Chapter 10, Problem 71P

To determine

Prove the given reflection and transmission coefficients for oblique incidence.

Explanation of Solution

Calculation:

Consider the expression of reflection coefficient in the parallel polarization of oblique incidence.

${\Gamma }_{\parallel }=\frac{{\eta }_2\cos \left({\theta }_t\right)-{\eta }_1\cos \left({\theta }_i\right)}{{\eta }_2\cos \left({\theta }_t\right)+{\eta }_1\cos \left({\theta }_i\right)}$        (1)

Here,

${\eta }_1$ is the intrinsic impedance of the medium-1,

${\eta }_2$ is the intrinsic impedance for medium-2,

${\theta }_t$ is the angle of transmission, and

${\theta }_i$ is the angle of incidence.

Consider the expression for intrinsic impedance for medium-1 (nonmagnetic medium).

${\eta }_1=\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r1}}}$

Consider the expression for intrinsic impedance for medium-2 (nonmagnetic medium).

${\eta }_2=\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}}$

Substitute $\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r1}}}$ for ${\eta }_1$ and $\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}}$ for ${\eta }_2$ in Equation (1).

$\begin{array}{c}{\Gamma }_{\parallel }=\frac{\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}}\cos \left({\theta }_t\right)-\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r1}}}\cos \left({\theta }_i\right)}{\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}}\cos \left({\theta }_t\right)+\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r1}}}\cos \left({\theta }_i\right)}\\ =\frac{\sqrt{\frac{1}{{\epsilon }_{r2}}}\cos \left({\theta }_t\right)-\sqrt{\frac{1}{{\epsilon }_{r1}}}\cos \left({\theta }_i\right)}{\sqrt{\frac{1}{{\epsilon }_{r2}}}\cos \left({\theta }_t\right)+\sqrt{\frac{1}{{\epsilon }_{r1}}}\cos \left({\theta }_i\right)}\\ =\frac{\cos \left({\theta }_t\right)-\frac{\sqrt{{\epsilon }_{r2}}}{\sqrt{{\epsilon }_{r1}}}\cos \left({\theta }_i\right)}{\cos \left({\theta }_t\right)+\frac{\sqrt{{\epsilon }_{r2}}}{\sqrt{{\epsilon }_{r1}}}\cos \left({\theta }_i\right)}\\ =\frac{\cos \left({\theta }_t\right)-\frac{\sin \left({\theta }_i\right)}{\sin \left({\theta }_t\right)}\cos \left({\theta }_i\right)}{\cos \left({\theta }_t\right)+\frac{\sin \left({\theta }_i\right)}{\sin \left({\theta }_t\right)}\cos \left({\theta }_i\right)}\left\{\because \frac{\sqrt{{\epsilon }_{r2}}}{\sqrt{{\epsilon }_{r1}}}=\frac{\sin \left({\theta }_i\right)}{\sin \left({\theta }_t\right)}\right\}\end{array}$

Simplify the expression.

$\begin{array}{c}{\Gamma }_{\parallel }=\frac{\sin \left({\theta }_t\right)\cos \left({\theta }_t\right)-\sin \left({\theta }_i\right)\cos \left({\theta }_i\right)}{\sin \left({\theta }_t\right)\cos \left({\theta }_t\right)+\sin \left({\theta }_i\right)\cos \left({\theta }_i\right)}\\ =\frac{2\sin \left({\theta }_t\right)\cos \left({\theta }_t\right)-2\sin \left({\theta }_i\right)\cos \left({\theta }_i\right)}{2\sin \left({\theta }_t\right)\cos \left({\theta }_t\right)+2\sin \left({\theta }_i\right)\cos \left({\theta }_i\right)}\\ =\frac{\sin \left(2{\theta }_t\right)-\sin \left(2{\theta }_i\right)}{\sin \left(2{\theta }_t\right)+\sin \left(2{\theta }_i\right)}\\ =\frac{\sin \left({\theta }_t-{\theta }_i\right)\cos \left({\theta }_t+{\theta }_i\right)}{\sin \left({\theta }_t+{\theta }_i\right)\cos \left({\theta }_t-{\theta }_i\right)}\end{array}$

Simplify the expression.

