Questionnaire 2.1.2 — Plane Waves & Interference

Answer: b) no.

Why, step by step:

1. Feed a single term $\sin(\mathbf{k}\cdot\mathbf{r} - \omega t)$ into the equation: $\nabla^2$ returns $-\lvert\mathbf{k}\rvert^2$ and $\partial^2/\partial t^2$ returns $-\omega^2$, leaving $\bigl(-\lvert\mathbf{k}\rvert^2 + \omega^2/v^2\bigr)$ times the term. Each term therefore demands $v = \lvert\omega\rvert/\lvert\mathbf{k}\rvert$.
2. First term: the note fixes the coefficient of $x$ at $1$, so $\lvert\mathbf{k}\rvert = 1$ and $\omega = 0.2$ — it needs $v = 0.2$.
3. Second term: again $\lvert\mathbf{k}\rvert = 1$, with $\omega = -2.2$ (a left-mover) — it needs $v = 2.2$.
4. The two terms are linearly independent functions of $x$ and $t$, so nothing can cancel between them. The full residual is$$\left(-1 + \frac{0.04}{v^2}\right)\sin(x - 0.2t) + \left(-1 + \frac{4.84}{v^2}\right)\sin(x + 2.2t),$$and it vanishes for all $x, t$ only if $v^2 = 0.04$ and $v^2 = 4.84$ simultaneously. No number is both $0.2$ and $2.2$.
5. Eleven-to-one is the mismatch here, but any mismatch at all would do: one medium, one $v$.

Why the tempting options fail:

• a) is baited by the sum-to-product identity: $\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$ turns the sum into $2\sin(x + t)\cos(1.2t)$, which looks like the (space) × (time) form of a legitimate standing wave. It is not: in a standing wave the first factor is a function of position alone, and here $\sin(x+t)$ still mixes $x$ and $t$. Had the second term been $\sin(x + 0.2t)$, the identity would have given the genuine standing wave $2\sin(x)\cos(0.2t)$ at $v = 0.2$.
• a) is also baited by "the equation is linear, so any sum of waves is a solution". Linearity superposes solutions of the same equation; these two solve different ones.
• c) fails because nothing is missing: each term wears its own $\lvert\mathbf{k}\rvert$ and $\omega$ on its sleeve, and the note settles the dimensions.

See it: the two speeds are $0.2$ and $2.2$ here; set the demo's $v_1 \neq v_2$ and no choice of the equation's $v$ flattens the residual underneath — the top trace looks like a perfectly good wave the whole time. Then set $v_1 = v_2$ and match $v$: the residual dies and what remains is the standing wave that option a) mistook this sum for.

Answer: a) yes — with $v = \tfrac{1}{3}$.

Why, step by step:

1. The function is a pure-time factor times a pure-space factor, so each derivative hits only one of them. Time first: $\partial^2/\partial t^2$ acting on $\cos(t)$ gives $-1$, so $\partial^2\phi/\partial t^2 = -\phi$, i.e. $\omega = 1$.
2. Space: $\sin(\mathbf{k}\cdot\mathbf{r})$ with $\mathbf{k} = (1, 2, 2)$. Differentiating twice in each direction and adding gives$$\nabla^2\phi = -\bigl(1^2 + 2^2 + 2^2\bigr)\phi = -9\phi,$$so $\lvert\mathbf{k}\rvert^2 = 9$ and $\lvert\mathbf{k}\rvert = 3$.
3. The equation becomes $-9\phi - \frac{1}{v^2}(-\phi) = 0$, i.e. $-9 + 1/v^2 = 0$, so $v^2 = \tfrac{1}{9}$ and $v = \tfrac{1}{3}$. A wave velocity exists, so the answer is yes.
4. The same conclusion without any calculus: product-to-sum gives $2\cos(t)\sin(\mathbf{k}\cdot\mathbf{r}) = \sin(\mathbf{k}\cdot\mathbf{r} + t) + \sin(\mathbf{k}\cdot\mathbf{r} - t)$ — two counter-propagating plane waves, each with $\lvert\omega\rvert/\lvert\mathbf{k}\rvert = 1/3$. This is a three-dimensional standing wave, and standing waves are always built this way.

