Interference — constructive, destructive conditions: Question bank

True or false — justify

Two waves always reinforce when their crests meet

True — crest meeting crest means displacements add with the same sign, so the resultant is larger. This is exactly the constructive case, path difference a whole number of wavelengths.

Interference requires two waves of the same amplitude

False — unequal amplitudes still interfere (Example 3 in the parent), but they can never cancel to zero; the minimum resultant amplitude is

∣ a_{1} - a_{2} ∣

(a nonzero displacement), not

0

Two independent light bulbs can produce a stable interference pattern

False — independent sources drift in phase randomly, so the phase difference isn't constant; the pattern washes out. Stable fringes need coherent sources. See Coherence and Path Difference.

At a destructive point the energy of the two waves is destroyed

False — energy is only redistributed; the excess above the incoherent sum

2 I_{0}

at bright points (an extra

2 I_{0}

, since

4 I_{0} - 2 I_{0} = 2 I_{0}

) exactly makes up for the

2 I_{0}

deficit at dark points. The spatial average stays

2 I_{0}

. See Energy in Waves.

If you add the two intensities you always get the true intensity

False — intensities add only for incoherent sources. For coherent waves you must add amplitudes first, then square, giving the cross term

2 a_{1} a_{2} cos ϕ

A phase difference of

2 π

gives the same result as a phase difference of

0

True — phase is cyclic with period

2 π

, so

ϕ = 2 π

and

ϕ = 0

both mean "perfectly in step" and both give maximum intensity.

Constructive interference violates conservation of energy because

4 I_{0} > I_{0} + I_{0}

False — that's the intensity at one bright point only; averaged over the whole pattern the intensity is

2 I_{0}

, exactly the sum of the two sources.

Zero path difference gives a dark fringe

False — zero path difference means perfectly in step, the brightest central maximum (

n = 0

Δ x = 0

). This is the classic "

λ

is the condition" trap.

Interference and superposition are the same thing

Mostly false as stated — superposition is the underlying addition rule at every instant; interference is that rule watched at a fixed pattern of maxima and minima. Interference is a special, observable consequence of Principle of Superposition.

Spot the error

"Destructive interference needs

Δ x = λ

Wrong —

Δ x = λ

is constructive (whole wavelength = back in step). Destructive needs a half-integer extra:

Δ x = (n + \frac{1}{2}) λ

with

n = 0, \pm 1, \pm 2, \dots

"Since

I = 4 I_{0} cos^{2} (ϕ /2)

, the maximum intensity is

2 I_{0}

Wrong —

cos^{2}

maxes at

1

, giving

I_{m a x} = 4 I_{0}

. The

2 I_{0}

is the average, reached at

ϕ = π /2

(that is

9 0^{\circ}

), not the maximum.

"Two waves with amplitudes

3 a

and

4 a

in antiphase cancel to zero."

Wrong — full cancellation needs equal amplitudes. Here the minimum resultant amplitude is

∣4 a - 3 a ∣ = a

(so the minimum intensity is

\propto a^{2}

, not zero).

"For interference we add the frequencies of the two waves."

Wrong — we add displacements, and the two waves must have the same frequency for a stable pattern. Adding different frequencies gives Beats, a time pattern, not a fixed spatial one.

ϕ = (2 π / λ) Δ x

, so a large wavelength gives a large phase difference."

Wrong direction —

λ

is in the denominator, so a larger

λ

gives a smaller phase for the same path. One wavelength of path always equals

2 π

of phase, whatever

λ

is.

"At

ϕ = 9 0^{\circ}

the waves fully cancel."

Wrong —

cos^{2} (4 5^{\circ}) = \frac{1}{2}

, so

I = 2 I_{0}

, the neutral average. Full cancellation needs

ϕ = 18 0^{\circ}

(that is

π

radians).

Why questions

Why do we add amplitudes and not intensities for coherent waves?

