Drag the sources S₁, S₂ or the probe P
Controls
View
Readings
- Distance from S₁, r₁
- 26.8cm
- Distance from S₂, r₂
- 24.0cm
- Path difference, Δ
- 2.83cm
- Path difference, Δ/λ
- 0.94λ
- Phase difference at P, δ
- 20°
- Resultant amplitude (A₁ = 1)
- 1.97
- Interference at P
- Partial
- Fringe width measured on screen
- 11.5cm
- β = λD/d (assumes D ≫ d)
- 11.0cm
How to use this simulation
- Drag S₁ and S₂, or use the separation slider: a larger d packs the fringes closer together.
- Switch the view to "time-averaged intensity" to replace the moving ripples with the fixed bright and dark fringes.
- Put the probe P on a green line and the path difference is a whole number of wavelengths; on a red dashed line it is a half-integer.
- Set the phase difference φ to 180°: every fringe shifts, and the centre line turns dark.
Two stones in a pond
Drop two stones into a still pond a short distance apart. Each one sends out rings of ripples, and soon the two sets of rings run through each other. Look closely and you will see a pattern: along some lines the water heaves up and down wildly, while along others it barely moves at all.
That pattern of loud and quiet places is interference. Where the two waves arrive in step they help each other; where they arrive out of step they cancel. The pattern is not random either. You can work out exactly where every calm line and every rough line will be before the stones even hit the water.
You carry an example of interference in your pocket if you own noise-cancelling headphones. They listen to the noise around you and play its mirror image into your ear, so the two sound waves cancel. Same physics, different wave.
Wave basics from scratch
A few words first. The highest point of a water wave is a crest and the lowest is a trough. The distance from one crest to the next is the wavelength, written λ (lambda). The biggest distance the water moves away from its resting level is the amplitude, A.
Phase describes where a particle is in its cycle at a given moment: at a crest, at a trough, rising or falling. Two points that rise and fall together are in phase. Two points where one is at a crest while the other is at a trough are completely out of phase, half a cycle apart.
The superposition principle: waves simply add
When two or more waves pass through the same point at the same time, the resulting displacement is the sum of the displacements each wave would cause on its own. That is the principle of superposition.
Think of two friends pushing a swing. Push together, in time, and the swing flies high. Push from opposite sides with equal force at the same moment and the swing hardly moves. Waves do exactly this, with up counted as positive and down as negative.
y = y₁ + y₂resulting displacement = sum of the separate displacements
Key words at a glance
These definitions come up again and again, so here they are in one place.
| Term | In plain words |
|---|---|
| Superposition | Waves meeting at a point add their displacements |
| Interference | The steady pattern of strong and weak spots that superposition creates |
| Coherent sources | Two sources with the same frequency and a phase difference that stays constant |
| Path difference (Δ) | How much farther one wave travels than the other: |r₁ − r₂| |
| Phase difference (δ) | How far one wave is ahead of the other in its cycle, as an angle |
| Fringe | One of the alternating bright or dark bands on a screen |
| Fringe width (β) | Distance between two neighbouring bright (or dark) fringes |
Conditions for constructive and destructive interference
What two waves do at a point depends on how they arrive there, and that depends on the path difference: how much farther one wave had to travel than the other.
Constructive: crest meets crest
If the path difference is a whole number of wavelengths (0, λ, 2λ, …), one wave has simply travelled some extra complete cycles. Crests land on crests and troughs on troughs. The amplitude becomes A₁ + A₂: a bright fringe for light, a loud spot for sound.
Δ = nλ, n = 0, 1, 2, …resultant amplitude A₁ + A₂
Destructive: crest meets trough
If the path difference is an odd number of half wavelengths (λ/2, 3λ/2, …), the crest of one wave lands on the trough of the other. The displacements point opposite ways and cancel. The amplitude is |A₁ − A₂|, which is zero when the two are equal: a dark fringe.
Δ = (2n + 1)λ/2resultant amplitude |A₁ − A₂|
From path difference to phase difference
One extra wavelength of path is one full cycle, which is 2π radians or 360° of phase. So any path difference converts to a phase difference with a simple ratio, and for any phase difference there is a general formula for the resultant amplitude.
