AICE 9702 · OPTICS

INTERFERENCE OF LIGHT

Explore Young's double-slit experiment, manipulate light waves, calculate fringe spacing, and discover why bright and dark interference patterns form. Every calculation shows its formula, its substitution and its units.

FRINGE SPACING

w = λD / d

PATH DIFFERENCE

Δ = d sin θ

BRIGHT FRINGE

Δ = mλ

DARK FRINGE

Δ = (m + ½)λ

SECTION 1

Virtual Double-Slit Laboratory

Manipulate the experiment itself: wavelength, slit separation, slit width, screen distance, light intensity, relative phase and viewing scale. The pattern, the diagram, the strip, the curve and the tables all recalculate from the same physics, and every quantity is labelled with its units.

Virtual Double-Slit Laboratory

Change the wavelength, the slit separation, the slit width, the screen distance, the light intensity, the relative phase of the sources, and the viewing scale. The screen, the strip and every table below recalculate immediately. You can also drag directly on the diagram: the cyan ruler moves the screen, the purple handle moves the upper slit, and the marker on the screen moves the probe.

Wavelength λ (nm)

380–780 nm is visible light; the drawing changes colour with it.

Screen distance D (m)

Slit separation d (mm)

Slit width a (mm)

Sets the diffraction envelope that decides which fringes are visible.

Light intensity (units)

Peak intensity of a bright fringe (4I₀ for equal sources).

Relative source phase φ₀ (°)

360° shifts the whole pattern by exactly one fringe.

Viewing scale (screen width shown) (mm)

Screen position (centre offset) (mm)

Pan the viewing window along the screen.

Slit-width envelope

Ray markers

FRINGE SPACING w

2.4 mm

HIGHEST ORDER

m = 833

FRINGES IN VIEW

FRINGES IN CENTRAL MAX

The drawing is deliberately not to scale: d and λ are magnified so you can see them. Every number printed on the diagram and in the tables, however, is the real calculated value in SI units. Angles use θ = arctan(y / D) exactly; the small-angle result y = mλD/d is shown alongside for comparison. Light intensity is in arbitrary units and the two slits are treated as equally bright, ideal, coherent sources.

Wavefronts are drawn in blue, the overlap region in purple. Green rays mark the bright fringes, red dashed rays the dark ones. Everything is recomputed from the physics each time a value changes.

FRINGES IN CENTRAL MAXIMUM

d / a

HIGHEST POSSIBLE ORDER

m = 833

LOCAL VISIBILITY V

100%

BRIGHT FRINGES IN THE VIEWING WINDOW

Order	sin θ	θ	y exact	y ≈ mλD/d	% of peak
m = -4	-0.0048	-0.275°	-9.6 mm	-9.6 mm	5.5%
m = -3	-0.0036	-0.2063°	-7.2 mm	-7.2 mm	25.5%
m = -2	-0.0024	-0.1375°	-4.8 mm	-4.8 mm	57.3%
m = -1	-0.0012	-0.0688°	-2.4 mm	-2.4 mm	87.5%
m = 0	0	0°	0 mm	0 mm	100%
m = 1	0.0012	0.0688°	2.4 mm	2.4 mm	87.5%
m = 2	0.0024	0.1375°	4.8 mm	4.8 mm	57.3%
m = 3	0.0036	0.2063°	7.2 mm	7.2 mm	25.5%
m = 4	0.0048	0.275°	9.6 mm	9.6 mm	5.5%

SUMMARY

Fringe spacing ww = λD / d (small angle)

2.4 mm

Small-angle error at m = 3exact D tan θ against mλD/d

0.00065 %

Missing ordersm = p · d / a

-10, -5, 5, 10

Nearest envelope zerofirst diffraction minimum, sin θ = λ / a

12 mm (p = 1)

Number of bright orders possible

1667

Reading the pattern: the fringe spacing is uniform to better than one part in ten thousand across the whole central maximum, which is why w = λD/d is safe to use in calculations. The brightness of each fringe is set by the diffraction envelope — a fringe at 20% of the peak is still there, just much dimmer, and a fringe that falls exactly on an envelope zero disappears altogether.

The Pattern on the Screen

The observation screen unrolled: brightness across the strip is the calculated intensity, brightest at the centre of each bright fringe and zero where the two waves cancel. Place the probe anywhere in the pattern to read the path difference behind what you see.

Fringe spacing 2.4 mm

Bright fringes in view 9 of 1667 possible

Light 600 nm · orange

Local visibility 100% — drag anywhere on the strip to move the probe.