Why the tempting options fail:

• b) is where the arithmetic traps land. Reading $\lvert\mathbf{k}\rvert$ off the $x$ coefficient alone gives $1$; adding the components gives $1 + 2 + 2 = 5$; stopping at $\lvert\mathbf{k}\rvert^2 = 9$ and calling it $\lvert\mathbf{k}\rvert$ gives $9$. Each produces a different $v$ ($1$, $1/5$, $1/9$) — and any of them, noticed as "that doesn't look right", can push you to answer no. But nothing has to come out round: every positive $v$ is an allowed medium speed, so an ugly number is never grounds for rejection.
• b) is also tempting if a product of a space factor and a time factor doesn't feel like a wave. Every standing wave has exactly that shape, and step 4 unfolds this one into two travelling waves.
• c) fails because both $\omega$ and $\mathbf{k}$ are fully specified; the only unknown, $v$, is what we are allowed to solve for.

See it: press the first preset, $\mathbf{k} = (1, 2, 2)$ with $\omega = 1$. Two right triangles close the components into $\lvert\mathbf{k}\rvert = 3$ — never the sum $5$ — and the lanes underneath run the correct $v = \omega/\lvert\mathbf{k}\rvert = 0.333$ against the two mistaken speeds, so you can watch how far off each one drifts.

Answer: a) yes — for the specific velocity $v = 1/k$, where $\omega = 1$.

Why, step by step:

1. Read the exponent carefully: there is no $i$. This is a real exponential, so each spatial derivative brings down a factor without a sign flip: $\partial^2 e^{k_x x}/\partial x^2 = +k_x^2 e^{k_x x}$, and summing the three axes gives$$\nabla^2\phi = +k^2\phi \quad\text{(not } -k^2\phi\text{)}.$$
2. The time factor is $e^{-\omega^2 t}$, so $\partial\phi/\partial t = -\omega^2\phi$ and$$\frac{\partial^2\phi}{\partial t^2} = \omega^4\phi.$$
3. The wave equation therefore demands $k^2 - \omega^4/v^2 = 0$, i.e. $k^2 v^2 = \omega^4$.
4. Now impose the stated dispersion relation $\omega = vk$, so $\omega^4 = v^4k^4$:$$k^2 v^2 = v^4 k^4 \;\Longrightarrow\; 1 = v^2k^2 \;\Longrightarrow\; vk = 1.$$That fixes $v = 1/k$ and hence $\omega = vk = 1$. Such a velocity exists, so the answer is yes.
5. Sanity check with the explicit numbers: at $\omega = 1$ the function is $\phi = e^{\mathbf{k}\cdot\mathbf{r} - t}$, giving $\nabla^2\phi = k^2\phi$, $\partial^2\phi/\partial t^2 = \phi$, and $k^2\phi - \frac{1}{v^2}\phi = 0$ at $v = 1/k$. ✓
6. The structural reason: write $s = \hat{\mathbf{k}}\cdot\mathbf{r}$ for distance along the propagation direction. Then$$\phi = \exp\!\left(k\left(s - \frac{\omega^2}{k}t\right)\right),$$a rigidly translating profile — d'Alembert's $f(s - ut)$ — for every $\omega$, with translation speed $u = \omega^2/k$. The equation's own speed is $v = \omega/k$. Being a solution just means $u = v$, i.e. $\omega^2 = \omega$, i.e. $\omega = 1$.

Why the tempting options fail:

• b) is usually reached by "no $i$, so it doesn't oscillate, so it isn't a wave". The wave equation is a partial differential equation, and its general solution is $f(s - vt) + g(s + vt)$ for arbitrary twice-differentiable profiles. Real exponentials qualify. Their growth as $s\to\infty$ would only matter if boundary conditions were imposed, and none are.
• b) is also reached by a sign slip: carrying over $\nabla^2 \to -k^2$ from the familiar complex plane wave leaves $-k^2 - \omega^4/v^2$, which is strictly negative and can never vanish. The missing $i$ flips that sign to $+$ — and that flip is precisely what makes a solution possible.
• b) is reached a third way by noticing that $\omega^2 t$ breaks the usual pairing. It does change the arithmetic, but rather than killing the solution it pins $\omega$ to $1$.
• c) fails because the condition solves in closed form: two equations ($k^2v^2 = \omega^4$ and $\omega = vk$) in the two unknowns $v, \omega$, given $k$.
• A note on reading the question: if you instead read it as asking whether the function solves the equation for every $\omega$ obeying $\omega = vk$, the answer would be no — only $\omega = 1$ works. The question asks about existence ("for some value of the wave velocity"), and such a velocity exists.

