Kvant Physics Problem 18

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Problem

If one looks at light through two combs with different tooth densities, superimposed on each other, bright regions will alternate with dark ones. At what speed will these regions move if one of the combs is moved relative to the other at a speed of 1 cm/s? The stationary comb has 5 teeth per centimeter, and the moving one has 6.

G. L. Kotkin

Setup and Assumptions

Two one-dimensional periodic structures represent the combs. The stationary comb has spatial frequency $k_1 = 5,\text{cm}^{-1}$, corresponding to tooth spacing $d_1 = 1/k_1 = 0.2,\text{cm}$. The moving comb has spatial frequency $k_2 = 6,\text{cm}^{-1}$, corresponding to spacing $d_2 = 1/k_2 = 1/6,\text{cm} \approx 0.1667,\text{cm}$. The second comb translates uniformly with velocity $v = 1,\text{cm/s}$ relative to the first.

The observed bright and dark regions are modeled as a moiré pattern produced by superposition of two ideal periodic transmissions. Each comb is assumed infinite, perfectly periodic, and unaffected by optical or mechanical distortions. The intensity maxima correspond to constructive overlap of the comb structures, which is equivalent to phase alignment of their periodic tooth distributions.

The motion is treated in one spatial dimension along the direction of translation, and relativistic effects are neglected.

Physical Principles

The spatial structure of a periodic grating with spatial frequency $k$ is represented by a phase $\phi = kx$.

A uniformly translating grating with velocity $v$ has phase dependence $\phi(x,t) = k(x - vt)$.

Moiré maxima occur when the phase difference between the two gratings remains constant in space and time, which implies that the combined phase

$$\Phi(x,t) = k_1 x - k_2(x - vt)$$

is constant along a moving fringe.

A moving pattern of constant phase satisfies $d\Phi = 0$, which leads to a relation between spatial and temporal derivatives defining the fringe velocity.

Derivation

The phase difference governing the brightness pattern is

$$\Phi(x,t) = k_1 x - k_2(x - vt).$$

Expanding the expression gives

$$\Phi(x,t) = k_1 x - k_2 x + k_2 v t = (k_1 - k_2)x + k_2 v t.$$

A bright fringe corresponds to a fixed value of $\Phi$, so its total differential vanishes:

$$d\Phi = (k_1 - k_2),dx + k_2 v,dt = 0.$$

Solving for the fringe velocity $u = \frac{dx}{dt}$ yields

$$(k_1 - k_2)\frac{dx}{dt} + k_2 v = 0,$$

so

$$u = \frac{dx}{dt} = -\frac{k_2}{k_1 - k_2}v.$$

Substituting $k_1 = 5,\text{cm}^{-1}$ and $k_2 = 6,\text{cm}^{-1}$ gives

$$u = -\frac{6}{5 - 6}v = -\frac{6}{-1}v = 6v.$$

Result

The fringe velocity is

$$u = 6v.$$

Substituting $v = 1,\text{cm/s}$ yields

$$u = 6 \cdot 1,\text{cm/s} = 6,\text{cm/s}.$$

$$\boxed{u = 6,\text{cm/s}}$$

Sanity Checks

The result depends only on the ratio of spatial frequencies, giving a dimensionless amplification factor $k_2/(k_2 - k_1)$, consistent with dimensional analysis since both $k_1$ and $k_2$ have units $\text{cm}^{-1}$ and cancel in the ratio.

If the two combs had equal densities, $k_1 = k_2$, the denominator would vanish and the moiré pattern speed would diverge, matching the physical expectation that identical periodicities produce a stationary infinite-scale interference pattern with no well-defined drift.

If the moving comb had a much smaller density than the stationary one, $k_2 \ll k_1$, the factor $k_2/(k_1 - k_2)$ would become small, indicating slow drift of the envelope relative to the motion, consistent with weak beating between nearly uniform and slowly varying structures.

The final velocity $6,\text{cm/s}$ exceeds the driving speed $1,\text{cm/s}$, which is physically consistent because moiré fringes represent a large-scale envelope whose motion is governed by phase accumulation rather than material transport, allowing apparent speeds greater than the underlying translation speed.