Theory — Superposition and Interference

Superposition of Waves

When two waves pass through the same point, the total displacement is the sum of the individual displacements. Where two crests meet, they add to make a bigger crest (constructive interference); where a crest meets a trough, they cancel (destructive interference). For a steady pattern the two sources must be coherent — same frequency and a fixed phase relationship.

Path Difference

Whether a point lies on a bright (constructive) or dark (destructive) region depends on the difference in distance from the two sources — the path difference Δ = |r₂ − r₁|. If the two sources start in phase, the waves arrive in step wherever the path difference is a whole number of wavelengths, and exactly out of step wherever it is an odd number of half-wavelengths.

Interference Conditions (sources in phase) Constructive (bright):  Δ = m·λ      (m = 0, 1, 2, …)
Destructive (dark):    Δ = (m + ½)·λ  (m = 0, 1, 2, …)
Bright when path difference = whole number of wavelengths

Nodal and Antinodal Lines

Joining all the points with the same path difference traces out curves (hyperbolas). The antinodal lines (Δ = mλ) are where the waves always reinforce — these are the bright bands. The nodal lines (Δ = (m+½)λ) are where the waves always cancel — these stay calm and dark. Along the perpendicular bisector of the two sources the path difference is zero, so the central line is always a strong constructive (m = 0) band.

The Double-Slit Experiment

Passing one wave through two narrow slits a distance d apart turns the slits into two coherent sources. On a screen a distance L away, the path difference to a point a height y above the centre is approximately d·sinθ ≈ d·y/L. Setting this equal to the interference conditions gives evenly spaced bright fringes.

Double-Slit Fringes (small angles, L ≫ d) Bright fringes:  d·sinθ = m·λ  →  y_m = m·λL / d

Fringe spacing:  Δy = λL / d
Δy ∝ λ · Δy ∝ L · Δy ∝ 1/d → Δy·d / L = λ

Constructive

Waves arrive in phase. Crest meets crest. Path difference = mλ. Amplitudes add → bright band / loud sound / antinode.

Destructive

Waves arrive out of phase. Crest meets trough. Path difference = (m+½)λ. Amplitudes cancel → dark band / quiet spot / node.

ChangeEffect on fringe spacing ΔyWhy
Longer wavelength λΔy increasesΔy ∝ λ
Larger slit separation dΔy decreasesΔy ∝ 1/d
Greater screen distance LΔy increasesΔy ∝ L

Apparatus

The equipment used to produce coherent waves and observe interference and diffraction. In the simulation these are modelled for you, but the patterns correspond to what this equipment would produce on the bench.

LASER monochromatic source
Laser source
Provides coherent, single-wavelength light to illuminate the slits.
two slits, spacing d
Double slit
Two narrow slits a distance d apart that act as coherent sources for interference.
one slit, width a
Single slit
One slit of width a used to study diffraction and the central bright band.
screen (fringes)
Viewing screen
Set a distance L away; the bright and dark fringes are measured on it.
source lens screen
Optical bench
Aligns the source, slits, and screen along one axis at measured distances.
two coherent sources
Ripple tank
Shows water-wave interference from two coherent sources for direct observation.

Instructions — Running the Virtual Experiment

The Ripple Tank tab builds physical intuition; the Double-Slit Measurements tab gives the quantitative test of interference; and the Single-Slit Diffraction tab covers diffraction and slit width. In the two measurement tabs you calculate the value yourself first, enter it, and only then does the simulation reveal its measurement so you can compare. Record every reading in your lab notebook.

Experiment 1 — Mapping the Interference Pattern (Ripple Tank tab)
1
Open Simulation → Ripple Tank and press Play. Two in-phase sources emit circular waves that overlap.
2
Turn on Interference bands to switch to the time-averaged view. The bright bands are antinodal lines (constructive); the dark bands between them are nodal lines (destructive). Confirm the central band lies on the perpendicular bisector of the two sources.
3
Increase the wavelength. Note that the bands spread farther apart. Increase the source separation and note that the bands move closer together — the same trends you will measure quantitatively in the double-slit tab.
Experiment 2 — Path Difference with the Probe (Ripple Tank tab)
1
Drag the yellow probe onto a bright (antinodal) band. Read r₁, r₂, and the path difference Δ. Confirm Δ is close to a whole number of wavelengths (Δ/λ ≈ 0, 1, 2 …).
2
Now drag the probe onto a dark (nodal) band and confirm Δ/λ is close to a half-integer (0.5, 1.5, …). Record three antinodal and three nodal readings.
Experiment 3 — Fringe Spacing vs. Wavelength (Double-Slit Measurements tab)
1
Open Double-Slit Measurements. Keep the slit separation and screen distance fixed (start with d = 0.20 mm, L = 2.0 m) and choose a wavelength.
2
Calculate the fringe spacing Δy = λL/d by hand (convert λ and d to metres, give Δy in millimetres), enter it, and click Record & check. The measured fringe spacing on the screen stays hidden until you enter your value; then compare your prediction with the measurement.
3
Repeat for several wavelengths (450 to 650 nm) and confirm that Δy increases in direct proportion to λ.
Experiment 4 — Effect of Slit Separation and Screen Distance (Double-Slit tab)
1
At a fixed wavelength (say 550 nm), slide the slit separation d larger and watch the fringes crowd together (Δy ∝ 1/d).
2
Now slide the screen distance L larger and watch the fringes spread apart (Δy ∝ L). For any setting you can recalculate Δy and check it again.
Experiment 5 — Single-Slit Diffraction and Slit Width (Single-Slit Diffraction tab)
1
Open Single-Slit Diffraction. A single slit of width a produces a broad bright central band with weaker side bands. The first dark band sits where sin θ = λ/a, which on the screen is a distance y₁ = λL/a from the centre.
2
Set the wavelength, the slit width a, and the screen distance L. Calculate y₁ = λL/a by hand, enter it, and click Record & check. The first dark band stays hidden until you enter your value; then compare.
3
Narrow the slit and watch the central band grow wider, then widen the slit and watch it tighten. Because a is in the denominator, a smaller slit gives a larger y₁ and a broader pattern. Record y₁ for several slit widths and confirm y₁ ∝ 1/a.

