Interference — constructive, destructive conditions: Worked examples

This is a companion to the parent note. There we derived the two master laws. Here we drill every kind of question those laws can produce — every sign of $n$ , the zero and degenerate cases, the limits, a word problem, and an exam twist. Nothing on the parent page is contradicted; we only go wider and slower.

Before we start, one reminder in plain words so no symbol is unearned:

Recall The two tools we reuse everywhere

$ϕ$ (phase difference) = "how many radians out of step the two waves are." One full step-around a circle is $2 π$ radians.
$Δ x$ (path difference) = "how much farther one wave travelled than the other," measured in metres.
$λ$ (wavelength) = "the length of one full wave," in metres.
$I_{0}$ = intensity (brightness/loudness) of one wave alone. Intensity is how much energy the wave delivers, and it scales as amplitude squared.
The bridge between the two worlds: $ϕ = \frac{2 π}{λ} Δ x$ .
Two equal sources: $I = 4 I_{0} cos^{2} \frac{ϕ}{2}$ . Unequal sources $a_{1}, a_{2}$ : $A = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} cos ϕ$ .

The scenario matrix

Every exam question about interference lands in one of these cells. Each example below is tagged with its cell.

#	Cell (case class)	What is special about it	Example
A	Phase given, ordinary angle	plug $ϕ$ into $4 I_{0} cos^{2} (ϕ /2)$	Ex 1
B	Path given, integer ratio	$Δ x / λ$ is whole → constructive	Ex 2
C	Path given, half-integer ratio	$Δ x / λ$ is half → destructive	Ex 2
D	The zero case ( $Δ x = 0$ , $ϕ = 0$ )	central maximum, brightest, sign trap	Ex 3
E	Negative order $n$	source on the other side; sign of $n$	Ex 4
F	In-between value (neither max nor min)	answer is a fraction of $4 I_{0}$	Ex 5
G	Degenerate: unequal amplitudes	cannot fully cancel; law of cosines	Ex 6
H	Degenerate limit: $a_{2} \to 0$	one wave dies → no fringes	Ex 6b
I	Real-world word problem	speakers / thin film, extract $Δ x$	Ex 7
J	Exam twist: intensity ratio → amplitude ratio	invert $I \propto a^{2}$ first	Ex 8

Example 1 — Cell A: phase given, ordinary angle

Example 2 — Cells B & C: path given, integer vs half-integer

Worked example Two speakers emit sound of

$λ = 0.40$ m. Listener P has path difference $Δ x = 1.20$ m; listener Q has $Δ x = 1.00$ m. Loud or quiet for each? Forecast: Divide each $Δ x$ by $λ$ . Whole → loud, half → quiet. Guess for P and Q before computing.

Step 1 — P: $\frac{Δ x}{λ} = \frac{1.20}{0.40} = 3.0$ . Why this step? Counting how many whole wavelengths fit into the extra path tells us if the waves re-line-up. A whole number means they re-line-up exactly. $3.0 = 3 λ$ → integer → constructive → P is loud ( $n = 3$ ).

Step 2 — Q: $\frac{Δ x}{λ} = \frac{1.00}{0.40} = 2.5 = (2 + \frac{1}{2})$ . Why this step? A half-extra wavelength flips one wave crest-to-trough against the other. Half-integer → destructive → Q is quiet ( $n = 2$ ).

Verify: Convert to phase to double-check. $ϕ_{P} = \frac{2 π}{0.40} (1.20) = 6 π = 2 π \cdot 3$ → multiple of $2 π$ → constructive ✓. $ϕ_{Q} = \frac{2 π}{0.40} (1.00) = 5 π = (2 \cdot 2 + 1) π$ → odd multiple of $π$ → destructive ✓. Both routes agree — see Coherence and Path Difference for why steady $Δ x$ is what makes this stable.

Example 3 — Cell D: the zero case (the classic trap)

Example 4 — Cell E: negative order $n$ (the other side)

Worked example In a fringe pattern (see

Young's Double Slit Experiment) a bright fringe sits where $Δ x = - 2 λ$ . Is it valid? Which order? Is it as bright as $+ 2 λ$ ? Forecast: Does the minus sign change brightness? Guess.

Step 1 — Read the sign geometrically.

A negative

Δ x

just means the other slit is the farther one — the point lies on the opposite side of the centre. Look at the figure: the two coral rays have swapped which is longer. Why this step?

