Riemann surface
matlab
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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。
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The index of refraction of glass can be increased by diffusing in impurities. It is then possible to make a lens of constant thickness. Given a disk of radius $a$ and thickness $d$, find the radial variation of the index of refraction $n(r)$ which will produce a lens with focal length $F$. You may assume a thin lens $(d<a)$.
Solution:
Let the refractive index of the material of the disk be $n$ and the radial distribution of the refractive index of the impurity-diffused disk be represented by $n(r)$, with $n(0)=n_0$. Incident plane waves entering the lens refract and converge at the focus $F$ as shown in Fig. 1.10. We have
$$
\left|n(r)-n_0\right| d=-\sqrt{F^2+r^2}+F
$$
i.e.,
$$
n(r)=n_0-\frac{\sqrt{F^2+r^2}-F}{d} .
$$
For $F>r$, we obtain
$$
n(r)=n_0-\frac{r^2}{2 d F}
$$
A line object $5 \mathrm{~mm}$ long is located $50 \mathrm{~cm}$ in front of a camera lens. The image is focussed on the film plate and is $1 \mathrm{~mm}$ long. If the film plate is moved back $1 \mathrm{~cm}$ the width of the image blurs to $1 \mathrm{~mm}$ wide. What is the $F$-number of the lens?
(Wisconsin)
Solution:
Substituting $u=50 \mathrm{~cm}$ and $\frac{v}{4}=\frac{1}{5}$ in the Gaussian lens formula
$$
\frac{1}{u}+\frac{1}{v}=\frac{1}{f}
$$
gives $f=8.33 \mathrm{~cm}, v=10 \mathrm{~cm}$. From the similar triangles in Fig. 1.14 we have
$$
\frac{D}{v}=\frac{0.1}{1}
$$
or $D=0.1 \mathrm{v}=1 \mathrm{~cm}$. Therefore, $F=f / D=8.33$.
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