.90\right)\tan \beta l}{\tan \beta l-2.10}\right]\end{array}$

For $l=\frac{\lambda }{8},$ we will have, $\beta l=\frac{\pi }{4}$

And so, we have, $\varphi \left(\frac{\lambda }{8}\right)={\tan }^{-1}\left[\frac{2.90}{1-2.10}\right]=-69.2°\text{or}(360°-69.2°\approx 291°).$