Riemann surface
matlab

Problem 1: Wanda’s world
a) The geometry for this problem is shown in Figure 1. For part (a), the object (Wanda) is located inside the bowl and we are interested to find where the image is formed. We start by using the matrix formulation to analyze the given system,
\begin{aligned} {\left[\begin{array}{c} \alpha_i \ x_i \end{array}\right] } & =\left[\begin{array}{ll} 1 & 0 \ s & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -\frac{(1-n)}{-R} \ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \ \frac{R}{n} & 1 \end{array}\right]\left[\begin{array}{c} n \alpha_o \ x_o \end{array}\right] \ & =\left[\begin{array}{cc} \frac{1}{n} & \frac{(1-n)}{R} \ \frac{R}{n}+\frac{s}{n} & 1+\frac{s(1-n)}{R} \end{array}\right]\left[\begin{array}{c} n \alpha_o \ x_o \end{array}\right] . \end{aligned}
To find the imaging condition, we note that all the rays, regardless of their departure angle $\alpha_o$, arrive at the image point $x_i$ (i.e. $\partial x_i / \partial \alpha_o=0$ ),
\begin{aligned} \frac{R}{n}+\frac{s}{n} & =0 \Rightarrow \ s & =-R . \end{aligned}
We see that the image is formed at the center of the bowl and is virtual. Using this result in equation 1 ,
$$\left[\begin{array}{l} \alpha_i \ x_i \end{array}\right]=\left[\begin{array}{cc} \frac{1}{n} & \frac{(1-n)}{R} \ 0 & n \end{array}\right]\left[\begin{array}{c} n \alpha_o \ x_o \end{array}\right] .$$
The lateral magnification is,
$$M_L=\frac{x_i}{x_o}=n$$
and therefore, the image is erect.
b) For this part, the object (Olive) is located outside the bowl and we are interested to find where the image is formed. Again, we solve this part using the matrix formulation,
\begin{aligned} {\left[\begin{array}{l} \alpha_i \ x_i \end{array}\right] } & =\left[\begin{array}{ll} 1 & 0 \ \frac{s^{\prime}}{n} & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -\frac{(n-1)}{R} \ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \ s & 1 \end{array}\right]\left[\begin{array}{c} n \alpha_o \ x_o \end{array}\right] \ & =\left[\begin{array}{cc} 1-\frac{s(n-1)}{R} & -\frac{(n-1)}{R^{\prime}(n-1)} \ \frac{s^{\prime}}{n}+s-\frac{s^{\prime}(n-1)}{n R} & 1-\frac{s^{\prime}(n R}{n R} \end{array}\right]\left[\begin{array}{c} n \alpha_o \ x_o \end{array}\right] . \end{aligned}

From the imaging condition $\left(\partial x_i / \partial \alpha_o=0\right)$, we solve for $s^{\prime}$,
\begin{aligned} \frac{s^{\prime}}{n}+s-\frac{s s^{\prime}(n-1)}{n R} & =0 \Rightarrow \ s^{\prime} & =\frac{s n R}{s(n-1)-R} . \end{aligned}
The lateral magnification is (for an on-axis ray, $\alpha_o=0$ ),
\begin{aligned} M_L & =\frac{x_i}{x_o}=1-\frac{s^{\prime}(n-1)}{n R} \ & =1-\frac{\operatorname{snR}(n-1)}{(s(n-1)-R) n R}=-\frac{R}{s(n-1)-R} \end{aligned}
From equations 6 and 7 , the following cases arise:

If $R>s(n-1) \rightarrow s^{\prime}<0$, the image is virtual, erect and is located at a distance $\left|s^{\prime}\right|$ outside the bowl.

If $R>s(n-1) \rightarrow s^{\prime}>0$, the image is real, inverted and is located at a distance $\left|s^{\prime}\right|$ inside the bowl.

c) If we were to consider the glass container of thickness $t$ as well as the inner, $R_1$, and outer, $R_2$, radii, the matrix formulation becomes,
\begin{aligned} {\left[\begin{array}{c} \alpha_i \ x_i \end{array}\right] } & =\left[\begin{array}{ll} 1 & 0 \ \frac{s^{\prime}}{n} & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -\frac{\left(n-n_g\right)}{R_1} \ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \ \frac{t}{n_g} & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -\frac{\left(n_g-1\right)}{R_2} \ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \ s & 1 \end{array}\right]\left[\begin{array}{c} n \alpha_o \ x_o \end{array}\right] \ & =\left[\begin{array}{ll} 1 & 0 \ \frac{s^{\prime}}{n} & 1 \end{array}\right]\left[\begin{array}{cc} 1-\frac{t\left(n-n_g\right)}{R_1 n_g} & P \ \frac{t}{n_g} & 1-\frac{t\left(n_g-1\right)}{R_2 n_g} \end{array}\right]\left[\begin{array}{ll} 1 & 0 \ s & 1 \end{array}\right]\left[\begin{array}{c} n \alpha_o \ x_o \end{array}\right] \ & =\left[\begin{array}{cc} 1+P s-\frac{t\left(n-n_g\right)}{R_1 n_g} & P \ \frac{s^{\prime}}{n}\left(1-\frac{t\left(n-n_g\right)}{R_1 n_g}\right)+\frac{t}{n_g}+s\left(\frac{s_{s^{\prime}} P}{n}+1-\frac{t\left(n_g-1\right)}{R_2 n_g}\right) & 1+\frac{s^{\prime} P}{n}-\frac{t\left(n_g-1\right)}{R_2 n_g} \end{array}\right]\left[\begin{array}{c} n \alpha_o \ x_o \end{array}\right], \end{aligned}
where,
$$P=-\left[\frac{\left(n-n_g\right)}{R_1}+\frac{\left(n_g-1\right)}{R_2}-\frac{t}{n_g R_1 R_2}\left(n-n_g\right)\left(n_g-1\right)\right] .$$
For the case of uniform glass: $R_1=R_2=R$. The imaging condition is,
$$\frac{s^{\prime}}{n}+s-\frac{s s^{\prime}(n-1)}{n R}-\frac{s^{\prime} t\left(n-n_g\right)}{R n_g n}+\frac{t}{n_g}-\frac{s t\left(n_g-1\right)}{R n_g}+\frac{s s^{\prime}}{n R}\left(\frac{t\left(n-n_g\right)\left(n_g-1\right)}{R n_g}\right)=0$$

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