Grade by Grade
E. The Wave Nature of Light · The Wave Nature of Light

Interference in Two Dimensions: The Two-Point-Source Pattern

E2.1 E2.2 E2.4 E3.1 A1.12 A1.13About 75 minutes
Learning goals
  • I can describe the interference pattern of two in-phase point sources in a ripple tank, with nodal and antinodal lines.
  • I can use |PₙS₁ − PₙS₂| = (n − ½)λ to find the wavelength from a point on a nodal line.
  • I can use sin θₙ = (n − ½)λ/d and xₙ/L = (n − ½)λ/d to analyse the pattern.
  • I can predict how changing the frequency, the source separation or the phase changes the pattern.

Words to know

interference
The combining of two or more waves that overlap in the same place, following the principle of superposition.
nodal line
A line of points where the waves from two sources always interfere destructively, so the medium stays (nearly) still.
antinodal line
A line of points where the waves from two sources interfere constructively, so the medium vibrates the most.
in phase
Two sources are in phase when they make crests at the same time.
path difference
The difference in distance from a point to the two sources, |PS₁ − PS₂|.
right bisector
The line through the midpoint of the two sources, perpendicular to the line joining them. In an in-phase pattern it is the central antinodal line.

A pattern that stands still

Tap two fingers into a pond at the same steady rhythm and something surprising happens. The circles overlap, yet the surface does not become a mess. Instead, a fixed pattern appears: calm lines fanning outward, with bands of choppy water in between. That pattern is the fingerprint of interference, and in the next lessons it will be the evidence that light is a wave.

Remember from Grade 11: in Waves and Sound you learned the principle of superposition and that, for two in-phase sources, a point is loud (constructive) when its path difference is a whole number of wavelengths and quiet (destructive) when it is a whole number plus a half. Now we make that quantitative in two dimensions.

The two-point-source pattern

Two dippers S₁ and S₂ vibrate in phase with the same frequency in a ripple tank. Each sends out circular waves.

  • Where a crest from S₁ meets a crest from S₂ (or trough meets trough), the water moves with a large amplitude: constructive interference. These points line up along antinodal lines.
  • Where a crest meets a trough, the waves cancel and the water stays almost still: destructive interference. These points line up along nodal lines.

The nodal lines are curves called hyperbolas, but far from the sources they are almost straight lines from the midpoint. The right bisector (the central line) is always antinodal for in-phase sources because every point on it is the same distance from both sources.

S₁S₂thin solid arcs: creststhin dotted arcs: troughsheavy dashed: nodal lines
Two in-phase sources seen from above. Where solid circles cross solid circles (crest on crest) the water vibrates strongly. The heavy dashed curves are nodal lines, where crests meet troughs; here the sources are 3λ apart, so there are three nodal lines on each side of the central antinodal line (solid).

Numbering the nodal lines

Count nodal lines outward from the centre: n = 1, 2, 3, … on each side. For any point Pₙ on the n-th nodal line:

|PₙS₁ − PₙS₂| = (n − ½)λ

So the first nodal line has a path difference of ½λ, the second 1½λ, the third 2½λ, and so on. (Antinodal lines have path differences 0, 1λ, 2λ, ….)

This equation is the most reliable one because it uses exact distances. Measure both distances from the point to the sources, subtract, and divide by (n − ½) to get λ.

Using angles and distances far from the sources

Far from the sources (distance L much greater than d), the two paths to P are almost parallel. Drop a perpendicular from S₁ to the path from S₂, as in the diagram: the extra length is the path difference = d sin θ, where θ is the angle between the right bisector and the line to the point. Setting this equal to (n − ½)λ gives:

sin θₙ = (n − ½)λ / d

Also, sin θ ≈ x/L when L is large, where x is the perpendicular distance from the right bisector to the point and L is the distance from the midpoint of the sources:

xₙ / L = (n − ½)λ / d

These say the nodal lines spread out (larger θ) when λ is larger or d is smaller.

S₁S₂Pextra path = d sin θdθxL (distance to the point, measured from the midpoint)
Geometry of the pattern (not to scale). Far away, the paths from S₁ and S₂ to P are nearly parallel, so the extra distance from S₂ is d sin θ. The point P is a distance x from the central line (dashed) and about L from the sources.
Worked example

Example 1: Wavelength from distances

In a ripple tank, point P is on the second nodal line. It is 12.0 cm from S₁ and 9.0 cm from S₂. The sources vibrate at 6.0 Hz. Find the wavelength and the speed of the waves.

  1. Path difference: |12.0 − 9.0| = 3.0 cm.
  2. Second nodal line, so n = 2: 3.0 = (2 − ½)λ = 1.5λ.
  3. λ = 3.0 ÷ 1.5 = 2.0 cm.
  4. v = fλ = 6.0 × 2.0 = 12 cm/s.

λ = 2.0 cm and v = 12 cm/s.

Worked example

Example 2: Wavelength from an angle

Two sources 5.0 cm apart vibrate in phase at 8.0 Hz. The first nodal line makes an angle of 14° with the right bisector. Find λ and v.

  1. sin θ₁ = (1 − ½)λ/d, so λ = d sin θ₁ / 0.5.
  2. λ = 5.0 × sin 14° ÷ 0.5 = 5.0 × 0.2419 ÷ 0.5 = 2.4 cm.
  3. v = fλ = 8.0 × 2.42 = 19 cm/s (keep 2.42 in the calculator).

λ ≈ 2.4 cm and v ≈ 19 cm/s.

Worked example

Example 3: How many nodal lines?

Two in-phase sources are 6.0 cm apart and make waves with λ = 2.5 cm. How many nodal lines are there in total?

