Kvant Physics Problem 193

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Verdicts: UNKNOWN + UNKNOWN
Solve time: 12m24s
Source on kvant.digital

Problem

There are two lens systems with identical focal lengths. The optical axes of the lenses coincide (Fig. 1). The first system consists of converging lenses; in the second, converging lenses alternate with diverging lenses. Find the trajectories of rays in each system if the distance between the lenses is much smaller than the focal length.

Figure 1

Setup and Assumptions

Two infinite sequences of thin lenses are placed along a common optical axis $z$. The distance between neighboring lenses is $d$, with $d \ll f$, where $f$ is the magnitude of the focal length of each lens.

In the first system all lenses are converging, each having focal length $f>0$. In the second system converging and diverging lenses alternate, with focal lengths $+f$ and $-f$ respectively, while the spacing remains the same.

A ray is described by its height $y(z)$ above the optical axis and its small angle $\theta(z)$ relative to the axis, with $\theta \approx \frac{dy}{dz}$. Paraxial optics is assumed, so all angles are small and higher-order terms in angles are neglected. Each lens is treated as a thin lens producing an instantaneous change in slope, while propagation between lenses is free space propagation over distance $d$.

The goal is to determine the form of the ray trajectories $y(z)$ in both periodic systems in the limit $d \ll f$.

Physical Principles

Thin lens action in the paraxial regime is described by the relation $\theta^{+} = \theta^{-} - \frac{y}{f}$.

Free-space propagation over a distance $d$ is governed by $y \to y + d\theta$, while the angle remains unchanged during propagation.

The paraxial relation between ray slope and trajectory is $\theta = \frac{dy}{dz}$.

In the limit of a rapidly repeating optical element with period $d$, a discrete map can be replaced by a continuous differential equation by expanding differences as derivatives with respect to $z$.

Derivation

Consider first a single cell consisting of a lens followed by propagation over distance $d$. Let $(y_n,\theta_n)$ denote the ray parameters immediately before the $n$-th lens. After the lens, the angle becomes $\theta_n - \frac{y_n}{f}$ while the height remains $y_n$. After propagation to the next lens,

$$y_{n+1} = y_n + d\left(\theta_n - \frac{y_n}{f}\right),$$

$$\theta_{n+1} = \theta_n - \frac{y_n}{f}.$$

Introducing a continuous coordinate $z = nd$, the finite differences are written as

$$\frac{y_{n+1} - y_n}{d} = \theta_n - \frac{y_n}{f}, \qquad \frac{\theta_{n+1} - \theta_n}{d} = -\frac{y_n}{fd}.$$

In the continuous limit this becomes

$$\frac{dy}{dz} = \theta - \frac{y}{f},$$

$$\frac{d\theta}{dz} = -\frac{y}{fd}.$$

Eliminating $\theta$ by differentiating the first equation with respect to $z$ gives

$$\frac{d^2 y}{dz^2} = \frac{d\theta}{dz} - \frac{1}{f}\frac{dy}{dz}.$$

Substituting the expression for $\frac{d\theta}{dz}$ yields

$$\frac{d^2 y}{dz^2} = -\frac{y}{fd} - \frac{1}{f}\frac{dy}{dz}.$$

Thus the ray equation for the first system is

$$\frac{d^2 y}{dz^2} + \frac{1}{f}\frac{dy}{dz} + \frac{1}{fd}y = 0.$$

Because $d \ll f$, the coefficient $\frac{1}{fd}$ dominates over $\frac{1}{f^2}$ scale variations, producing a rapidly restoring transverse force. The characteristic equation

$$\lambda^2 + \frac{1}{f}\lambda + \frac{1}{fd} = 0$$

has complex conjugate roots with a large imaginary part of order $\sqrt{\frac{1}{fd}}$, so the transverse coordinate oscillates with weak exponential damping. The trajectory is therefore a rapidly oscillating function with slowly varying envelope.

For the second system, lenses alternate between $+f$ and $-f$. Over one period consisting of two lenses separated by equal distances, the net angular change after passing through a converging lens and then a diverging lens is

$$\Delta \theta = -\frac{y}{f} + \frac{y}{f} = 0$$

to leading order in $y$, since the ray height changes only by terms of order $d\theta$, which are small compared with $f$.

Over a full period the accumulated change in angle is therefore of higher order in $d/f$, so the leading-order continuous limit satisfies

$$\frac{d^2 y}{dz^2} = 0.$$

Hence the trajectory is linear in $z$, corresponding to rectilinear propagation with only weak corrections of order $d/f$.

Result

For the first system of identical converging lenses the ray trajectory satisfies

$$\frac{d^2 y}{dz^2} + \frac{1}{f}\frac{dy}{dz} + \frac{1}{fd}y = 0,$$

and therefore

$$y(z) = e^{-\frac{z}{2f}}\left(A\cos\left(\sqrt{\frac{1}{fd} - \frac{1}{4f^2}},z\right) + B\sin\left(\sqrt{\frac{1}{fd} - \frac{1}{4f^2}},z\right)\right),$$

with constants $A,B$ set by initial conditions.

For the second system with alternating $+f$ and $-f$ lenses the leading-order trajectory is

$$\frac{d^2 y}{dz^2} = 0,$$

so

$$y(z) = Az + B.$$

Sanity Checks

Dimensional consistency holds since $[f]=[d]=\mathrm{m}$ implies $\frac{1}{fd}$ has units $\mathrm{m^{-2}}$, matching $d^2y/dz^2$. The damping term $\frac{1}{f}\frac{dy}{dz}$ has units $\mathrm{m^{-2}}$ as well.

In the limit $d \to \infty$ (very sparse lenses), the focusing term vanishes and the equation reduces toward free propagation, consistent with diminishing lens influence.

In the limit $d \to 0$ at fixed $f$, the focusing term $\frac{1}{fd}$ becomes dominant, producing increasingly rapid transverse oscillations, consistent with a continuous medium of strong focusing power.

For alternating lenses, exact cancellation of lens powers occurs in the lowest order, so straight-line propagation follows as expected for a system with zero net optical power per period.