Riemann surface
matlab
还在面临光学的学习挑战吗？别担心！我们的optics-guide团队专业为您解决光的传播、折射、反射等光学问题。我们拥有深厚的专业背景和丰富的经验，能帮您完成高水平的作业和论文，让您的学习之路一帆风顺！
以下是一些我们可以帮助您解决的问题：
光的传播与波动：光的传播方式、光的波动性质和光的干涉与衍射现象等。
光的折射与反射：光的折射定律、反射定律和光的全反射等。
透镜和光学仪器：透镜的类型、性质和光学成像原理，以及光学仪器的设计和应用。
光的色散与偏振：光的色散现象和光的偏振性质，如色散衍射和偏振光的传播等。
光的干涉与衍射：光的干涉现象、衍射现象和干涉、衍射实验的分析和应用。
光的相干与激光：光的相干性质和激光的原理、特性和应用。
无论您面临的光学问题是什么，我们都会尽力为您提供专业的帮助，确保您的学习之旅顺利无阻！
Problem 1: Wiper speed control
Figure 1 shows an example of an optical system designed to detect the amount of water present on the windshield of a car to adjust the wiper speed. As shown in this figure, we can use the windshield as a waveguide to guide the light from a source located at one end (bottom of the windshield) to a detector located in the opposite end. The light suffers total-internal reflection (TIR) at the glass-air interface. However, when rain drops are present, some of the light will suffer frustrated TIR escaping outside the waveguide. Since we know the power of the light source, a given drop in power can be correlated to the amount of water present and used to adjust the wiper speed.
To quantify the feasibility of this design we start by computing the critical angles of the glass-air and glass-water interfaces,
$$
\begin{aligned}
& \theta_{c_{g-a}}=\arcsin \left(\frac{n_{\text {air }}}{n_{\text {glass }}}\right)=41.8^{\circ}, \
& \theta_{c_{g-w}}=\arcsin \left(\frac{n_{\text {water }}}{n_{\text {glass }}}\right)=62.5^{\circ} .
\end{aligned}
$$
The incidence angle of a given ray propagating inside this waveguide is restricted to the following cases:
For $\theta<\theta_{c_{g-a}}$ : The light will suffer frustrated TIR and escape out of the waveguide regardless of whether the interface is glass-air or glass-water.
For $\theta_{c_{g-a}}<\theta<\theta_{c_{g-w}}$ : The light will suffer TIR at the glass-air interface and frustrated TIR at the glass-water interface.
For $\theta_{c_{g-w}}<\theta$ : The light will suffer TIR at both interfaces.
Problem 4: Immersion lens
a) We begin by computing the composite matrix of the optical system,
$$
\begin{gathered}
{\left[\begin{array}{c}
n_2 \alpha_i \
x i
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \
\frac{s_i}{n_2} & 1
\end{array}\right]\left[\begin{array}{cc}
1 & -\left(\frac{n_2-n_0}{R_2}\right) \
0 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & -\left(\frac{n_0-n_1}{R_1}\right) \
0 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & 0 \
\frac{s_o}{n_1} & 1
\end{array}\right]\left[\begin{array}{c}
n_1 \alpha_o \
x_o
\end{array}\right]} \
=\left[\begin{array}{cc}
1+\frac{s_o}{n_1} P & P \
\frac{s_i}{n_2}+\frac{s_o}{n_1}\left(1+\frac{s_i}{n_2} P\right) & 1+\frac{s_i}{n_2} P
\end{array}\right]\left[\begin{array}{c}
n_1 \alpha_o \
x_o
\end{array}\right],
\end{gathered}
$$
where,
$$
P=-\frac{1}{f}=-\left[n_0\left(\frac{1}{R_1}-\frac{1}{R_2}\right)-\left(\frac{n_1}{R_1}-\frac{n_2}{R_2}\right)\right]
$$
and we see that for the case of $n_1=n_2=1$, equation 16 reduces to the “Lens Maker’s ” equation that we saw in class. Equation 16 is a more generalized version of this equation.
