91Ó°ÊÓ

(Assume${\mathrm{μ}}_{\text{1}}={\mathrm{μ}}_{\text{2}}={\mathrm{μ}}_{\text{0}}$ .)

The "crossover" angle, at which the reflected and transmitted amplitudes are equal.

Write the formula of amplitude at the normal incidence.

$\frac{{\mathbf{E}}_{\mathbf{0}\mathbf{T}}}{{\mathbf{E}}_{\mathbf{0}\mathbf{I}}}=\frac{\mathbf{2}}{\mathrm{Î\pm }+\mathrm{β}}$ …… (1)

Here,$\mathrm{Î\pm }$ is purely imaginary, $\mathrm{β}$is real.

Write the formula ofBrewster’s angle.

${\mathrm{θ}}_{\mathbf{B}}={\mathbf{tan}}^{-}\frac{{\mathbf{n}}_{\mathbf{2}}}{{\mathbf{n}}_{\mathbf{1}}}$ …… (2)

Here, role="math" localid="1658736610477" ${\mathbf{n}}_{\mathbf{1}}$ is the refractive index of the first object and ${\mathbf{n}}_{\mathbf{2}}$ is the refractive index of the second object.

Write the formula of crossover angle.

${\mathrm{Î\pm }}^{\mathbf{2}}\left(\mathbf{1}-{\mathbf{sin}}^{\mathbf{2}}\mathrm{θ}\right)=\mathbf{1}-{\left(\frac{{\mathbf{n}}_{\mathbf{1}}}{{\mathbf{n}}_{\mathbf{2}}}\right)}^{\mathbf{2}}{\mathbf{sin}}^{\mathbf{2}}\mathrm{θ}$ ….. (3)

Here, $\mathrm{Î\pm }$ is purely imaginary, ${\mathbf{n}}_{\mathbf{1}}$ is the refractive index of the first object and is the refractive index of the second object.

From Fresnel’s equation

$$\begin{gathered} \mathrm{β}=\frac{{\mathrm{μ}}_1{\mathrm{ν}}_1}{{\mathrm{μ}}_2{\mathrm{ν}}_2} \\ =\frac{{\mathrm{μ}}_1}{{\mathrm{μ}}_2}\frac{n_2}{n_1} \end{gathered}$$

Here, $n_1$ is the refractive index of the first object, $n_2$ is the refractive index of the second object, ${\mathrm{ν}}_1$ is the velocity of the first object, and ${\mathrm{ν}}_2$is the velocity of the second object.

Substitute 2.42 for n₂ and 1 for n₁ in the above equation $\mathrm{β}$.

$$\begin{gathered} \mathrm{β}=\frac{2.42}{1} \\ =2.42 \end{gathered}$$

Since${\mathrm{μ}}_1={\mathrm{μ}}_2={\mathrm{μ}}_0$ and the refractive index of the air is 1.

Determine the reflected amplitude wave is

${\stackrel{→}{E}}_{0R}=\left(\frac{\mathrm{Î\pm }-\mathrm{β}}{\mathrm{Î\pm }+\mathrm{β}}\right){\stackrel{→}{E}}_{0I}$

Determine the transmitted wave is

${\stackrel{→}{E}}_{0T}=\left(\frac{2}{\mathrm{Î\pm }+\mathrm{β}}\right){\stackrel{→}{E}}_{0T}$

Here amplitudes of reflected and transmitted waves depend on angle of incidence.

Thus, the amplitude of the reflected and transmitted waves is

$$\begin{gathered} \mathrm{Î\pm }=\frac{\sqrt{1-{\sin }^2{\mathrm{θ}}_T}}{\cos {\mathrm{θ}}_1} \\ =\frac{\sqrt{1-{\left(\frac{n_1}{n_2}\right)}^2{\sin }^2{\mathrm{θ}}_1}}{\cos {\mathrm{θ}}_1} \end{gathered}$$ …… (4)

The graph that represents the air diamond interaction is displayed in the following illustration.

Electric field ratio is represented on the y axis, while incidence angle is represented on the -x axis.