See it: slide $\omega$ and watch the two fronts. The purple front is the wall itself, travelling at $\omega^2/k$; the blue one is where the equation says a disturbance of speed $\omega/k$ should be. They separate everywhere except at $\omega = 1$, where the residual $k^2(1 - \omega^2)$ dies — and then $k$ can be dragged anywhere at all without breaking the lock, since $v$ simply follows as $1/k$.

Answer: a) yes — with $v = 0.1/\sqrt{14} \approx 0.0267$.

Why, step by step:

1. Each term is a plane wave $e^{i(\mathbf{k}_n\cdot\mathbf{r} - \omega_n t)}$, so each demands $v = \lvert\omega_n\rvert/\lvert\mathbf{k}_n\rvert$, and the sum is a solution exactly when those two demands agree.
2. First wavevector $\mathbf{k}_1 = (2, 1, 3)$: $\lvert\mathbf{k}_1\rvert^2 = 4 + 1 + 9 = 14$.
3. Second wavevector $\mathbf{k}_2 = (1, 2, 3)$: $\lvert\mathbf{k}_2\rvert^2 = 1 + 4 + 9 = 14$. The components have merely been permuted, and a sum of squares does not care about order — the magnitudes are identical, $\sqrt{14}$.
4. Both terms carry the same $\omega = 0.1$, so both need $v = 0.1/\sqrt{14} \approx 0.0267$. One speed serves both, and the superposition solves that wave equation.
5. Geometrically: the two waves travel in genuinely different directions — their fronts cross at an angle — but they cross at the same speed, which is all the equation ever asks.

Why the tempting options fail:

• b) is the over-correction from E1. There, two visibly different terms killed the solution, so "the wavevectors are different, therefore no" feels like the same lesson. It is not: E1's terms differed in the ratio $\lvert\omega\rvert/\lvert\mathbf{k}\rvert$, while these differ only in direction. $\nabla^2$ is rotationally invariant and returns $-\lvert\mathbf{k}\rvert^2$; the individual components never appear.
• b) also tempts anyone who compares the vectors component by component — $2 \neq 1$, $1 \neq 2$ — instead of comparing the one scalar the equation actually forms.
• c) fails because both wavevectors and both frequencies are given explicitly.
• Worth noting: the "add the components" mistake gives $6$ for both wavevectors here, so it happens to reach the right verdict by luck. E6 is where it stops being lucky.

See it: swing $\angle\mathbf{k}_2$ through the whole half-circle at fixed $\lvert\mathbf{k}_2\rvert$ — the fronts cross at every angle you like and the two speeds never move off each other, so the verdict stays green. Then nudge the length slider by one notch: the speeds split at once. Length is the only thing the equation compares.

Answer: a) yes — with $v = 1/\sqrt{3} \approx 0.577$.

Why, step by step:

1. First term: $\mathbf{k}_1 = (1,1,1)$, so $\lvert\mathbf{k}_1\rvert = \sqrt{3}$, and the time dependence $e^{-it}$ gives $\omega_1 = 1$. Its required speed is$$v_1 = \frac{\lvert\omega_1\rvert}{\lvert\mathbf{k}_1\rvert} = \frac{1}{\sqrt{3}} \approx 0.577.$$
2. Second term: $\mathbf{k}_2 = (1,0,0)$, so $\lvert\mathbf{k}_2\rvert = 1$. Matching the standard form $e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)}$ against $e^{i(\mathbf{k}_2\cdot\mathbf{r} + t/\sqrt{3})}$ gives $\omega_2 = -1/\sqrt{3}$. Its required speed is$$v_2 = \frac{\lvert\omega_2\rvert}{\lvert\mathbf{k}_2\rvert} = \frac{1/\sqrt{3}}{1} = \frac{1}{\sqrt{3}} \approx 0.577.$$
3. $v_1 = v_2$, so both terms solve the same wave equation and their sum does too, with $v = 1/\sqrt{3}$.
4. Check it directly if you prefer: the residual coefficients are $-\lvert\mathbf{k}\rvert^2 + \omega^2/v^2$, namely $-3 + 1/(1/3) = 0$ and $-1 + (1/3)/(1/3) = 0$. Both vanish at the same $v$.