Because the physical quantity that superposes is the displacement (amplitude), instant by instant; intensity (

\propto A^{2}

) is a derived, squared quantity and squaring doesn't distribute over a sum.

Why does one wavelength of path difference correspond to exactly

2 π

of phase?

One full wavelength is one complete cycle of the wave, and one complete cycle is

2 π

radians of phase — the master link

ϕ = (2 π / λ) Δ x

just scales path into that cycle count.

Why is intensity proportional to amplitude squared and not amplitude itself?

Energy carried by a wave depends on the square of displacement (like kinetic energy

\propto v^{2}

in a swinging particle), so brightness/loudness scales as

A^{2}

. See Energy in Waves.

Why does the cross term

2 a_{1} a_{2} cos ϕ

carry all the "interference"?

The

a_{1}^{2}

and

a_{2}^{2}

pieces are just the two waves' own intensities; only the

cos ϕ

term depends on their relative phase, swinging positive (reinforce) or negative (cancel). Kill coherence and

cos ϕ

averages to zero, leaving plain

I_{0} + I_{0}

Why must sources be coherent for a visible pattern?

The eye/detector averages over time; if

ϕ

jitters randomly, the

cos^{2} (ϕ /2)

term averages to

\frac{1}{2}

everywhere, giving a uniform

2 I_{0}

with no bright and dark fringes.

Why does the average intensity across the whole pattern equal

2 I_{0}

Because energy is conserved: the excess at bright points (

4 I_{0}

) exactly balances the deficit at dark points (

0

), and the mean of

cos^{2}

over a full cycle is

\frac{1}{2}

, giving

4 I_{0} \times \frac{1}{2} = 2 I_{0}

Why can a standing wave be seen as interference?

Two identical waves travelling in opposite directions superpose; their fixed pattern of nodes (always destructive) and antinodes (always constructive) is interference frozen in space. See Standing Waves.

Edge cases

What is the intensity when the two waves have a phase difference of exactly

π /2

(that is

9 0^{\circ}

I = 4 I_{0} cos^{2} (π /4) = 2 I_{0}

— the neutral "average" point, neither a maximum nor a minimum.

What happens to interference if one source is switched off?

There's no second wave to superpose, so no pattern forms — you get the plain, uniform intensity

I_{0}

of the single source everywhere.

What is the resultant when two equal waves are exactly

18 0^{\circ}

(that is

π

radians) out of phase?

They cancel completely:

A = 2 a cos (9 0^{\circ}) = 0

, so

I = 0

. This is perfect destructive interference, only possible with equal amplitudes.

At the central point of Young's Double Slit Experiment, what is the path difference and what fringe appears?

The path difference is zero (equal distances to both slits), so it's the central bright maximum,

n = 0

If two coherent waves have a very small amplitude difference, can you ever get total darkness?

No — the minimum resultant amplitude is

∣ a_{1} - a_{2} ∣

, which is nonzero however small the difference. Only exactly equal amplitudes give true zero.

What does the interference pattern look like for two waves of slightly different frequency?

No fixed spatial pattern; instead the resultant amplitude rises and falls in time — these are Beats, the temporal cousin of interference.

Recall check

Recall Which quantity superposes — amplitude or intensity?

Amplitude (displacement). Intensity is derived afterward by squaring the resultant amplitude.

Recall Fastest way to classify a path difference

Compare $Δ x$ to $λ$ : if $Δ x = nλ$ (a whole number of wavelengths) → constructive (Wow); if $Δ x = (n + \frac{1}{2}) λ$ (a half-integer, $n = 0, \pm 1, \pm 2, \dots$ ) → destructive (Hush).

Connections

Interference — constructive, destructive conditions — the parent this bank drills.
Principle of Superposition — the add-displacements rule every trap tests.
Coherence and Path Difference — why independent sources fail.
Young's Double Slit Experiment — where the central-fringe edge case lives.
Standing Waves — interference of counter-propagating waves.
Beats — interference of different frequencies in time.
Energy in Waves — why $I \propto A^{2}$ and energy is conserved.