At the normal incidence ${\mathrm{θ}}_1=0°$.

Substitute 2.42 for n₂, 0⁰ for${\mathrm{θ}}_1$, and 1 for n₁ into above equation (4).

$$\begin{gathered} \mathrm{Î\pm }=\frac{\sqrt{1-{\left(\frac{1}{2.42}\right)}^2\sin 0}}{\cos 0} \\ =1 \end{gathered}$$

Then, the ratio of the reflected and incident amplitudes is

$\frac{E_{0R}}{E_{0I}}=\frac{\mathrm{Î\pm }-\mathrm{β}}{\mathrm{Î\pm }+\mathrm{β}}$ or $\mathrm{β}$ into equation (1).

Substitute 1 for $\mathrm{Î\pm }$ and 2.42 for $\mathrm{β}$ into above equation.

$$\begin{gathered} \frac{E_{0R}}{E_{0I}}=\frac{1-2.42}{1+2.42} \\ =-0.415 \end{gathered}$$

Determine the ratio of transmitted and incident amplitude is

Substitute 1 for and 2.42

$$\begin{gathered} \frac{E_{0T}}{E_{0I}}=\frac{2}{1+2.42} \\ =0.585 \\ {E}_{0T}=0.585E_{0I} \end{gathered}$$

Therefore, the value of amplitude at the normal incidence is $E_{0T}=0.585E_{0I}$.

At Brewster’s angle

$\tan {\mathrm{θ}}_Bâ‰\dots \frac{n_2}{n_1}$

Determine the Brewster’s angle.

Substitute 1 for $\mathrm{Î\pm }$ and 2.42 for $\mathrm{β}$into equation (2).

$$\begin{gathered} {\mathrm{θ}}_B={\tan }^{-1}\left(2.42\right) \\ =67.55° \end{gathered}$$

Therefore, the Brewster’s angle is 67.55⁰.

As given reflected and transmitted amplitudes are equal.

${\stackrel{→}{E}}_{0R}={\stackrel{→}{E}}_{0T}$

Then,

$$\begin{gathered} \mathrm{Î\pm }-\mathrm{β}=2 \\ \mathrm{Î\pm }=\mathrm{β}+2 \end{gathered}$$

Substitute 2.42 for $\mathrm{β}$ into above equation.

Then we know that

$$\begin{gathered} \mathrm{Î\pm }=\frac{\sqrt{1-{\left(\frac{n_1}{n_2}\right)}^2{\sin }^2{\mathrm{θ}}_1}}{\cos {\mathrm{θ}}_1} \\ \mathrm{Î\pm }=\frac{1-{\left(\frac{n_1}{n_2}\right)}^2{\sin }^2\mathrm{θ}}{{\cos }^2\mathrm{θ}} \\ {\mathrm{Î\pm }}^2{\cos }^2\mathrm{θ}=1-{\left(\frac{n_1}{n_2}\right)}^2{\sin }^2\mathrm{θ} \\ {\mathrm{Î\pm }}^2\left(1-{\sin }^2\mathrm{θ}\right)=1-{\left(\frac{n_1}{n_2}\right)}^2{\sin }^2\mathrm{θ} \end{gathered}$$

Determine the crossover angle.

Substitute 1 for n1 and 2.42 for n₂ into equation (3).

$$\begin{gathered} {\mathrm{Î\pm }}^2\left(1-{\sin }^2\mathrm{θ}\right)=1-{\left(\frac{1}{2.42}\right)}^2{\sin }^2\mathrm{θ} \\ {\left(4.42\right)}^2\left(1-{\sin }^2\mathrm{θ}\right)=1-\frac{1}{{\left(2.42\right)}^2}{\sin }^2\mathrm{θ} \\ \sin \mathrm{θ}=0.979 \end{gathered}$$

From the above equation, the angle is

$$\begin{gathered} \mathrm{θ}={\sin }^{-1}\left(0.979\right) \\ =78.24° \end{gathered}$$

Therefore, the value of crossover angle is 78.24⁰ .