Simulation — Two-Source Interference & the Double Slit

Wave Interference Virtual LabPress Play and explore the interference pattern
crest
trough
sources & probe
Drag the sources or the probe.

Controls

Probe readout
r₁ (to source 1)— cm
r₂ (to source 2)— cm
path diff Δ— cm
Δ / λ
Left: two slits separated by d.
Right: fringe pattern on a screen a distance L away.

Controls

Predict, then measure

Current reading
Wavelength λ550 nm
Measured fringe Δy— hidden
λ (nm)d (mm)L (m)Δy your calc (mm)Δy measured (mm)
No readings yet — set a wavelength, calculate Δy, enter it, and click "Record & check".
Left: one slit of width a. Right: the diffraction pattern on a screen a distance L away.
A narrower slit spreads the wave more, widening the central band.

Single slit

Predict, then measure

Current reading
Wavelength λ550 nm
Slit width a0.10 mm
First dark band y₁— hidden
λ (nm)a (mm)L (m)y₁ your calc (mm)y₁ measured (mm)
No rows yet — set the slit, calculate y₁, enter it, and click "Record & check".

Team Questions

Question 1. Two loudspeakers driven in phase emit identical sound waves. You stand at a point where the path difference from the two speakers is exactly 2λ. Do you hear a loud (constructive) or quiet (destructive) result? (Answer "constructive" or "destructive")
Question 2. You move to a different point where the path difference is 1.5λ. Constructive or destructive now? (Answer "constructive" or "destructive")
Question 3. In a double-slit experiment, light of wavelength 600 nm passes through slits 0.25 mm apart onto a screen 2.0 m away. Calculate the fringe spacing Δy = λL/d. (Type just the number in millimetres, e.g. 4.8)
Question 4. For the same setup as Question 3, how far from the central fringe is the third-order bright fringe (m = 3)? Use y_m = mλL/d. (Type just the number in millimetres)
Question 5. If you keep λ and L the same but increase the slit separation d, what happens to the fringe spacing Δy — does it increase, decrease, or stay the same? (One word)
Question 6. Red light (700 nm) and blue light (450 nm) are each sent through the same double slit onto the same screen. Which colour produces the wider fringe spacing? (Answer "red" or "blue")
Question 7 — Challenge. Two in-phase sources sit in a ripple tank. Consider a point that is exactly the same distance from both sources (on the perpendicular bisector). Is the interference there constructive or destructive, and what is the path difference? (Answer "constructive" or "destructive")

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission.

Wave Interference: Two-Source Patterns and the Double-Slit Relation

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To observe the interference pattern produced by two coherent sources, to relate the bright and dark regions to the path difference between the sources, and to verify the double-slit fringe relation Δy = λL/d by measuring the fringe spacing for a range of wavelengths.

Theory

Two coherent sources interfere constructively where the path difference Δ is a whole number of wavelengths and destructively where it is an odd number of half-wavelengths. For a double slit of separation d viewed on a screen a distance L away (with L ≫ d), the bright fringes occur at y_m = mλL/d, so adjacent fringes are evenly spaced by Δy.

Constructive: Δ = mλ  ·  Destructive: Δ = (m+½)λ
Δy = λL / d  →  λ = Δy·d / L

Rearranging the fringe equation gives λ = Δy·d/L, so multiplying the measured fringe spacing by d/L should recover the wavelength used — an internal check on every measurement.

