Why can’t two torches produce an interference pattern?

Their light is not coherent. The phase difference between them changes randomly many times a second, so any pattern moves too quickly to observe.

Why does a grating give sharper maxima than a double slit?

With many slits, even a tiny departure from the exact angle causes the contributions to cancel, so light is only bright at precisely the maxima.

Do the waves destroy each other in destructive interference?

No. They pass through unchanged; the energy is redistributed to the regions of constructive interference.

Why did Young’s experiment support the wave model?

Dark fringes appear where light from both slits overlaps. Only waves, adding in antiphase, can combine to give darkness.

Back to course

65–66, 83–85WavesCore

Superposition and Interference

When two waves meet, their displacements simply add. That is the $\textbf{principle of superposition}$ , and when the waves are coherent it produces a stable pattern of loud and quiet, or bright and dark, called $\textbf{interference}$ .

40 min Topic 5 · Waves and the Particle Nature of Light

What you'll be able to do

State the principle of superposition
Explain constructive and destructive interference using path difference
State the conditions for observable interference, including coherence
Use $\lambda = ax/D$ for double slits and $d\sin\theta = n\lambda$ for gratings

Superposition

When two or more waves meet at a point, the resultant displacement is the $\textbf{vector sum}$ of the individual displacements.

Two crests meeting give a larger crest — $\textbf{constructive}$ superposition. A crest meeting an equal trough gives zero displacement — $\textbf{destructive}$ superposition.

The waves pass through each other unchanged; superposition describes only what happens at the point where they overlap.

Path difference and phase

Two waves from separate sources reaching a point have travelled different distances. That $\textbf{path difference}$ determines their phase difference on arrival.

If the path difference is a whole number of wavelengths, $n\lambda$ , the waves arrive in phase and interfere constructively. If it is an odd number of half-wavelengths, $(n + \tfrac12)\lambda$ , they arrive in antiphase and interfere destructively.

This assumes the sources themselves are in phase. If they start in antiphase, the conditions swap.

\text{path difference} = n\lambda \Rightarrow \text{constructive}

In phase on arrival.

\text{path difference} = (n+\tfrac12)\lambda \Rightarrow \text{destructive}

Antiphase on arrival.

Example 1Two loudspeakers emit in phase at wavelength 0.50 m. A point is 3.25 m from one and 4.00 m from the other. Is it loud or quiet?

1Path difference:

4.00 - 3.25 = 0.75

2In wavelengths:

0.75/0.50 = 1.5

— an odd number of halves.

AnswerQuiet — destructive interference.

Tip — Always convert path difference into a number of wavelengths. Whole numbers mean constructive, halves mean destructive.

Coherence and Young’s double slit

A stable interference pattern needs $\textbf{coherent}$ sources: the same frequency and a constant phase difference. Two separate bulbs are not coherent, so their patterns shift too quickly to see. Splitting light from one source through two slits guarantees coherence.

In $\textbf{Young’s double-slit}$ experiment, light through two narrow slits a distance $a$ apart produces bright and dark fringes on a screen a distance $D$ away. The fringe spacing $x$ satisfies $\lambda = ax/D$ , valid when $D \gg a$ .

The result was historically decisive: particles could not produce dark fringes where two beams overlap, so the pattern established light as a wave.

\lambda = \dfrac{ax}{D}

Double-slit fringe spacing,

D \gg a

Example 1Slits 0.40 mm apart, screen 1.6 m away, fringes 2.4 mm apart. Find the wavelength.

\lambda = \dfrac{(0.40\times10^{-3})(2.4\times10^{-3})}{1.6}

Answer

6.0\times10^{-7}\ \mathrm{m}

— orange light.

Tip — Measure across many fringes and divide to find $x$ . A single fringe spacing carries a large percentage uncertainty.

The diffraction grating

A $\textbf{diffraction grating}$ has many closely spaced slits. Light from adjacent slits reaching a maximum at angle $\theta$ has path difference $d\sin\theta$ , where $d$ is the slit spacing; constructive interference requires that to be a whole number of wavelengths.

With so many slits contributing, the maxima are extremely sharp and bright, and everything between is effectively dark — which makes gratings far better than double slits for measuring wavelength precisely.

Gratings are specified in lines per millimetre: $d = 1/N$ . The highest order visible is the largest $n$ with $\sin\theta \leq 1$ .

d\sin\theta = n\lambda

Grating equation.

n_{\max} \leq \dfrac{d}{\lambda}

Highest order, since

\sin\theta \leq 1

Example 1Light of wavelength 550 nm falls on a grating with 500 lines per mm. Find the angle of the second-order maximum and the highest order.

d = \dfrac{1\times10^{-3}}{500} = 2.0\times10^{-6}

\sin\theta = \dfrac{2 \times 550\times10^{-9}}{2.0\times10^{-6}} = 0.55

, so

\theta = 33.4^{\circ}

3Highest order:

d/\lambda = 3.64

, so

n = 3

AnswerSecond order at

33.4^{\circ}

; maximum order 3.

Equation recap

n\lambda

Path difference for constructive.

(n+\tfrac12)\lambda

Path difference for destructive.

\lambda = ax/D

Double slit.

d\sin\theta = n\lambda

Diffraction grating.

Common mistakes to avoid

Saying coherent means same amplitude.

Coherent means same frequency and constant phase difference.

Using lines per mm directly as

d

d

is the reciprocal, in metres.

Rounding the maximum order up.

Take the largest whole number with

\sin\theta \leq 1

Applying

\lambda = ax/D

when the screen is close.

It needs

D \gg a

so the small-angle approximation holds.

Key takeaways

Superposition: displacements add where waves meet.
Path difference $n\lambda$ gives constructive, $(n+\tfrac12)\lambda$ destructive interference.
Observable interference needs coherent sources.
Double slit: $\lambda = ax/D$ ; grating: $d\sin\theta = n\lambda$ .
Gratings give sharp bright maxima; the highest order is rounded down.

Test yourself

Answer a few questions, then review anything you missed.

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