Why the tempting options fail:

• b) is baited by how unalike the two terms look: different magnitudes ($\sqrt{3}$ against $1$), different frequencies ($1$ against $1/\sqrt{3}$), different amplitudes ($2$ against $4$), and one written with $+t$ where the other has $-t$. Every one of those differences is invisible to the equation. What it compares is the single ratio $\lvert\omega\rvert/\lvert\mathbf{k}\rvert$, and $\omega$ and $\lvert\mathbf{k}\rvert$ here shrink by exactly the same factor $\sqrt{3}$.
• b) is also reached by treating the $+t$ as fatal. A positive sign means $\omega < 0$: the wave runs the other way. The wave equation contains $v^2$ and $\omega^2$ only, so it cannot see that sign at all — as E1 and E2 already showed, counter-propagating waves happily share one equation (that is what a standing wave is).
• b) by way of the amplitudes $2$ and $4$: a linear equation is indifferent to how loud each term is.
• c) fails because everything needed is written down; only $v$ is unknown, and it is exactly what we solve for.

See it: load the "unlike wavevectors, one speed" preset — this pair, rounded. The two points sit at completely different places in the $\omega$–$k$ plane and still land on one line through the origin, so the strip below shows their sum translating rigidly. Now drag $\omega_1$ negative: the point does not move at all, because the plane is drawn in $\lvert\omega\rvert$. Drag $k_2$ instead and the sum starts to churn as it travels.

Answer: $v_1 \approx 2.67$.

Why, step by step:

1. Read the wavevector straight off the exponent: $\mathbf{k}_1 = (0.1,\, 0.2,\, 0.3)$.
2. Combine the components in quadrature:$$\lvert\mathbf{k}_1\rvert = \sqrt{0.1^2 + 0.2^2 + 0.3^2} = \sqrt{0.01 + 0.04 + 0.09} = \sqrt{0.14} \approx 0.374166.$$
3. Read the angular frequency off the time term: $-t$ means $-1t$, so $\omega_1 = 1$.
4. Divide:$$v_1 = \frac{\lvert\omega_1\rvert}{\lvert\mathbf{k}_1\rvert} = \frac{1}{0.374166} = 2.6726\ldots \approx 2.67.$$

Why the tempting wrong values fail:

• $1.67$ comes from adding the components, $0.1 + 0.2 + 0.3 = 0.6$, and dividing $1/0.6$. The Laplacian produces $-\bigl(k_x^2 + k_y^2 + k_z^2\bigr)$; there is no arrangement of derivatives that yields a plain sum.
• $7.14$ stops one step early at $\lvert\mathbf{k}_1\rvert^2 = 0.14$ and divides by that. Since $0.14 < 1$, forgetting the square root here makes the speed too large — the same slip on a wavevector longer than $1$ would make it too small.
• $0.374$ is $\lvert\mathbf{k}_1\rvert/\lvert\omega_1\rvert$, the ratio inverted. Dimensionally, speed is (rad/s) per (rad/m), so $\omega$ belongs on top.
• $0.425$ or $16.8$ come from importing a $2\pi$: $\omega$ is already an angular frequency in rad/s, and $\lvert\mathbf{k}\rvert$ is already $2\pi/\lambda$. The ratio needs no conversion.

See it: press the second preset, $\mathbf{k} = (0.1, 0.2, 0.3)$ with $\omega = 1$. The box redraws with these components and reads out $\lvert\mathbf{k}\rvert = 0.374$ and $v = 2.673$. Watch the "forgot the square root" lane in particular: with $\lvert\mathbf{k}\rvert < 1$ it now runs ahead of the truth, the opposite of what it did on E2's preset.

Answer: $v_2 \approx 2.28$.

Why, step by step:

1. $\mathbf{k}_2 = (0.4,\, 0.5,\, 0.6)$, so$$\lvert\mathbf{k}_2\rvert = \sqrt{0.16 + 0.25 + 0.36} = \sqrt{0.77} \approx 0.877496.$$
2. The time term is $-2t$, so $\omega_2 = 2$ — twice E6's frequency.
3. $$v_2 = \frac{2}{0.877496} = 2.2792\ldots \approx 2.28.$$
4. Note what did not happen: $\lvert\mathbf{k}_2\rvert$ is only $2.35$ times $\lvert\mathbf{k}_1\rvert$ while $\omega$ doubled, so the two speeds land close together — $2.67$ against $2.28$ — without being equal. Close is not equal, and E8 turns on exactly that.

