PhysicsGrade 10–13

Interference of two circular waves

Two circular waves overlap: in some places they reinforce each other, in others they cancel out.

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What is it about?

Two sources E1E_1 and E2E_2 at a distance dd dip into the water of a ripple tank in step and send out circular waves of wavelength λ=cf\lambda = \frac{c}{f}. Where they meet, the displacements add up (superposition). Whether they reinforce or cancel at a point PP depends on the path difference Δs=s1−s2\Delta s = s_1 - s_2:

  • maximum (constructive): Δs=k⋅λ\Delta s = k \cdot \lambda, amplitude 2y^2\hat y,
  • minimum (destructive): Δs=(k+12)⋅λ\Delta s = \left(k + \tfrac{1}{2}\right) \cdot \lambda, the water stays at rest.

In general A=2y^⋅∣cos⁡δ2∣A = 2\hat y \cdot \left|\cos\frac{\delta}{2}\right| with δ=2π⋅Δsλ+Δφ\delta = 2\pi \cdot \frac{\Delta s}{\lambda} + \Delta\varphi (Δφ\Delta\varphi: phase difference of the sources). All points with the same Δs\Delta s lie on a hyperbola with foci E1E_1 and E2E_2. Since ∣Δs∣≤d|\Delta s| \le d, for sources in phase there are only the orders ∣k∣≤dλ|k| \le \frac{d}{\lambda} – for d=3λd = 3\lambda the 77 antinodal lines k=−3,…,3k = -3, \dots, 3.

Far away the hyperbolas approach straight lines through the centre: sin⁡αk=k⋅λd\sin\alpha_k = k \cdot \frac{\lambda}{d} – the same formula as for the double slit.

The snapshot shows crests light and troughs dark, the amplitude pattern shows how strongly the water oscillates on average. The waves are shown greatly slowed down; damping is neglected.

Try it

  1. Press “Restart the waves” and watch the pattern build up.
  2. Drag the cork PP onto a calm line and read off Δs\Delta s. Repeat on another calm line.
  3. Switch to the amplitude pattern with all lines. Load example
  4. Drag E2E_2 outwards. How many antinodal lines are there for d=2λd = 2\lambda, 3λ3\lambda, 4λ4\lambda?
  5. Set Δφ=180°\Delta\varphi = 180°. What happens on the perpendicular bisector? Antiphase
  6. Show the far field. Far field

Tasks

Task 1: Maximum or minimum?

Load task

d=6 cmd = 6\,\text{cm}, λ=2 cm\lambda = 2\,\text{cm}, in phase. PP is 8 cm8\,\text{cm} straight above E2E_2. Find s1s_1, s2s_2, Δs\Delta s and decide.

Show solution

s2=8 cms_2 = 8\,\text{cm}, s1=10 cms_1 = 10\,\text{cm}, Δs=2 cm=1⋅λ\Delta s = 2\,\text{cm} = 1 \cdot \lambda: antinodal line of order 1, amplitude 2y^2\hat y. Show in the simulation

Task 2: Counting lines

Load task

How many antinodal and nodal lines are there for d=7 cmd = 7\,\text{cm} and λ=2 cm\lambda = 2\,\text{cm}?

Show solution

Antinodal: ∣k∣≤3.5|k| \le 3.5, so 77 lines. Nodal: ∣Δs∣=1,3,5,7 cm|\Delta s| = 1, 3, 5, 7\,\text{cm}, so 88 lines; the two with ∣Δs∣=d|\Delta s| = d lie on the line through the sources, outside them. Show

Task 3: Far field

For d=4 cmd = 4\,\text{cm} and λ=1.2 cm\lambda = 1.2\,\text{cm}, find the directions of the antinodal lines far away.

Show solution

sin⁡αk=0.3⋅k\sin\alpha_k = 0.3 \cdot k: α1≈17.5°\alpha_1 \approx 17.5°, α2≈36.9°\alpha_2 \approx 36.9°, α3≈64.2°\alpha_3 \approx 64.2°; 77 antinodal lines in total. Show

Ideas for the classroom

  • Use: upper secondary, interference of mechanical waves, as a bridge to the double slit; a good companion to a real ripple tank, since here everything can be paused, slowed down and measured. With val=0 the values in the picture are hidden for measuring tasks.
  • Model: equal amplitudes, no damping, ideal point sources.
  • Common stumbling blocks:
    • “There are no waves on the nodal lines.” – Both waves pass through; they just cancel.
    • “The path difference is the distance between the sources.” – It is the difference of the paths to PP; only on the axis outside the sources is ∣Δs∣=d|\Delta s| = d.
    • “A stable pattern needs any two waves.” – It needs equal frequency and a fixed phase difference (coherence).