$\begin{array}{c}{\Gamma }_{\parallel }=\left[\frac{\sin \left({\theta }_t-{\theta }_i\right)}{\cos \left({\theta }_t-{\theta }_i\right)}\right]\left\{\frac{1}{\left[\frac{\sin \left({\theta }_t+{\theta }_i\right)}{\cos \left({\theta }_t+{\theta }_i\right)}\right]}\right\}\\ =\frac{\tan \left({\theta }_t-{\theta }_i\right)}{\tan \left({\theta }_t+{\theta }_i\right)}\end{array}$

Consider the expression of transmission coefficient in the parallel polarization of oblique incidence.

${\tau }_{\parallel }=\frac{2{\eta }_2\cos \left({\theta }_i\right)}{{\eta }_2\cos \left({\theta }_t\right)+{\eta }_1\cos \left({\theta }_i\right)}$

Substitute $\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r1}}}$ for ${\eta }_1$ and $\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}}$ for ${\eta }_2$.

$\begin{array}{c}{\tau }_{\parallel }=\frac{2\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}}\cos \left({\theta }_i\right)}{\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}}\cos \left({\theta }_t\right)+\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r1}}}\cos \left({\theta }_i\right)}\\ =\frac{2\sqrt{\frac{1}{{\epsilon }_{r2}}}\cos \left({\theta }_i\right)}{\sqrt{\frac{1}{{\epsilon }_{r2}}}\cos \left({\theta }_t\right)+\sqrt{\frac{1}{{\epsilon }_{r1}}}\cos \left({\theta }_i\right)}\\ =\frac{2\cos \left({\theta }_i\right)}{\cos \left({\theta }_t\right)+\frac{\sqrt{{\epsilon }_{r2}}}{\sqrt{{\epsilon }_{r1}}}\cos \left({\theta }_i\right)}\\ =\frac{2\cos \left({\theta }_i\right)}{\cos \left({\theta }_t\right)+\frac{\sin \left({\theta }_i\right)}{\sin \left({\theta }_t\right)}\cos \left({\theta }_i\right)}\end{array}$

Simplify the expression.

$\begin{array}{c}{\tau }_{\parallel }=\frac{2\sin \left({\theta }_t\right)\cos \left({\theta }_i\right)}{\sin \left({\theta }_t\right)\cos \left({\theta }_t\right)\left[{\sin }^2\left({\theta }_i\right)+{\cos }^2\left({\theta }_i\right)\right]+\sin \left({\theta }_i\right)\cos \left({\theta }_i\right)\left[{\sin }^2\left({\theta }_t\right)+{\cos }^2\left({\theta }_t\right)\right]}\\ =\frac{2\sin \left({\theta }_t\right)\cos \left({\theta }_i\right)}{\left[\sin \left({\theta }_t\right)\cos \left({\theta }_i\right)+\sin \left({\theta }_i\right)\cos \left({\theta }_t\right)\right]\left[\sin \left({\theta }_i\right)\sin \left({\theta }_t\right)+\cos \left({\theta }_i\right)\cos \left({\theta }_t\right)\right]}\\ =\frac{2\cos \left({\theta }_i\right)\sin \left({\theta }_t\right)}{\sin \left({\theta }_t+{\theta }_i\right)\cos \left({\theta }_t-{\theta }_i\right)}\end{array}$

Consider the expression of reflection coefficient in the perpendicular polarization of oblique incidence.

${\Gamma }_{\perp }=\frac{{\eta }_2\cos \left({\theta }_i\right)-{\eta }_1\cos \left({\theta }_t\right)}{{\eta }_2\cos \left({\theta }_i\right)+{\eta }_1\cos \left({\theta }_t\right)}$

Substitute $\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r1}}}$ for ${\eta }_1$ and $\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}}$ for ${\eta }_2$.