  1. A nodal line exists only if sin θₙ ≤ 1 (strictly, below 1): (n − ½)λ < d.
  2. Divide by λ: n − ½ < 6.0 ÷ 2.5 = 2.4, so n < 2.9.
  3. n = 1 and n = 2 work; n = 3 needs 2.5λ = 6.25 cm, which is more than d.
  4. That gives 2 nodal lines on each side of the central line.

4 nodal lines in total (2 on each side).

Changing the pattern

  • Increase the frequency (shorter λ): the nodal lines move closer together, and there are more of them.
  • Move the sources farther apart (larger d): the same thing happens: more nodal lines, closer together.
  • Put the sources out of phase (one makes a crest when the other makes a trough): the whole pattern shifts so that the central line becomes nodal. The pattern stays symmetric but nodal and antinodal lines swap places.

So measuring the pattern lets you work out the wavelength, which is exactly what Young did with light around 1801 to 1803 (next lessons).

Just for fun

Two dippers in a ripple tank were asked why they never agree. "We do agree," they said, "just not everywhere. Along the nodal lines we cancel each other out completely."

Hands-on activity

Two-source interference at home

You need: the baking-dish ripple tank from the last lesson (clear dish, about 1 cm of water, lamp above, white paper below) or a bathtub with a few centimetres of water, and two pencils held together with an elastic band so their tips are about 3 cm apart.

Grown-up note: keep the lamp and its cord away from the water.

  1. Hold the pencils upright and tap both tips into the water together, steadily, about 4 to 6 times per second.
  2. Look for the calm nodal lines fanning out between the choppy regions. Count how many you can see on each side of the centre.
  3. Tap faster. Do the nodal lines move closer together or farther apart?
  4. Spread the pencil tips to about 5 cm and repeat.
  5. Optional: use a free ripple-tank simulation (for example, PhET Wave Interference, two drips). Measure the distances from a point on the 2nd nodal line to each source and calculate λ with |PₙS₁ − PₙS₂| = (n − ½)λ. Compare with the wavelength you measure directly.

What to look for: faster tapping and wider spacing both give more, closer nodal lines. In the simulation your calculated λ should match the measured λ within a few percent.

Your turn

1

What is a nodal line?

2

What is the path difference to any point on the first nodal line?

3

Two in-phase sources are 3.0 cm apart and make waves with λ = 1.2 cm. Find the angle between the right bisector and the first nodal line.

4

For the same sources, find the angle to the second nodal line.

5

A point on the second nodal line is 10.0 cm from S₁ and 7.5 cm from S₂. Find λ.

6

What happens to the nodal lines when the frequency increases?

7

Why is the right bisector an antinodal line for in-phase sources?

More practice: warm-up

1

Crest meets crest. Which kind of interference?

2

What is the path difference to a point on the third nodal line?

3

Write the equation for the angle to the n-th nodal line.

4

Does moving the sources closer together give more or fewer nodal lines?

5

What shape are nodal lines close to the sources?

More practice: on your own

1

Sources 8.0 cm apart make waves with λ = 2.0 cm. How many nodal lines are there in total?

2

For the sources in the last question, find the angle to the second nodal line.

3

Sources 4.0 cm apart make waves with λ = 1.6 cm. A ruler is placed 25 cm from the midpoint of the sources. How far from the central line does it cross the first nodal line?

4

Two sources 6.0 cm apart vibrate in phase at 10 Hz. The second nodal line is at 30° from the right bisector. Find λ and v.

5

The two dippers are now made to vibrate out of phase. Describe what happens at the right bisector and explain.

Challenge

1

Show that if two in-phase sources are less than ½λ apart, there are no nodal lines at all.

2

Point P is 15.0 cm from S₁ and 12.0 cm from S₂. The waves travel at 18 cm/s. The frequency is slowly increased from 1.0 Hz. Find the two lowest frequencies at which P is on a nodal line, and say which nodal line it is on each time.

3

In a ripple tank, a student measures x₃ = 9.0 cm at L = 30 cm for sources 4.0 cm apart. She uses xₙ/L = (n − ½)λ/d to get λ = 0.48 cm. Her partner measures the distances from that same point to the sources directly and gets a different answer. Explain which method is more trustworthy here and why.

Check your understanding

Quiz

1.

A point on the third nodal line has a path difference of…

2.

Two in-phase sources make waves with λ = 2.0 cm. P is 11.0 cm from S₁ and 8.0 cm from S₂. P lies on…

3.

What happens to a two-source pattern when the frequency of both sources increases?

4.

Sources 5.0 cm apart; first nodal line at 30°. What is λ?

5.

If two sources vibrate exactly out of phase, the right bisector is…

6.

Which condition is needed to use xₙ/L = (n − ½)λ/d accurately?

7.

Sources are 6.0 cm apart and λ = 2.5 cm. How many nodal lines are there on each side of the central line?

For parents and teachers

Big idea: expectations E3.1 (describe and explain interference of water waves in two dimensions), E2.2 (ripple-tank or simulation inquiry) and E2.4 (solving two-point-source problems). Every formula in this lesson comes back in Young's double-slit experiment, with light instead of water.

How to help: if the baking-dish version is hard to see, the free PhET Wave Interference simulation shows the pattern perfectly and lets your teen measure distances on screen.

Common difficulties:

  • Using n instead of (n − ½) for nodal lines. Antinodal lines use nλ; nodal lines use (n − ½)λ.
  • Counting nodal lines from 0. The first nodal line is n = 1.
  • Using x/L close to the sources, where it is inaccurate. Encourage the exact-distance method when L is not much bigger than d.
  • Calculator in radian mode.