Exercises — Interference — constructive, destructive conditions

WHAT we do: compare $\Delta x$ to $\lambda$. WHY: the parent condition says a whole number of wavelengths means the waves line up crest-on-crest. $\frac{\Delta x}{\lambda }=\frac{3\lambda }{\lambda }=3(\text{a whole number, order }n=3)$ Whole number $\Rightarrow$ constructive. (Mnemonic: Whole = Wow.)

WHAT/WHY: in phase means $\varphi =0$, so feed it into $I=4I_0{\cos }^2(\varphi /2)$. $I=4I_0{\cos }^2(0)=4I_0\cdot 1=4I_0$ This is the brightest possible point — four times a single wave, not two. The "extra" energy is borrowed from the dark points, as Energy in Waves insists.

WHAT we do: compare $\Delta x$ to $\lambda$ and read off the fractional part. WHY: a half extra wavelength of path means one wave is exactly half a cycle behind — crest arrives where the other's trough arrives, so they cancel (this is the $\Delta x=(n+\frac{1}{2})\lambda$ destructive condition from the parent note). $\frac{\Delta x}{\lambda }=2.5=\left(2+\frac{1}{2}\right)$ A half-integer (order $n=2$, plus a half) $\Rightarrow$ destructive. (Half = Hush.)

Units first: $90^{\circ }=\pi /2\text{ rad}$. Either form is fine, but your calculator's mode must match — here we work in degrees, so set it to DEG (or convert and use RAD). Then halve the phase and square the cosine. $I=4I_0{\cos }^2\left(\frac{90^{\circ }}{2}\right)=4I_0{\cos }^{2}45^{\circ }=4I_0{\left(\frac{1}{\sqrt{2}}\right)}^2=4I_0\cdot \frac{1}{2}=2I_0$ Exactly the average intensity — $90^{\circ }$ ($=\pi /2$) is the "neutral" phase, neither brightening nor darkening.

WHAT: use the master link to turn distance into phase. WHY: the intensity law speaks in $\varphi$. The $2\pi$ in the link means $\varphi$ comes out in radians. $\varphi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{0.40}\times 0.60=2\pi \times 1.5=3\pi \text{ rad}$ $3\pi =(2\cdot 1+1)\pi$, an odd multiple of $\pi$ $\Rightarrow$ destructive — a quiet spot. (Using periodicity, $3\pi$ reduces to $\pi$ — same result.) Check by ratio: $\Delta x/\lambda =1.5$, a half-integer. ✓

Units: the phase is quoted in degrees, $\varphi =120^{\circ }$, so we keep the calculator in DEG and work entirely in degrees below — no need to convert. (If you prefer radians, use $\varphi =2\pi /3$ and ${\cos }^2(\pi /3)$ in RAD mode; same answer.) $I=4I_0{\cos }^2\left(\frac{120^{\circ }}{2}\right)=4(5){\cos }^{2}60^{\circ }=20{\left(\frac{1}{2}\right)}^2=20\cdot \frac{1}{4}=5{\text{W m}}^{-2}$ Same as one source alone — a nice reminder that overlapping does not guarantee brightening.

WHY the general formula — build it from the picture. Each wave is drawn as a phasor: a little arrow whose length is the amplitude and whose direction is the phase. Look at the figure: the first arrow $a_1$ lies flat; the second arrow $a_2$ is turned by the phase angle $\varphi$ and placed tip-to-tail on the first. The resultant $A$ (the red arrow) is the arrow from the very start to the very end. To find its length we use the triangle formed by the two arrows. The angle inside the triangle, between $a_1$ (extended) and $a_2$, is $180^{\circ }-\varphi$, so the law of cosines gives $A^2=a_1^2+a_2^2-2a_1{a}_2\cos (180^{\circ }-\varphi )=a_1^2+a_2^2+2a_1{a}_2\cos \varphi ,$ because $\cos (180^{\circ }-\varphi )=-\cos \varphi$. Hence $A=\sqrt{a_1^2+a_2^2+2a_1{a}_2\cos \varphi }.$ Now read off the extremes from the arrows:

• Max at $\varphi =0$ ($\cos \varphi =1$): the arrows point the same way and simply add — $A_{\max }=a_1+a_2=8$.
• Min at $\varphi =\pi$ ($\cos \varphi =-1$): the arrows point opposite ways and partly cancel — $A_{\min }=|a_1-a_2|=2$, not zero, because a short arrow cannot erase a long one. $\frac{I_{\max }}{I_{\min }}={\left(\frac{A_{\max }}{A_{\min }}\right)}^2={\left(\frac{8}{2}\right)}^2=16.$

WHAT: $I=\frac{1}{4}{I}_{\max }=\frac{1}{4}(4I_0)=I_0$. Set the law equal and solve for $\varphi$. $4I_0{\cos }^2\frac{\varphi }{2}=I_0\Rightarrow {\cos }^2\frac{\varphi }{2}=\frac{1}{4}\Rightarrow \cos \frac{\varphi }{2}=\frac{1}{2}$ $\frac{\varphi }{2}=60^{\circ }\Rightarrow \varphi =120^{\circ }=\frac{2\pi }{3}\text{ rad}.$ Why we took the positive root: the smallest positive $\varphi$ comes from the positive cosine branch. (Every other solution differs from this by a whole $2\pi$, which by periodicity gives the same intensity.)

