Question:The index of refraction of diamond is 2.42. Construct the graph analogous to Fig. 9.16 for the air/diamond interface. (Assume .) In particular, calculate (a) the amplitudes at normal incidence, (b) Brewster's angle, and (c) the "crossover" angle, at which the reflected and transmitted amplitudes are equal.

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

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

Write the formula of amplitude at the normal incidence.

$\frac{{\mathbf{E}}_{\mathbf{0}\mathbf{T}}}{{\mathbf{E}}_{\mathbf{0}\mathbf{I}}}=\frac{\mathbf{2}}{\alpha +\beta }$ …… (1)

Here,$\alpha$ is purely imaginary, $\beta$is real.

Write the formula ofBrewster’s angle.

${\theta }_{\mathbf{B}}={\mathbf{tan}}^{-}\frac{{\mathbf{n}}_{\mathbf{2}}}{{\mathbf{n}}_{\mathbf{1}}}$ …… (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.

${\alpha }^{\mathbf{2}}\left(\mathbf{1}-{\mathbf{sin}}^{\mathbf{2}}\theta \right)=\mathbf{1}-{\left(\frac{{\mathbf{n}}_{\mathbf{1}}}{{\mathbf{n}}_{\mathbf{2}}}\right)}^{\mathbf{2}}{\mathbf{sin}}^{\mathbf{2}}\theta$ ….. (3)

Here, $\alpha$ 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

$$\begin{gathered} \beta =\frac{{\mu }_1{\nu }_1}{{\mu }_2{\nu }_2} \\ =\frac{{\mu }_1}{{\mu }_2}\frac{n_2}{n_1} \end{gathered}$$

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

Substitute 2.42 for n₂ and 1 for n₁ in the above equation $\beta$.

$$\begin{gathered} \beta =\frac{2.42}{1} \\ =2.42 \end{gathered}$$

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

Determine the reflected amplitude wave is

${\overrightarrow{E}}_{0R}=\left(\frac{\alpha -\beta }{\alpha +\beta }\right){\overrightarrow{E}}_{0I}$

Determine the transmitted wave is

${\overrightarrow{E}}_{0T}=\left(\frac{2}{\alpha +\beta }\right){\overrightarrow{E}}_{0T}$

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

Thus, the amplitude of the reflected and transmitted waves is

$$\begin{gathered} \alpha =\frac{\sqrt{1-{\sin }^2{\theta }_T}}{\cos {\theta }_1} \\ =\frac{\sqrt{1-{\left(\frac{n_1}{n_2}\right)}^2{\sin }^2{\theta }_1}}{\cos {\theta }_1} \end{gathered}$$ …… (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 ${\theta }_1=0°$.

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

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

Then, the ratio of the reflected and incident amplitudes is

$\frac{E_{0R}}{E_{0I}}=\frac{\alpha -\beta }{\alpha +\beta }$ or $\beta$ into equation (1).

Substitute 1 for $\alpha$ and 2.42 for $\beta$ 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

$\tan {\theta }_B\cong \frac{n_2}{n_1}$

Determine the Brewster’s angle.

Substitute 1 for $\alpha$ and 2.42 for $\beta$into equation (2).

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

Therefore, the Brewster’s angle is 67.55⁰.

As given reflected and transmitted amplitudes are equal.

${\overrightarrow{E}}_{0R}={\overrightarrow{E}}_{0T}$

Then,

$$\begin{gathered} \alpha -\beta =2 \\ \alpha =\beta +2 \end{gathered}$$

Substitute 2.42 for $\beta$ into above equation.

Then we know that

$$\begin{gathered} \alpha =\frac{\sqrt{1-{\left(\frac{n_1}{n_2}\right)}^2{\sin }^2{\theta }_1}}{\cos {\theta }_1} \\ \alpha =\frac{1-{\left(\frac{n_1}{n_2}\right)}^2{\sin }^2\theta }{{\cos }^2\theta } \\ {\alpha }^2{\cos }^2\theta =1-{\left(\frac{n_1}{n_2}\right)}^2{\sin }^2\theta \\ {\alpha }^2\left(1-{\sin }^2\theta \right)=1-{\left(\frac{n_1}{n_2}\right)}^2{\sin }^2\theta \end{gathered}$$

Determine the crossover angle.

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

$$\begin{gathered} {\alpha }^2\left(1-{\sin }^2\theta \right)=1-{\left(\frac{1}{2.42}\right)}^2{\sin }^2\theta \\ {\left(4.42\right)}^2\left(1-{\sin }^2\theta \right)=1-\frac{1}{{\left(2.42\right)}^2}{\sin }^2\theta \\ \sin \theta =0.979 \end{gathered}$$

From the above equation, the angle is

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

Therefore, the value of crossover angle is 78.24⁰ .