δ = (2π/λ)·Δphase difference
A² = A₁² + A₂² + 2A₁A₂cosδresultant amplitude for any phase
Why the sources must be coherent
Shine two ordinary bulbs at a wall and you will never see fringes. Each bulb's light changes its phase randomly millions of times a second, so at any point the interference flips from constructive to destructive faster than any eye or camera can follow. All you see is the average.
A steady pattern needs coherent sources: the same frequency and a phase difference that stays fixed. The easy way to get them is to split one source in two, which is exactly what Young did with a pair of narrow slits.
- Same frequency (and so the same wavelength).
- Constant phase difference.
- Similar amplitudes, so the dark fringes are truly dark and the pattern is sharp.
- Narrow sources, close together.
Young's double-slit experiment, step by step
Around 1801 Thomas Young used a simple setup to show that light behaves as a wave. It is still the centrepiece of the topic in every physics syllabus.
The setup
Monochromatic light passes through a single narrow slit S, then reaches two very close narrow slits S₁ and S₂, a distance d apart. A screen stands a distance D away. Because S₁ and S₂ are lit by the same wavefront, they act as coherent sources.
On the screen you see evenly spaced bright and dark bands. The centre, where the path difference is zero, is bright as long as the two slits are in phase.
Fringe position and fringe width
At a distance y from the centre of the screen, the path difference is approximately Δ = yd/D when D is large. Setting Δ = nλ gives the position of the nth bright fringe, and the gap between neighbours is the fringe width.
yₙ = nλD/dposition of the nth bright fringe
β = λD/dfringe width
- Longer wavelength, wider fringes: red fringes are wider than blue ones.
- Move the screen back (larger D) and the fringes spread out.
- Bring the slits closer (smaller d) and the fringes widen.
- Put the whole apparatus under water and λ shrinks, so the fringes get narrower.
Interference vs diffraction
Both produce bright and dark bands, which is why they get mixed up. The real difference is where the overlapping waves come from.
| Feature | Interference | Diffraction |
|---|---|---|
| Source of waves | Two coherent sources | Different parts of the same wavefront |
| Fringe width | Usually all equal | Central band widest, others narrower |
| Brightness | Bright fringes roughly equal | Falls off quickly away from the centre |
| Dark fringes | Can be completely dark | Never completely dark |
| Example | Young's double slit | Light through one narrow slit or past a sharp edge |
Experiments to try in the simulation
The ripple tank above is a small laboratory. Use it to check everything on this page for yourself.
- Increase the wavelength λ: the ripples lengthen and the green and red lines spread apart, just as β = λD/d predicts.
- Reduce the separation d: fewer lines, wider fringes.
- Sweep the phase difference φ from 0° to 360°: the whole pattern slides sideways and returns home at 360°.
- Lower the amplitude ratio: the dark fringes turn grey instead of black.
- Drag the probe P around and read the path difference in wavelengths from the readings panel.
Solved problems
Each problem is worked step by step. Convert everything to SI units first.
Problem 1: fringe width
In a double-slit experiment λ = 600 nm, the slits are d = 0.5 mm apart and the screen is D = 1 m away. Find the fringe width.
β = λD/d = (600 × 10⁻⁹ × 1) / (0.5 × 10⁻³) = 1.2 mm. The third bright fringe then sits y₃ = 3β = 3.6 mm from the centre.
Problem 2: wavelength from a measured fringe width
A student measures a fringe width of 0.75 mm with D = 1.5 m and d = 1 mm. What is the wavelength?
Rearrange: λ = βd/D = (0.75 × 10⁻³ × 1 × 10⁻³) / 1.5 = 500 nm, which is green light.
Problem 3: halving the slit separation
In problem 1 the separation is halved to 0.25 mm. The new fringe width is β′ = λD/(d/2) = 2.4 mm, exactly double. Fringe width is inversely proportional to d.
Problem 4: bright or dark?
With λ = 600 nm, two points have path differences of 1.5 µm and 1.8 µm. What happens at each?