Δ x

carries a sign only to tell us which wave leads; brightness cannot care about a label.

Step 2 — Apply the condition. $Δ x = nλ$ with $n = - 2$ . Since $- 2$ is an integer, it is constructive — a valid bright fringe, order $n = - 2$ . Why this step? The constructive condition allows $n = 0, \pm 1, \pm 2, \dots$ — negatives included.

Step 3 — Brightness. $ϕ = \frac{2 π}{λ} (- 2 λ) = - 4 π$ ; $cos^{2} (- 4 π /2) = cos^{2} (- 2 π) = 1$ , so $I = 4 I_{0}$ . Why this step? $cos$ is an even function: $cos (- θ) = cos θ$ . The sign washes out.

Verify: $cos^{2} (- 2 π) = cos^{2} (2 π) = 1$ ✓. Same brightness as the $+ 2 λ$ fringe — the pattern is symmetric about the centre, exactly as figures of double-slit fringes show.

Example 5 — Cell F: an in-between value

Example 6 — Cell G: unequal amplitudes (degenerate — can't fully cancel)

Worked example Sources of amplitude

$a_{1} = 5 a$ and $a_{2} = 2 a$ . Find the maximum and minimum resultant amplitude, and the max/min intensity relative to $I_{0}$ (of a single unit- $a$ wave). Forecast: With unequal waves, can destructive interference reach exactly zero? Guess yes/no first.

Step 1 — General resultant (law of cosines).

A = (5 a)^{2} + (2 a)^{2} + 2 (5 a) (2 a) cos ϕ

Why this step? When amplitudes differ we can no longer use

sin A + sin B

; instead we add the two waves as phasors — little arrows of length

5 a

and

2 a

with angle

ϕ

between them (see figure). The length of the sum is the law of cosines.

Step 2 — Maximum ( $ϕ = 0$ , arrows aligned). $cos 0 = 1$ : $A_{m a x} = 25 a^{2} + 4 a^{2} + 20 a^{2} = 49 a^{2} = 7 a .$ Why this step? Aligned arrows just add their lengths: $5 a + 2 a = 7 a$ .

Step 3 — Minimum ( $ϕ = π$ , arrows opposed). $cos π = - 1$ : $A_{m i n} = 25 a^{2} + 4 a^{2} - 20 a^{2} = 9 a^{2} = 3 a = ∣5 a - 2 a ∣.$ Why this step? Opposed arrows subtract: the leftover is the difference, never zero when lengths differ.

Step 4 — Intensities. With one unit- $a$ wave having $I_{0} \propto a^{2}$ : $I_{m a x} \propto (7 a)^{2} = 49 a^{2} = 49 I_{0}, I_{m i n} \propto (3 a)^{2} = 9 a^{2} = 9 I_{0} .$

Verify: $A_{m i n} = 3 a \neq = 0$ ✓ — the forecast "cannot fully cancel" is correct: full cancellation needs equal amplitudes. Sanity: $A_{m a x} = 7 a = 5 a + 2 a$ and $A_{m i n} = 3 a = ∣5 a - 2 a ∣$ , the two extremes of a triangle side ✓.

Example 6b — Cell H: the limit $a_{2} \to 0$

Worked example In the setup above, let

$a_{2} \to 0$ (the second source is switched off). What happens to the fringes? Forecast: No second wave — can there still be bright/dark alternation? Guess.

Step 1 — Take the limit. With $a_{2} = 0$ : $A = a_{1}^{2} + 0 + 0 = a_{1}$ for every $ϕ$ . Why this step? The $cos ϕ$ term is multiplied by $a_{2}$ ; kill $a_{2}$ and the phase dependence vanishes entirely.

Step 2 — Interpret. $A$ no longer depends on $ϕ$ , so $I = a_{1}^{2}$ everywhere — uniform, no maxima, no minima. Why this step? Interference is the variation of $A$ with $ϕ$ . Remove one wave and there is nothing to interfere with; the pattern flattens (see Energy in Waves — with one source the energy is just spread evenly).

Verify: $A_{m a x} = A_{m i n} = a_{1}$ so the fringe contrast $(A_{m a x} - A_{m i n}) = 0$ ✓ — no visible fringes, exactly the degenerate limit.