To find $f_i$, we consider a plane wave coming from the left such that $\alpha_o=0$, and $x_i=0$ (i.e. for an object at infinity, $s_o=\infty$ ),
$$
\begin{aligned}
0 & =\left(1+\frac{s_i}{n_2} P\right) x_o \
& \Rightarrow f_i=n_2 f .
\end{aligned}
$$
| {:[R_(1) > 0],[R_(2) < 0]:} | n_(1)=1 | n_(2)=1 | n_(1) > 1 | n_(2) > 1 | {:[R_(1) < 0],[R_(2) > 0]:} | n_(1)=1 | n_(2)=1 | n_(1) > 1 | n_(2) > 1 |
|---|---|---|---|---|---|---|---|---|---|
| n_(1)=1 | B | C^(****) | n_(1)=1 | B | A^(****) | ||||
| n_(2)=1 | B | C^(**) | . | n_(2)=1 | B | A^(**) | . | ||
| n_(1) > 1 | C^(**) | . | D | n_(1) > 1 | A^(**) | D | |||
| n_(2) > 1 | C^(****) | D | . | n_(2) > 1 | A^(****) | . | D | . | |
| {:[R_(1) > 0],[R_(2) > 0]:} | bar(n_(1)=1) | n_(2)=1 | bar(n_(1) > 1) | n_(2) > 1 | {:[R_(1) < 0],[R_(2) < 0]:} | n_(1)=1 | n_(2)=1 | bar(n_(1) > 1) | n_(2) > 1 |
| n_(1)=1 | B | A^(****) | n_(1)=1 | B | C^(****) | ||||
| n_(2)=1 | B | C^(**) | . | n_(2)=1 | B | A^(**) | |||
| n_(1) > 1 | . | C^(**) | D | n_(1) > 1 | . | A^(**) | . | D | |
| n_(2) > 1 | A^(****) | D | n_(2) > 1 | C^(****) | D |
$$
\begin{aligned}
M_A & =\frac{\partial \alpha_i}{\partial \alpha_o} \
& =\frac{n_1}{n_2}\left(1-\frac{s_o}{n_1 f}\right) \
& =\frac{n_1}{n_2}\left(\frac{n_1 f-s_o}{n_1 f}\right) .
\end{aligned}
$$
From the imaging condition of equation 18 , we solve for $s_i$,
$$
\frac{1}{s_i}=\frac{s_o-n_1 f}{n_2 f s_o} .
$$
Using equation 20 in equation 19 we get,
$$
M_A=-\frac{s_o}{s_i}
$$
Similarly, we derive the lateral magnification using the equation for $x_i$ for an on-axis ray (i.e. $\alpha_o=0$ ),
$$
\begin{gathered}
x_i=\left(1-\frac{s_i}{n_2 f}\right) x_o \
M_L=\frac{x_i}{x_o}=\left(1-\frac{s_i}{n_2 f}\right)=-\frac{n_1 s_i}{n_2 s_o} .
\end{gathered}
$$
d) We consider the triangle formed by a ray that originates a the object’s tip and aims at the optical center of the lens. The angle of the ray respect to the optical axis is,
$$
\tan \alpha_o \cong \alpha_o=-\frac{x_o}{s_o} \text {. }
$$
From equation 15 we see that the object and image angles are related according to,
$$
\begin{aligned}
n_2 \alpha_i & =\left(1-\frac{s_o}{n_1 f}\right) n_1 \alpha_o-\frac{x_o}{f} \
& =-\left(1-\frac{s_o}{n_1 f}\right) \frac{n_1 x_o}{s_o}-\frac{x_o}{f} \
& =-\frac{x_o}{s_o} n_1=\alpha_o n_1,
\end{aligned}
$$
which is Snell’s law!
\end{prob}
E-mail: help-assignment@gmail.com 微信:shuxuejun
help-assignment™是一个服务全球中国留学生的专业代写公司
专注提供稳定可靠的北美、澳洲、英国代写服务
专注于数学，统计，金融，经济，计算机科学，物理的作业代写服务