Destructive points sit at $\Delta x=(n+\frac{1}{2})\lambda$, where the order $n$ steps up by one from one dark point to the next. Consecutive ones differ by exactly one $\lambda$: $\Delta x_{n+1}-\Delta x_n=\left[(n+1)+\frac{1}{2}\right]\lambda -\left[n+\frac{1}{2}\right]\lambda =\lambda =600\text{ nm}.$ The spacing between successive dark points is one wavelength — a fact Young's Double Slit Experiment turns into fringe spacing.

WHAT: find each speaker's distance to the listener, subtract for $\Delta x$. WHY: the condition depends only on the difference in path length. Distance from $S_1$ (straight in front): $r_1=L=4.0\text{ m}$. Distance from $S_2$: it forms a right triangle with legs $L=4.0$ and $d=3.0$: $r_2=\sqrt{L^2+d^2}=\sqrt{4.0^2+3.0^2}=\sqrt{16+9}=\sqrt{25}=5.0\text{ m}.$ $\Delta x=r_2-r_1=5.0-4.0=1.0\text{ m}=1\lambda .$ One whole wavelength (order $n=1$) $\Rightarrow$ constructive — a loud spot ($I=4I_0$).

The geometry is unchanged, so $\Delta x=1.0\text{ m}$ still. Now $\frac{\Delta x}{\lambda }=\frac{1.0}{2.0}=0.5=\left(0+\frac{1}{2}\right),$ a half-integer (order $n=0$) $\Rightarrow$ destructive. Formally, $\varphi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{2.0}(1.0)=\pi \text{ rad},I=4I_0{\cos }^2\frac{\pi }{2}=0.$ Same room, same speakers — doubling the wavelength flipped loud into silent. The pattern is glued to $\lambda$, not to the geometry alone.

WHAT: average $I=4I_0{\cos }^2(\varphi /2)$ over a full cycle of $\varphi$. WHY: a slow uniform drift means every phase is equally likely, so we take the mean value — and by periodicity one full turn of $2\pi$ captures every possible behaviour. The mean of ${\cos }^2\theta$ over a full cycle is $\frac{1}{2}$, so $⟨I⟩=4I_0\cdot \frac{1}{2}=2I_0.$ This is exactly $I_0+I_0$ — the incoherent result: intensities simply add and no pattern survives. Compared to the coherent maximum $4I_0$, the drifting case is half as bright at best, and shows no fringes. This is precisely why Coherence and Path Difference demands a steady $\varphi$.

WHAT/WHY: use the equal-amplitude resultant $A=2a\cos (\varphi /2)$ from the parent note at each fixed phase.

• At $\varphi =\pi$: $A=2a\cos (\pi /2)=2a\cdot 0=0$. The point stays at zero displacement for all time — a node.
• At $\varphi =0$ (or $2\pi$ — the same by periodicity): $A=2a\cos 0=2a$, the largest possible swing — an antinode.

Because the two waves travel toward each other, this pattern of permanently-still nodes and maximally-swinging antinodes is fixed in space and does not move along the string. That frozen interference pattern is a standing wave.

WHY amplitudes: intensity $\propto$ amplitude${}^2$, so $\frac{a_1}{a_2}=\sqrt{\frac{I_1}{I_2}}=\sqrt{4}=2$. Take $a_2=1$, $a_1=2$. $A_{\max }=a_1+a_2=3,A_{\min }=|a_1-a_2|=1.$ $\frac{I_{\max }}{I_{\min }}={\left(\frac{A_{\max }}{A_{\min }}\right)}^2={\left(\frac{3}{1}\right)}^2=9.$ So $I_{\max }:I_{\min }=9:1$. (General formula: ${\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)}^2$.)

Because the frequencies differ, the phase difference $\varphi$ is not constant — it grows steadily with time, sweeping through constructive ($\varphi =2\pi n$, loud) and destructive ($\varphi =(2n+1)\pi$, silent) as the clock ticks. The result is loudness that waxes and wanes: Beats. Spatial interference needs equal frequencies; temporal beats come from unequal ones.