Riemann surface
matlab

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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