Subfield: Optics


Question: Consider an optical system made of many identical ideal (negligible-thickness) halflenses with focal length $f>0$, organized so that they share the same center and are angular-separated equally at density $n$ (number of lenses per unit-radian). Define the length-scale $\lambda=f / n$. A light-ray arrives perpendicular to the first lens at distance $\lambda / 2$ away from the center, then leaves from the last lens at distance $2 \lambda$ away from the center. Estimate the total deflection angle (in rad) of the light-ray by this system in the limit $n \rightarrow \infty$.
![](https://cdn.mathpix.com/cropped/2023_12_21_31037f3a3fcf7f3d1745g-1.jpg?height=556&width=669&top_left_y=1695&top_left_x=728)

Solution:
We define the angles as in Fig. A. The light-path inside the optical system is $r(\theta)$, and the angle between the first and last lens is $\Theta$ (which is an unknown but can be uniquely determined from know information).


![](https://cdn.mathpix.com/cropped/2023_12_21_1dc8e68f97c9d182f754g-1.jpg?height=694&width=1484&top_left_y=225&top_left_x=316)

Consider two consecutive lens at angle $\theta$ and $\theta+d \theta$, in which $d \theta=1 / n \rightarrow 0$ in the continuum limit $n \rightarrow \infty$. From the ideal-lens' equation, using the approximation that $f$ is very large compare to other relevant length-scales in this optical setting:

$$
\frac{1}{f}=\frac{1}{r \tan \phi}+\frac{1}{r \tan (\pi-\phi-\delta \phi)} \approx \frac{\delta \phi}{r \sin ^{2} \phi} \Rightarrow \delta \phi \approx \frac{r}{f} \sin ^{2} \phi
\tag{20}
$$

the differential equation for the angle of arrival $\phi$ can be written as:

$$
d \phi=\delta \phi-d \theta \Rightarrow \frac{d \phi}{d \theta}=\frac{r}{f / n} \sin ^{2} \phi-1=\frac{r}{\lambda} \sin ^{2} \phi-1
\tag{21}
$$

We also have the differential relation between radial position $r(\theta)$ of the light-path and the angle of arrival $\phi$ as followed:

$$
\frac{d r}{d \theta}=r \cot \phi
\tag{22}
$$

From Eq. (21) and Eq. (22), we arrive at:

$$
\frac{d \phi}{d r}=\frac{\frac{r}{\lambda}-1}{r \cot \phi}
\tag{23}
$$

Define $\zeta=\tan \phi$, then Eq. (23) becomes:

$$
\frac{d \phi}{d r}=\frac{1}{1+\zeta^{2}} \frac{d \zeta}{d r}=\frac{\frac{r}{\lambda} \frac{\zeta^{2}}{1+\zeta^{2}}-1}{r / \zeta} \Rightarrow-\frac{d \zeta}{\zeta^{3} d r}-\frac{1}{\zeta^{2} r}=\frac{1}{r}-\frac{1}{\lambda}
\tag{24}
$$

Define $\xi=1 / \zeta^{2}=1 / \tan ^{2} \phi$, then Eq. (23) gives:

$$
\frac{d \zeta}{\zeta^{2} d r}=-\frac{1}{2} \frac{d \xi}{d r} \Rightarrow \frac{d \xi}{d r}-\frac{2}{r} \xi=2\left(\frac{1}{r}-\frac{1}{\lambda}\right) \Rightarrow \frac{d}{d r}\left(\frac{\xi}{r^{2}}\right)=\frac{2}{r^{2}}\left(\frac{1}{r}-\frac{1}{\lambda}\right)
\tag{25}
$$

Integrating both sides, then up to a constant value $C$, Eq. (25) gives:

$$
\frac{\xi}{r^{2}}=-\frac{1}{r^{2}}+\frac{2}{\lambda r}+C \Rightarrow \xi=-1+2 \frac{r}{\lambda}+C \frac{r^{2}}{\lambda^{2}}
\tag{26}
$$



At $\theta=0, r=\lambda / 2$ and $\phi=\pi / 2$ (thus $\xi=0$ ), we can determine $C=0$. Hence:

$$
\cot \phi=\sqrt{2 \frac{r}{\lambda}-1}
\tag{27}
$$

Plug Eq. (27) into Eq. (22):

$$
\frac{d r}{d \theta}=\frac{r}{\lambda} \sqrt{2 \frac{r}{\lambda}-1} \Rightarrow \theta=2 \arctan \sqrt{2 \frac{r}{\lambda}-1}
\tag{28}
$$

At $\theta=\Theta, r=2 \lambda$ therefore we can use Eq. (28) to get:

$$
\Theta=2 \arctan \sqrt{3}=\frac{2 \pi}{3}
\tag{29}
$$

Using Eq. (27), the deflection angle $\Delta$ can be calculated to be:

$$
\Delta=\Theta-\left.\phi\right|_{r=\lambda / 2}+\left.\phi\right|_{r=2 \lambda}=\Theta-\frac{\pi}{2}+\operatorname{arccot} \sqrt{3}=\frac{2 \pi}{3}-\frac{\pi}{2}+\frac{\pi}{6}=\frac{\pi}{3} \approx 1.05 \mathrm{rad}
\tag{30}
$$

For the sake of completeness, we provide the simulated light-path inside the optical system where $n=1000$ using MatLab (which is in great agreement with our theoretical analysis).

Final answer: $1.05$ rad