Why the tempting wrong values fail:

• $2.67$ is E6's answer reused, from assuming both waves share a speed. They share a medium in the sense of living in the same problem, but nothing in the question says they obey one wave equation — deciding that is E8's job.
• $1.14$ keeps $\omega = 1$ from E6 and forgets that this exponent carries $-2t$. The frequency must be reread for every wave.
• $1.33$ adds the components again, $0.4 + 0.5 + 0.6 = 1.5$, then divides $2/1.5$.
• $2.60$ takes $\sqrt{0.77} \approx 0.877$ and rounds it to $0.77$ somewhere in the middle of the division; keep the square root to full precision until the final rounding.

See it: press the third preset, $\mathbf{k} = (0.4, 0.5, 0.6)$ with $\omega = 2$, then flip back to the second. The construction shows $\lvert\mathbf{k}\rvert$ going from $0.374$ to $0.877$ while $\omega$ only doubles — nearly, but not quite, cancelling — and the correct-speed lane changes pace between the two presets by just enough to matter.

Answer: b) no.

Why, step by step:

1. From E6, $\phi_1$ solves the wave equation with $v_1 \approx 2.67$; from E7, $\phi_2$ solves it with $v_2 \approx 2.28$. Different equations.
2. Substituting the sum into a trial equation of speed $v$ leaves$$\left(-\lvert\mathbf{k}_1\rvert^2 + \frac{\omega_1^2}{v^2}\right)\phi_1 + \left(-\lvert\mathbf{k}_2\rvert^2 + \frac{\omega_2^2}{v^2}\right)\phi_2 .$$
3. $\phi_1$ and $\phi_2$ are linearly independent (their wavevectors differ, so no multiple of one equals the other), so the residual vanishes for all $\mathbf{r}, t$ only if both brackets vanish separately.
4. That would need $v^2 = \omega_1^2/\lvert\mathbf{k}_1\rvert^2 = 1/0.14 \approx 7.14$ and $v^2 = \omega_2^2/\lvert\mathbf{k}_2\rvert^2 = 4/0.77 \approx 5.19$ at the same time. No.

Why the tempting options fail:

• a) rests on "the wave equation is linear, so sums of solutions are solutions". The theorem is narrower than that: it superposes solutions of one and the same equation. Each of these waves is a flawless solution — of a different equation.
• a) is also propped up by how close the two speeds are, $2.67$ against $2.28$. A near miss is a miss: the residual amplitude is set by the difference, not by whether it is small.
• c) fails because E6 and E7 already extracted both speeds; nothing further is needed. It would be the honest answer only if the medium were unspecified and the frequencies unreadable — here they are printed in the exponents.
• Worth keeping straight: this superposition is still a perfectly good function, and you can write it down and plot it. It just is not a solution of any classical wave equation. Contrast E4 and E5, where the ratios matched and the sum survived.

See it: load the "similar wavevectors, two speeds" preset — this pair, rounded to $(0.37, 1)$ and $(0.88, 2)$. The two points look near-neighbours, yet drag $v$ through its whole range and the line can be made to pass through one or the other, never both. The strip underneath shows the price: the sum's shape churns as it travels instead of holding together.

Answer: b) $2 + 2\cos(0.2y)$.

Why, step by step:

1. The two wavevectors share their $x$ part and differ only in the sign of their $y$ part: $\mathbf{k}_{3,4} = (0.6, \pm 0.1, 0)$. Factor the common phase out of the sum:$$\phi_3 + \phi_4 = e^{i(0.6x - t)}\left(e^{i0.1y} + e^{-i0.1y}\right).$$
2. The bracket is a real cosine by Euler's formula, $e^{i\theta} + e^{-i\theta} = 2\cos\theta$:$$\Phi_{tot} = 2\cos(0.1y)\,e^{i(0.6x - t)}.$$
3. Take the modulus squared. The travelling factor has modulus exactly $1$ — it is a pure phase — so it disappears completely:$$\lvert\Phi_{tot}\rvert^2 = \bigl(2\cos(0.1y)\bigr)^2 \cdot 1 = 4\cos^2(0.1y).$$
4. Convert with the double-angle identity $\cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta)$:$$4\cos^2(0.1y) = 2\bigl(1 + \cos(0.2y)\bigr) = 2 + 2\cos(0.2y).$$This is why the question points at the trigonometric identities: $4\cos^2(0.1y)$ is equally correct, but only its double-angle form appears in the list.

