Kvant Physics Problem 18
Two transparent combs are superimposed and viewed against a uniform source of light.
Verified: yes
Verdicts: PASS + PASS
Solve time: 11m50s
Source on kvant.digital
Problem
If one looks at light through two combs with different tooth spacings 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 transparent combs are superimposed and viewed against a uniform source of light. The teeth and gaps form periodic transmission gratings. One comb is fixed, while the other moves parallel to the teeth at a constant speed.
The stationary comb has a tooth density
$$n_1=5\ \text{cm}^{-1},$$
and the moving comb has
$$n_2=6\ \text{cm}^{-1}.$$
The relative speed of the moving comb is
$$v=1\ \text{cm/s}.$$
The unknown quantity is the speed $V$ of the alternating bright and dark regions, that is, the speed of the moiré pattern produced by the superposition of the two periodic structures.
The combs are treated as ideal one dimensional periodic gratings. Their teeth are assumed to be sufficiently narrow and regular that the visible bright and dark bands are determined only by the relative phase of the two periodic patterns. Edge effects and diffraction are neglected.
Physical Principles
A periodic structure with spatial period $d$ may be described by a phase that changes by $2\pi$ over each period.
If two periodic structures of periods $d_1$ and $d_2$ are superimposed, the large scale bright and dark bands arise from the slowly varying phase difference between the two structures.
The positions of the bright bands are determined by the condition that the phase difference be constant. The velocity of these bands is obtained by differentiating that condition with respect to time.
The spatial period of a comb is the reciprocal of its tooth density:
$$d=\frac{1}{n}.$$
Derivation
Let the coordinate along the combs be $x$.
The stationary comb has period
$$d_1=\frac{1}{n_1},$$
and the moving comb has period
$$d_2=\frac{1}{n_2}.$$
The phase of the stationary comb may be written as
$$\phi_1=\frac{2\pi x}{d_1}.$$
Since the second comb moves with speed $v$, its phase is
$$\phi_2=\frac{2\pi (x-vt)}{d_2}.$$
The bright regions correspond to a fixed value of the phase difference
$$\Delta\phi=\phi_1-\phi_2.$$
Substituting the expressions above gives
$$\Delta\phi = 2\pi \left( \frac{x}{d_1} - \frac{x-vt}{d_2} \right).$$
For a given bright band, $\Delta\phi$ remains constant. Hence
$$\frac{x}{d_1} - \frac{x-vt}{d_2} = \text{const}.$$
Expanding,
$$x\left(\frac1{d_1}-\frac1{d_2}\right) +\frac{v t}{d_2} = \text{const}.$$
Differentiating with respect to time,
$$V\left(\frac1{d_1}-\frac1{d_2}\right) +\frac{v}{d_2} =0,$$
where
$$V=\frac{dx}{dt}$$
is the speed of the moiré bands.
Solving for $V$,
$$V = -\frac{v/d_2}{(1/d_1)-(1/d_2)}.$$
Using $1/d_1=n_1$ and $1/d_2=n_2$,
$$V = -\frac{v,n_2}{n_1-n_2} = \frac{v,n_2}{n_2-n_1}.$$
The positive sign means that the bands move in the same direction as the moving comb.
Result
The speed of the bright and dark regions is
$$V=\frac{v,n_2}{n_2-n_1}.$$
Substituting the given values,
$$V = \frac{(1\ \text{cm/s})(6\ \text{cm}^{-1})} {6\ \text{cm}^{-1}-5\ \text{cm}^{-1}}.$$
Thus
$$V = \frac{6\ \text{cm/s}} {1} = 6\ \text{cm/s}.$$
Hence
$$\boxed{V=6\ \text{cm/s}}.$$
The bright and dark bands move in the same direction as the moving comb.
Sanity Checks
The factor multiplying $v$ is
$$\frac{n_2}{n_2-n_1},$$
which is dimensionless. The resulting unit is therefore cm/s, matching the required unit of velocity.
If the two combs had nearly equal tooth densities, $n_2-n_1$ would be very small. The moiré pattern would then have a very large spatial period and would move much faster than either comb. This behavior agrees with familiar moiré phenomena.
If the densities were exactly equal, the denominator would vanish. In that limit there is no finite moiré wavelength, and the pattern ceases to be a distinct moving set of bands. The singular behavior of the formula correctly signals this special case.
The most common place for an error is the differentiation of the constant phase condition. A sign mistake in the term containing $vt$ would predict motion opposite to the actual direction, while confusing tooth density $n$ with spacing $d$ would produce an incorrect factor in the final speed.