Lesson 1 of 5The Wave Nature of Light

Unit 4 · Lesson 1 of 5

Interference and Diffraction of Waves

Two sources in a ripple tank, the nodal lines they carve out, the path-difference rule that locates every one of them, and why waves bend through openings only when the opening is not much wider than the wavelength.

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Key things to remember

1Superposition: where waves overlap, the resulting displacement is the sum of the individual displacements.
2For two in-phase sources, a point is on an antinodal line if $∣ P S_{1} - P S_{2} ∣ = mλ$ and on a nodal line if $∣ P S_{1} - P S_{2} ∣ = (n - \frac{1}{2}) λ$ .
3For distant points: $sin θ_{n} = (n - \frac{1}{2}) \frac{λ}{d} \approx \frac{x _{n}}{L}$ .
4Shorter wavelength or larger source separation gives more nodal lines, closer together.
5Diffraction is significant when $λ / w$ is roughly 1 or larger, and negligible when the opening is many wavelengths wide.
6The wave equation still links the quantities: $v = f λ$ .

Before arguing about whether light is a wave, you need a clear picture of what waves do that particles never do. Throw two handfuls of pebbles through a pair of doorways and you get two piles. Send water waves through the same doorways and you get something stranger: a fan of calm stripes where the water barely moves, alternating with stripes of violent motion. That pattern — interference — together with the spreading of waves through gaps — diffraction — is the fingerprint of a wave. This topic studies both with water, where you can see them, so the rest of the unit can look for the same fingerprints in light.

Superposition in two dimensions

Fig 1In phase the displacements add; half a wavelength out of step they cancel.

In Grade 11 you added pulses on a string: where two waves overlap, the displacement is the sum of the two separate displacements. A ripple tank does the same thing across a whole surface. Two dippers tap the water in phase — at the same frequency, rising and falling together — so each sends out circular ripples with the same wavelength $λ$ . Wherever a crest from one source meets a crest from the other, the water rises twice as high (constructive interference). Wherever a crest meets a trough, the water stays nearly flat (destructive interference). Because the sources keep the same timing, these spots do not wander about. They line up into a stable pattern of curved lines called nodal lines (always calm) separated by antinodal lines (always strongly oscillating).

Path difference decides everything

Pick any point $P$ in the tank. It is a distance $P S_{1}$ from one source and $P S_{2}$ from the other. The waves leave in phase, so what matters is how much farther one wave travelled than the other: the path difference $∣ P S_{1} - P S_{2} ∣$ . If that difference is a whole number of wavelengths ( $0, λ, 2 λ, \dots$ ), a crest arrives with a crest and $P$ is on an antinodal line. If it is an odd number of half-wavelengths ( $\frac{1}{2} λ, \frac{3}{2} λ, \frac{5}{2} λ, \dots$ ), a crest arrives with a trough and $P$ is on a nodal line.

antinodal: ∣ P S_{1} - P S_{2} ∣ = mλ nodal: ∣ P S_{1} - P S_{2} ∣ = (n - \frac{1}{2}) λ

m = 0, 1, 2, ... counts antinodal lines from the centre; n = 1, 2, 3, ... counts nodal lines outward on each side.

The perpendicular bisector between the sources is equally far from both, so it is always the central antinodal line ( $m = 0$ ). The first nodal line on either side ( $n = 1$ ) has a path difference of half a wavelength, the second ( $n = 2$ ) of one and a half, and so on. For a point far away compared with the source separation $d$ , the path difference is very nearly $d sin θ$ , where $θ$ is the angle measured from the bisector. That gives a way to locate nodal lines by angle, or by their distance $x_{n}$ from the bisector at a distance $L$ from the sources:

sin θ_{n} = (n - \frac{1}{2}) \frac{λ}{d} \approx \frac{x _{n}}{L}

Valid for distant points, where the two paths to P are nearly parallel.

Change	Effect on the pattern	Reason
Increase the frequency of both sources	More nodal lines, packed closer together	Higher $f$ means shorter $λ$ , so $sin θ_{n}$ is smaller for each line
Move the sources farther apart	More nodal lines, packed closer together	Larger $d$ makes $λ / d$ smaller
Move the sources closer together	Fewer nodal lines, spread wider	Once $d < \frac{1}{2} λ$ there are no nodal lines at all
Make one source half a cycle out of phase	Nodal and antinodal lines swap places	The bisector now receives a crest and a trough together

How many nodal lines are there?