Example 7 — Cell I: real-world word problem (thin film / speakers)

Worked example A listener stands equidistant... then walks until the sound first goes silent. The two speakers (

$λ = 0.68$ m) now have path difference $Δ x$ . What is the smallest positive $Δ x$ giving the first silence, and what is $ϕ$ there? Forecast: "First silence" — which order $n$ of the destructive condition is that? $n = 0$ or $n = 1$ ? Guess.

Step 1 — Pick the right condition. Silence = destructive: $Δ x = (n + \frac{1}{2}) λ$ . Why this step? Silence is cancellation, which is the destructive (half-integer) family, not the integer one.

Step 2 — Smallest positive $Δ x$ . The smallest non-negative choice is $n = 0$ : $Δ x = (0 + \frac{1}{2}) (0.68) = 0.34 m .$ Why this step? Walking outward from the centre, the first dead spot is the $n = 0$ destructive one, half a wavelength of extra path.

Step 3 — Phase there. $ϕ = \frac{2 π}{0.68} (0.34) = π$ . Why this step? Half a wavelength of path is exactly half a cycle = $π$ radians = perfect opposition.

Verify: $cos^{2} (π /2) = 0$ → $I = 0$ ✓ genuine silence. Units: $0.34$ m is a physical distance, and it is exactly $λ /2$ as a dead spot should be ✓.

Example 8 — Cell J: exam twist (intensity ratio → amplitude ratio first)

Worked example Two coherent sources have

intensities in ratio $I_{1} : I_{2} = 9 : 1$ . Find the ratio $I_{m a x} : I_{m i n}$ of the interference pattern. Forecast: Tempting to write $9 + 1 = 10$ and $9 - 1 = 8$ . But intensities don't add like that. Guess whether the true answer is bigger or smaller than $10 : 8$ .

Step 1 — Convert intensity ratio to amplitude ratio. Since $I \propto a^{2}$ , $a \propto I$ : $\frac{a _{1}}{a _{2}} = \frac{I _{1}}{I _{2}} = 9 = 3 \Rightarrow a_{1} : a_{2} = 3 : 1.$ Why this step? Interference adds amplitudes, not intensities. We must drop back into amplitude-land before combining — the recurring lesson of this whole topic.

Step 2 — Extreme amplitudes. $A_{m a x} = a_{1} + a_{2} = 3 + 1 = 4, A_{m i n} = ∣ a_{1} - a_{2} ∣ = 3 - 1 = 2.$ Why this step? Aligned phasors add ( $ϕ = 0$ ); opposed phasors subtract ( $ϕ = π$ ) — same as Ex 6.

Step 3 — Back to intensities (square them). $\frac{I _{m a x}}{I _{m i n}} = \frac{A _{m a x}^{2}}{A _{m i n}^{2}} = \frac{4 ^{2}}{2 ^{2}} = \frac{16}{4} = 4.$ Why this step? Now we square to leave amplitude-land, because brightness is what the question asks.

Verify: $I_{m a x} : I_{m i n} = 4 : 1$ , not $10 : 8$ ✓ — the forecast trap avoided. General formula check: $\frac{( I _{1} + I _{2} ) ^{2}}{( I _{1} - I _{2} ) ^{2}} = \frac{( 3 + 1 ) ^{2}}{( 3 - 1 ) ^{2}} = 4$ ✓.

Recall

Recall Quick fire across the matrix

Q: $Δ x = 0$ — bright or dark? ::: Bright — central maximum ( $n = 0$ ), $I = 4 I_{0}$ . Q: Can unequal-amplitude waves reach $I = 0$ ? ::: No — minimum is $(a_{1} - a_{2})^{2} \neq = 0$ . Q: Given $I_{1} : I_{2} = 9 : 1$ , what is $I_{m a x} : I_{m i n}$ ? ::: $4 : 1$ (convert to $a$ -ratio $3 : 1$ first). Q: Smallest destructive path difference? ::: $λ /2$ (the $n = 0$ dead spot). Q: Does a negative order $n$ change brightness? ::: No — $cos^{2}$ is even, sign washes out.

Connections

Parent: conditions — the two laws drilled here.
Principle of Superposition — why amplitudes (not intensities) add.
Young's Double Slit Experiment — where $\pm n$ orders (Ex 4) live on a screen.
Coherence and Path Difference — why steady $Δ x$ makes Ex 2 stable.
Energy in Waves — the flat pattern of the $a_{2} \to 0$ limit (Ex 6b).
Standing Waves and Beats — sibling superposition patterns.