Riemann surface
matlab

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问题 1.

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)=1n_(2)=1n_(1) > 1n_(2) > 1{:[R_(1) < 0],[R_(2) > 0]:}n_(1)=1n_(2)=1n_(1) > 1n_(2) > 1
n_(1)=1BC^(****)n_(1)=1BA^(****)
n_(2)=1BC^(**).n_(2)=1BA^(**).
n_(1) > 1C^(**).Dn_(1) > 1A^(**)D
n_(2) > 1C^(****)D.n_(2) > 1A^(****).D.
{:[R_(1) > 0],[R_(2) > 0]:}bar(n_(1)=1)n_(2)=1bar(n_(1) > 1)n_(2) > 1{:[R_(1) < 0],[R_(2) < 0]:}n_(1)=1n_(2)=1bar(n_(1) > 1)n_(2) > 1
n_(1)=1BA^(****)n_(1)=1BC^(****)
n_(2)=1BC^(**).n_(2)=1BA^(**)
n_(1) > 1.C^(**)Dn_(1) > 1.A^(**).D
n_(2) > 1A^(****)Dn_(2) > 1C^(****)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}

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