Since

sin θ

can never exceed 1, a nodal line exists only if

(n - \frac{1}{2}) λ < d

. Count the whole numbers

n

that satisfy this: that is the number of nodal lines on each side of the bisector. The total is twice that.

Diffraction: waves bending through openings

Send straight wavefronts at a barrier with a gap and the waves that emerge are not a neat beam the width of the gap: they spread into the shadow region behind the barrier. This bending is diffraction, and it also happens around the edge of a single obstacle. Huygens' principle explains why: every point on a wavefront acts as a tiny source of new circular wavelets, and the next wavefront is where those wavelets overlap. Far from any barrier the sideways wavelets cancel each other out, so the wave carries straight on. At the edges of a gap there are no neighbours on one side to cancel them, so the wave curves around.

AWavelength similar to or larger than the gap

The ratio $λ / w$ is large (about 1 or more)
Waves fan out into nearly semicircular fronts
Strong bending into the shadow
Example: $1 m$ sound waves through a $0.9 m$ doorway

BWavelength much smaller than the gap

The ratio $λ / w$ is tiny
Waves pass through as a nearly straight beam
Only the edges of the beam curve slightly
Example: $550 nm$ light through the same doorway

The single most useful idea in the unit

Diffraction depends on the ratio of wavelength to opening width, not on either number alone. Halve the gap or double the wavelength and the spreading increases in the same way. This is why you can hear a conversation around a corner but cannot see the people: sound wavelengths are about a metre, the same size as a doorway, while visible light wavelengths are about two million times smaller. To see light diffract you need an opening a fraction of a millimetre wide, which is exactly what the next topic uses.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Two point sources in a ripple tank vibrate in phase at

8.0 Hz

. A point

P

on the second nodal line is

12.0 cm

from one source and

15.0 cm

from the other. Find the wavelength and the speed of the waves.

Example 2

Two in-phase sources are

6.0 cm

apart and produce waves of wavelength

1.5 cm

. (a) At what angle from the bisector is the first nodal line? (b) How many nodal lines are there in total?

Example 3

340 Hz

tone passes through a doorway

0.90 m

wide. Take the speed of sound as

343 m/s

. Compare how much the sound diffracts with how much green light of wavelength

550 nm

diffracts through the same doorway.

Where marks get lost

Using $mλ$ for a nodal line. Nodal lines need an odd number of half-wavelengths, $(n - \frac{1}{2}) λ$ ; whole wavelengths give antinodal lines.
Counting the central line as a nodal line. With in-phase sources the bisector is always antinodal, and the first nodal line is $n = 1$ on either side of it.
Forgetting to double when asked for the total number of nodal lines. The inequality counts the lines on one side only.
Thinking a bigger opening causes more diffraction. It is the reverse: the narrower the opening compared with the wavelength, the more the wave spreads.
Using $sin θ \approx x / L$ for points close to the sources. The approximation needs $L$ to be much larger than $d$ .

Flashcards 8

Nodal line

Antinodal line

Path difference for the nth nodal line

Angle of the nth nodal line

+ 4 more

Practice

Two in-phase sources produce waves with a wavelength of

1.2 cm

. What is the path difference, in centimetres, for a point on the third nodal line?

In a ripple tank, the frequency of two in-phase sources is increased while their separation stays the same. What happens to the interference pattern?

Two in-phase sources are

4.0 cm

apart and produce waves of wavelength

1.0 cm

. At what angle from the perpendicular bisector is the second nodal line? Give your answer in degrees to one decimal place.

What is the name for a line of points in an interference pattern where destructive interference occurs all the time?

Two in-phase sources are

4.5 cm

apart and produce waves of wavelength

2.0 cm

. How many nodal lines are there in the whole pattern?

Straight water waves of wavelength

2.0 cm

pass through a gap. Which gap width produces the most diffraction?

Two in-phase sources are

3.0 cm

apart. On a line

30 cm

from the sources, a point on the first nodal line is

2.5 cm

from the bisector. What is the wavelength, in centimetres?

Why can you hear someone talking in the next room through an open door, but not see them?

Match each change in a two-source ripple tank to its effect on the pattern.

Increase the source separation

Lower the frequency

Put one source half a cycle out of phase

Make the separation less than half a wavelength

Order these steps for finding the wavelength from a photograph of a two-source ripple-tank pattern.

1Solve for the wavelength and check it against v = f times wavelength
2Set the path difference equal to (n - 1/2) times the wavelength
3Subtract to find the path difference
4Identify which nodal line the chosen point lies on, counting outward from the bisector
5Measure the distances from the point to each source

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