Why the tempting options fail:

• a) $2$ is what you get by adding intensities instead of amplitudes: $\lvert\phi_3\rvert^2 + \lvert\phi_4\rvert^2 = 1 + 1 = 2$. That throws away the cross term $2\,\mathrm{Re}(\phi_3^{*}\phi_4)$, which is the interference. Note it is exactly the average of the true answer over $y$ — the pattern smeared flat — which is why it looks so plausible.
• c) $2 + 2\cos(1.2x - 2t)$ adds the two phases where the modulus subtracts them: $(\mathbf{k}_3 + \mathbf{k}_4)\cdot\mathbf{r} - 2\omega t = 1.2x - 2t$. But the common part is the factor that got pulled out and normalized away in step 3; only the difference $\mathbf{k}_3 - \mathbf{k}_4 = (0,\,0.2,\,0)$ survives. Physically, c) describes a pattern sweeping along $x$ at the wave speed, which no detector could resolve — real two-beam fringes stand still.
• d) fails because both waves are given completely; no medium property is needed, since the modulus squared is pure algebra.
• e) fails because b) is exactly right, in its double-angle disguise.

See it: drag the $y$ cut. The purple trace in the top panel — the actual wave along $x$ — races and swells and dies as you cross bright and dark fringes, but the $\lvert\Phi\rvert^2$ line beside it stays dead flat at every $y$ and every instant: that flatness is option c) failing. The bottom panel is what is left, $2 + 2\cos(2qy)$, standing still.

Answer: c) $31.4$.

Why, step by step:

1. From E9 the intensity is $\lvert\Phi_{tot}\rvert^2 = 2 + 2\cos(0.2y)$. Its maxima are where $\cos(0.2y) = 1$, i.e. $0.2y = 2\pi n$, so$$y_n = \frac{2\pi n}{0.2} = 10\pi n, \qquad n = 0, \pm 1, \pm 2,\dots$$
2. Adjacent maxima are therefore separated by$$\Delta y = 10\pi = 31.4159\ldots \approx 31.4.$$
3. The same number from the general rule: fringe spacing is $2\pi/\lvert\Delta\mathbf{k}\rvert$ with $\Delta\mathbf{k} = \mathbf{k}_3 - \mathbf{k}_4 = (0,\,0.2,\,0)$, giving $2\pi/0.2 = 31.4$.
4. Both premises the exotic options attack do hold. Both waves have $\lvert\mathbf{k}\rvert = \sqrt{0.36 + 0.01} = \sqrt{0.37} \approx 0.608$ and $\omega = 1$, hence the same speed $v \approx 1.644$ — so the superposition is a solution — and its modulus squared genuinely varies with $y$ — so a pattern is present.

Why the tempting options fail:

• d) $15.7$ is $5\pi$, exactly half the spacing: the distance from a bright fringe to the next dark one. Maximum-to-minimum is half a period, and a fringe spacing is a full period — maximum to next maximum, or minimum to next minimum, as the question itself specifies.
• a) $0.200$ is $\lvert\Delta\mathbf{k}\rvert$ itself, quoted as though it were a length. It has units of inverse metres; spacings go as its reciprocal, times $2\pi$.
• b) $5.00$ is $1/0.2$: the reciprocal taken without the factor $2\pi$ — the same slip as writing $k = 1/\lambda$ instead of $k = 2\pi/\lambda$. It is off by exactly $2\pi$ from the right answer.
• e) would need $\lvert\Phi_{tot}\rvert^2$ to be constant; step 1 shows it oscillates in $y$. It would be the right answer if the two wavevectors were identical, since then $\Delta\mathbf{k} = 0$ and the "spacing" is infinite.
• f) would need the two speeds to differ, as they did in E8; step 4 shows they do not.
• g) fails because everything follows from the two exponents. h) fails because c) is right.

See it: at the default $q = 0.1$ the bracket between the first two crests reads $31.4$, with the dashed half-bracket beside it marking option d)'s $15.7$ reaching only as far as the first dark line. Then open the beams up: raising $q$ packs the fringes tighter, spacing $\propto 1/q$ — the reciprocal relationship options a) and b) each mangle in their own way.