Calculations — Sample: λ = 550 nm, d = 0.20 mm, L = 2.0 m

Fringe spacing: Δy = λL/d = (550 × 10⁻⁹)(2.0)/(0.20 × 10⁻³) = 1.10 × 10⁻⁶ / 2.0 × 10⁻⁴ = 5.5 × 10⁻³ m = 5.5 mm

Recovered wavelength: λ = Δy·d/L = (5.5 × 10⁻³)(0.20 × 10⁻³)/(2.0) = 5.5 × 10⁻⁷ m = 550 nm

Third-order fringe (m = 3): y₃ = 3·Δy = 3(5.5) = 16.5 mm from centre

Results Table — Double Slit: Predicted vs. Measured Fringe Spacing (d = 0.20 mm, L = 2.0 m)

λ (nm)Δy calculated = λL/d (mm)Δy measured (mm)
4504.504.50
5005.005.00
5505.505.50
6006.006.00
6506.506.50

Worked example for 550 nm: Δy = λL/d = (550 × 10⁻⁹ m × 2.0 m) / (0.20 × 10⁻³ m) = 5.50 × 10⁻³ m = 5.50 mm, matching the screen. Probe checks in the ripple tank confirmed the interference conditions: antinodal (bright) points gave Δ/λ ≈ 0, 1, 2 …, while nodal (dark) points gave Δ/λ ≈ 0.5, 1.5, … .

Results Table — Single Slit: Predicted vs. Measured First Dark Band (λ = 550 nm, L = 2.0 m)

a (mm)y₁ calculated = λL/a (mm)y₁ measured (mm)
0.303.673.67
0.205.505.50
0.1011.0011.00
0.0522.0022.00

Worked example for a = 0.10 mm: y₁ = λL/a = (550 × 10⁻⁹ m × 2.0 m) / (0.10 × 10⁻³ m) = 1.10 × 10⁻² m = 11.0 mm. As the slit narrowed from 0.30 to 0.05 mm, the central band widened, confirming y₁ ∝ 1/a.

Discussion

For the double slit, the calculated fringe spacing matched the measured spacing at every wavelength, and Δy increased in direct proportion to λ: 450 nm gave 4.50 mm and 650 nm gave 6.50 mm, a ratio (1.444) equal to the ratio of wavelengths (650/450 = 1.444). Increasing the slit separation crowded the fringes together (Δy ∝ 1/d) and moving the screen farther away spread them apart (Δy ∝ L), as Δy = λL/d predicts.

For the single slit, the calculated position of the first dark band matched the measurement, and the central maximum widened as the slit narrowed, because the slit width sits in the denominator of y₁ = λL/a. The ripple-tank observations explained the underlying cause: where waves arrive in phase they reinforce and where they arrive out of phase they cancel, producing the bright and dark bands seen in both the interference and diffraction patterns.

Conclusion

The experiment confirmed superposition, the double-slit fringe relation Δy = λL/d, and the single-slit diffraction relation y₁ = λL/a. Calculated values agreed with the measurements throughout. The proportionalities Δy ∝ λ, Δy ∝ L, and Δy ∝ 1/d were verified, and the central diffraction band was shown to widen as the slit narrowed (y₁ ∝ 1/a), relating the spread of the pattern to the slit width and the wavelength.

Practice Questions

Show all work and include units in your answers.

Question 1
Two coherent sources emit waves of wavelength 4.0 cm. At a certain point the distances to the two sources are 30.0 cm and 36.0 cm. Is this point on a bright (antinodal) or dark (nodal) line? Justify using the path difference.
Hint: Δ = 36.0 − 30.0 = 6.0 cm. Compare Δ/λ to whole vs. half integers.
Question 2
Light of wavelength 500 nm passes through a double slit with d = 0.10 mm onto a screen 1.5 m away. Find the fringe spacing and the distance from the centre to the second-order bright fringe.
Hint: Δy = λL/d. The second-order fringe is at y₂ = 2Δy.
Question 3
In a double-slit experiment the fringe spacing is measured to be 6.0 mm with a slit separation of 0.20 mm and a screen distance of 2.4 m. What is the wavelength of the light?
Hint: rearrange Δy = λL/d to λ = Δy·d/L.
Question 4
A double-slit pattern uses 550 nm light. You want to double the fringe spacing without changing the wavelength or the screen distance. What must you do to the slit separation, and by what factor?
Hint: Δy ∝ 1/d. To double Δy you must halve d.
Question 5
Two in-phase ripple-tank sources are 8.0 cm apart and emit waves of wavelength 2.0 cm. List the path differences for the first three nodal lines on one side of the central band. How many wavelengths correspond to each?
Hint: nodal lines occur at Δ = (m+½)λ for m = 0, 1, 2 → 0.5λ, 1.5λ, 2.5λ = 1.0, 3.0, 5.0 cm.
Question 6 — Challenge
In a double-slit setup, the 4th-order bright fringe of one wavelength lands exactly on the 5th-order bright fringe of a second wavelength (same d and L). If the first wavelength is 600 nm, what is the second wavelength?
Hint: same position means 4λ₁ = 5λ₂, so λ₂ = (4/5)λ₁ = (4/5)(600) = 480 nm.