First point: Δ/λ = 2.5, a half-integer. The phase difference is δ = (2π/λ)Δ = 5π rad, an odd multiple of π, so the interference is destructive: dark.
Second point: Δ/λ = 3, a whole number. δ = 6π rad, a multiple of 2π, so the interference is constructive: bright.
Problem 5: unequal amplitudes
Two coherent waves have amplitudes 3 cm and 1 cm. The maximum amplitude is 3 + 1 = 4 cm and the minimum is 3 − 1 = 2 cm. Intensity goes as amplitude squared, so Imax : Imin = 4² : 2² = 4 : 1.
At a phase difference of 60°, A² = 3² + 1² + 2 × 3 × 1 × cos60° = 13, so A = 3.61 cm.
Common mistakes
Knowing these in advance saves marks.
- Mixing mm and nm in β = λD/d. Convert everything to metres first.
- Thinking energy is destroyed at dark fringes. It is only redistributed.
- Thinking a larger d gives wider fringes. It is the reverse.
- Expecting two separate bulbs to interfere. They are not coherent.
- Confusing interference with diffraction. Ask: two sources, or one wavefront?
Interference in everyday life
Interference is not just a textbook experiment. It is happening around you all the time.
- Soap bubbles and oil films: light reflected from the front and back surfaces of a thin film interferes. Different thicknesses reinforce different colours, which is where the rainbow swirls come from.
- Anti-reflection coatings on glasses and camera lenses: a very thin layer makes the reflections from its two surfaces cancel, so less light bounces off and more gets through.
- Noise-cancelling headphones: a microphone picks up outside noise and the headphones play the same sound 180° out of phase. The two interfere destructively, which works best on steady sounds like engine hum.
- Radio and Wi-Fi dead spots: signals reflected off walls interfere with the direct signal, so a few spots in a room get a weaker connection.
Exam corner
These question types come up most often.
- Define coherent sources and explain why two separate lamps cannot produce a steady interference pattern.
- State the conditions for constructive and destructive interference in terms of path difference and phase difference.
- Calculate the fringe width for given λ, D and d. With this page's numbers, β = 1.2 mm.
- Explain how the pattern changes when d is halved, the screen is moved back or the apparatus is placed in water.
- Distinguish between interference and diffraction.
Quick revision
Read this the night before a test and the main ideas will come back.
- Superposition: y = y₁ + y₂.
- Interference needs coherent sources.
- Constructive: Δ = nλ, amplitude A₁ + A₂.
- Destructive: Δ = (2n + 1)λ/2, amplitude |A₁ − A₂|.
- Phase difference δ = (2π/λ)Δ.
- Fringe width β = λD/d.
- Energy is redistributed, never destroyed.
Frequently asked questions
What is the principle of superposition?
When two or more waves overlap at a point, the resulting displacement equals the sum of the displacements the individual waves would produce there.
What is wave interference?
It is the steady pattern of reinforcement and cancellation that forms when waves from coherent sources overlap: large amplitude where they meet in phase, small or zero where they meet out of phase.
What are the conditions for constructive interference?
The path difference must be a whole number of wavelengths, Δ = nλ, which means a phase difference of 0, 2π, 4π and so on. The resultant amplitude is then A₁ + A₂.
Why must interfering sources be coherent?
Only sources with the same frequency and a constant phase difference keep constructive and destructive points fixed in place. Incoherent sources shift the pattern so fast that only an average brightness is seen.
What is the fringe width formula?
β = λD/d, where λ is the wavelength, D the distance from the slits to the screen and d the distance between the slits.
What is the difference between interference and diffraction?
Interference combines waves from two coherent sources and gives equally spaced, equally bright fringes. Diffraction combines waves from different parts of one wavefront and gives a wide central band with fainter, narrower bands beside it.
Where does the energy go in destructive interference?
It moves to the regions of constructive interference. Total energy is conserved; interference only redistributes it.
Keep studying this topic
The animation made the idea click; now turn it into marks. Syllabus, suggestions, textbooks and admission-test guides are below.
SSC
SSC Physics: syllabus and preparation
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প্রতিবেদন: বিজ্ঞান মেলা (in Bangla)
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SSC syllabus
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