### 物理代写|光学工程代写Optical Engineering代考|EGEE480

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|光学工程代写Optical Engineering代考|Interference and Diffraction

Consider two sinusoidal waves passing from point $\mathrm{A}$ to point $\mathrm{C}$ and from point $\mathrm{B}$ to point C. For simplicity, the frequencies of the two waves are the same. Vectors of AC and $\mathrm{BC}$ are denoted by $\boldsymbol{r}{A C}$ and $\boldsymbol{r}{B C}$, respectively, and their wave number vectors by $\boldsymbol{k}A$ and $\boldsymbol{k}_B$, respectively (see Fig. 2.1). The plane wave arriving at point $\mathrm{C}$ from point $A$ is given by $$u_A=A_A \exp \left[\mathbf{i}\left(\boldsymbol{k}_A \cdot \boldsymbol{r}{A C}+\phi_A-\omega t\right)\right],$$
and the plane wave arriving at point $\mathrm{C}$ from point $\mathrm{B}$ is
$$u_B=A_B \exp \left[\mathrm{i}\left(\boldsymbol{k}B \cdot \boldsymbol{r}{B C}+\phi_B-\omega t\right)\right] .$$
Suppose the position vectors of points A and C are $\boldsymbol{r}A\left(x_A, y_A, z_A\right)$ and $\boldsymbol{r}_C\left(x_C, y_C, z_C\right)$, we have $$r{A C}=r_C-r_A .$$
By using Eq. (1.30), we have
\begin{aligned} \boldsymbol{k}A \cdot \boldsymbol{r}{A C} & =\boldsymbol{k} \cdot\left(\boldsymbol{r}C-\boldsymbol{r}_A\right) \ & =\frac{2 \pi}{\lambda_0} n\left[\left(x_C-x_A\right) \cos \alpha_A+\left(y_C-y_A\right) \cos \beta_A+\left(z_C-z_A\right) \cos \gamma_A\right], \end{aligned} where the directional cosines of the vector $\boldsymbol{k}_A$ are $\left(\cos \alpha_A, \cos \beta_A, \cos \gamma_A\right)$. If $$\left(x_C-x_A\right) \cos \alpha_A+\left(y_C-y_A\right) \cos \beta_A+\left(z_C-z_A\right) \cos \gamma_A=l{A C},$$
then $l_{A C}$ gives the distance between points A and $\mathrm{C}$, and $n l_{A C}$ is called the optical distance, where $n$ is the refractive index of the medium. Equations (2.1) and (2.2) are rewritten as
$$u_A=A_A \exp \left[\mathrm{i}\left(\frac{2 \pi}{\lambda_0} n l_{A C}+\phi_A-\omega t\right)\right]$$ and
$$u_B=A_B \exp \left[\mathrm{i}\left(\frac{2 \pi}{\lambda_0} n l_{B C}+\phi_B-\omega t\right)\right],$$
respectively. Since the amplitude $u_C$ at point $\mathrm{C}$ is given by the superposition of two waves $u_A$ and $u_B$, we have
\begin{aligned} u_C & =A_A \exp \left[\mathrm{i}\left(\frac{2 \pi}{\lambda_0} n l_{A C}+\phi_A-\omega t\right)\right]+A_B \exp \left[\mathrm{i}\left(\frac{2 \pi}{\lambda_0} n l_{B C}+\phi_B-\omega t\right)\right] \ & =\left{A_A \exp \left[\mathrm{i}\left(\frac{2 \pi}{\lambda_0} n l_{A C}+\phi_A\right)\right]+A_B \exp \left[\mathrm{i}\left(\frac{2 \pi}{\lambda_0} n l_{B C}+\phi_B\right)\right]\right} \exp (-\mathrm{i} \omega t) \end{aligned}

## 物理代写|光学工程代写Optical Engineering代考|FRINGE VISIBILITY

As the measure of interference fringe clarity, the contrast or the visibility is defined by
$$V=\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }},$$
where $I_{\max }$ and $I_{\min }$ are the maximum and the minimum of the fringe intensity, respectively. We have
$$V=\frac{2 \sqrt{I_A I_B}}{I_A+I_B}$$ by using Eq. (2.9). The maximum contrast of $V=1$ is obtained when $I_A=I_B$. When either $I_A$ or $I_B$ is 0 , the minimum $V=0$ so that the fringe is invisible.

Whenever $A_A=A_B$, is the contrast always $V=1$ ? In reality, a special condition is necessary for $V=1$. The interference fringe exists stably in time, only when the difference $\phi_B-\phi_A$ of the initial phase of $\phi_A$ and $\phi_B$ at the point A and B is stable in time. The difference $\phi_B-\phi_A$ depends on the properties of the light sources, the distances from the light source to the point A and B, and so on. This means the contrast of the interference fringe depends not only on the path difference of $l_{B C}-l_{A C}$ but also the light source properties and the layout of the optical system.

The phase of wave from the light source in many cases is stable in less than $10^{-8}$ seconds. We can see the wave shape is sinusoidal only within this short time, where the amplitude and the phase are fixed within an appropriate time. Many wavelets with fixed duration generated from the light source form practical waves. From this concept, to generate a stable interference fringe at point $\mathrm{C}$, the waves A and B departing from the same source and at the same time should superimpose at point $\mathrm{C}$. The device used to perform such a superposition is called as an interferometer.
As described in Section 10.1, the stability of the phase difference $\phi_B-\phi_A$ depends on the properties of the light source. The measure is called the degree of coherence, $\gamma_{A B}$. We have $0 \leq \gamma_{A B} \leq 1$. In the case of $\gamma_{A B}=1$, two waves from the point $A$ and B are called coherent each other, and incoherent in the case of $\gamma_{A B}=0$. When considering the coherence, the fringe contract is rewritten as
$$V=\frac{2 \sqrt{I_A I_B}}{I_A+I_B} \gamma_{A B} .$$

# 光学工程代考

## 物理代写|光学工程代写Optical Engineering代考|Interference and Diffraction

$$u_A=A_A \exp \left[\mathrm{i}\left(\frac{2 \pi}{\lambda_0} n l_{A C}+\phi_A-\omega t\right)\right]$$

$$u_B=A_B \exp \left[\mathrm{i}\left(\frac{2 \pi}{\lambda_0} n l_{B C}+\phi_B-\omega t\right)\right],$$

## 物理代写|光学工程代写Optical Engineering代考|FRINGE VISIBILITY

$$V=\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }},$$

$$V=\frac{2 \sqrt{I_A I_B}}{I_A+I_B}$$

$$V=\frac{2 \sqrt{I_A I_B}}{I_A+I_B} \gamma_{A B}$$

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## MATLAB代写

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