$\begin{array}{c}{\Gamma }_{\perp }=\frac{\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}}\cos \left({\theta }_i\right)-\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r1}}}\cos \left({\theta }_t\right)}{\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}}\cos \left({\theta }_i\right)+\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r1}}}\cos \left({\theta }_t\right)}\\ =\frac{\sqrt{\frac{1}{{\epsilon }_{r2}}}\cos \left({\theta }_i\right)-\sqrt{\frac{1}{{\epsilon }_{r1}}}\cos \left({\theta }_t\right)}{\sqrt{\frac{1}{{\epsilon }_{r2}}}\cos \left({\theta }_i\right)+\sqrt{\frac{1}{{\epsilon }_{r1}}}\cos \left({\theta }_t\right)}\\ =\frac{\cos \left({\theta }_i\right)-\frac{\sqrt{{\epsilon }_{r2}}}{\sqrt{{\epsilon }_{r1}}}\cos \left({\theta }_t\right)}{\cos \left({\theta }_i\right)+\frac{\sqrt{{\epsilon }_{r2}}}{\sqrt{{\epsilon }_{r1}}}\cos \left({\theta }_t\right)}\\ =\frac{\cos \left({\theta }_i\right)-\frac{\sin \left({\theta }_i\right)}{\sin \left({\theta }_t\right)}\cos \left({\theta }_t\right)}{\cos \left({\theta }_i\right)+\frac{\sin \left({\theta }_i\right)}{\sin \left({\theta }_t\right)}\cos \left({\theta }_t\right)}\left\{\because \frac{\sqrt{{\epsilon }_{r2}}}{\sqrt{{\epsilon }_{r1}}}=\frac{\sin \left({\theta }_i\right)}{\sin \left({\theta }_t\right)}\right\}\end{array}$

Simplify the expression.

$\begin{array}{c}{\Gamma }_{\perp }=\frac{\sin \left({\theta }_t\right)\cos \left({\theta }_i\right)-\sin \left({\theta }_i\right)\cos \left({\theta }_t\right)}{\sin \left({\theta }_t\right)\cos \left({\theta }_i\right)+\sin \left({\theta }_i\right)\cos \left({\theta }_t\right)}\\ =\frac{\sin \left({\theta }_t-{\theta }_i\right)}{\sin \left({\theta }_t+{\theta }_i\right)}\end{array}$

Consider the expression of transmission coefficient in the perpendicular polarization of oblique incidence.

${\tau }_{\perp }=\frac{2{\eta }_2\cos \left({\theta }_i\right)}{{\eta }_2\cos \left({\theta }_i\right)+{\eta }_1\cos \left({\theta }_t\right)}$

Substitute $\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r1}}}$ for ${\eta }_1$ and $\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}}$ for ${\eta }_2$.

$\begin{array}{c}{\tau }_{\perp }=\frac{2\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}}\cos \left({\theta }_i\right)}{\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r2}}\cos \left({\theta }_i\right)}+\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}{\epsilon }_{r1}}\cos \left({\theta }_t\right)}}\\ =\frac{2\sqrt{\frac{1}{{\epsilon }_{r2}}}\cos \left({\theta }_i\right)}{\sqrt{\frac{1}{{\epsilon }_{r2}}\cos \left({\theta }_i\right)}+\sqrt{\frac{1}{{\epsilon }_{r1}}\cos \left({\theta }_t\right)}}\\ =\frac{2\cos \left({\theta }_i\right)}{\cos \left({\theta }_i\right)+\frac{\sqrt{{\epsilon }_{r2}}}{\sqrt{{\epsilon }_{r1}}}\cos \left({\theta }_t\right)}\\ =\frac{2\cos \left({\theta }_i\right)}{\cos \left({\theta }_i\right)+\frac{\sin \left({\theta }_i\right)}{\sin \left({\theta }_t\right)}\cos \left({\theta }_t\right)}\end{array}$

Simplify the expression.

$\begin{array}{c}{\tau }_{\perp }=\frac{2\cos \left({\theta }_i\right)\sin \left({\theta }_t\right)}{\sin \left({\theta }_t\right)\cos \left({\theta }_i\right)+\sin \left({\theta }_i\right)\cos \left({\theta }_t\right)}\\ =\frac{2\cos \left({\theta }_i\right)\sin \left({\theta }_t\right)}{\sin \left({\theta }_t+{\theta }_i\right)}\end{array}$

Conclusion:

Hence, the given reflection and transmission coefficients for oblique incidence are proved.