Diffraction envelope
Intensity on screen

Green dots are the exact positions of the bright fringes, red dots the dark ones. Click anywhere on the chart to move the probe to that position.

PROBE AT y = 0 mm

Angle θ = arctan(y / D)

0°

Path difference Δ = d sin θ

0 nm

In wavelengths Δ / λ

0 λ

Phase difference φ

0°

Intensity here

100% of a bright fringe

Bright (constructive)

The path difference is exactly 0λ — the two waves arrive in step and add constructively (order m = 0).

I / I_peak = cos²(π · Δ/λ) = cos²(π × 0) = 1

Move the probe to a bright fringe and Δ/λ is a whole number; move it halfway between two bright fringes and the waves cancel. That single ratio decides the entire pattern.

Path Difference Geometry

Why the path difference is d sin θ. Drop a perpendicular from the upper slit onto the ray from the lower slit: the extra distance the lower ray travels is the green segment, and it equals d sin θ. Drag the point on the screen and watch the two rays, the angle and the extra distance change together.

Extra distance d sin θ = 0 nm

In wavelengths 0 λ

Small-angle check 0 mm from d·y/D

Fringe Spacing Calculator

w = λD/d, rearranged for whichever quantity is unknown. Enter what the question gives; the answer comes with the formula, the substitution in scientific notation, and the unit conversion written out.

Which quantity are you finding?

w = λD / d ⇔ λ = wd / D ⇔ D = wd / λ ⇔ d = λD / w

Wavelength λ (nm)

380–780 nm is visible light

Screen distance D (m)

Slit separation d (mm)

Fringe spacing w (mm)

Being found

Fringe spacing w = 2.4 mm

1w = λD / d
2w = (6 × 10⁻⁷ m × 2 m) / 5 × 10⁻⁴ m
3w = 1.2 × 10⁻⁶ m / 5 × 10⁻⁴ m
4w = 2.4 × 10⁻³ m = 2.4 mm

600 nm is visible orange light.
Convert to SI first: nanometres to metres (×10⁻⁹) and millimetres to metres (×10⁻³).

WHERE THE FIRST THREE FRINGES LAND

Order	sin θ	θ	y exact (D tan θ)	y ≈ mλD/d
m = 1	0.0012	0.0688°	2.4 mm	2.4 mm
m = 2	0.0024	0.1375°	4.8 mm	4.8 mm
m = 3	0.0036	0.2063°	7.2 mm	7.2 mm

Current geometry: 600 nm light, slits 0.5 mm apart, screen 2 m away.

WHAT THE ANSWER MEANS

Fringe spacingfrom the values in the boxes

2.4 mm

Spacing in metres

0.0024 m

Slit separation in metres

0.0005 m

Wavelength in metres

6e-7 m

Every quantity in w = λD/d must be in metres before you substitute. The most common lost mark in this topic is a millimetre or a nanometre that never got converted, so write the conversion out in full — the working above does it for you.

The spacing is uniform for every pair of neighbouring fringes; only the exact position drifts slightly at high orders.

Interference & Phase Calculator

Move between path difference, phase difference, order, angle and screen position — including finding a wavelength from an observed order, or a slit separation from an observed fringe position.

Δ = d sin θ

Slit separation d (mm)

Screen position y (mm)

Positive above the centre

Path difference Δ = 600 nm

1Δ = d sin θ
2θ = arctan(y / D) = arctan(2.4 / 2) = 0.069°
3Δ = 6 × 10⁻⁷ m = 600 nm

The angle was taken from the screen position with θ = arctan(y / D).
In wavelengths: Δ = 1λ.

sin θ for the order

—

Angle from the geometry

—

Position on the screen

—

Work in one direction and check in the other. Every bright fringe on the screen has a whole-number path difference; the dark fringe exactly between two of them has Δ = (m + ½)λ. If a calculated order gives sin θ greater than 1, that order simply does not exist for those slits.

Bright & Dark Conditions

A live superposition diagram: drag the path difference and watch two coherent waves drift in and out of step, with the resultant amplitude and the intensity factor following them.

Path difference Δ / λ

Try 0, 0.5, 1, 1.5 and 2 to walk through the bright and dark conditions in order.

Bright (constructive)

The path difference is exactly 0λ — the two waves arrive in step and add constructively (order m = 0).

Path difference

0 λ

Phase difference

0°

Resultant amplitudeA = 2A₀ |cos(φ/2)|

2 A₀

Intensity factorI = I_peak cos²(φ/2)

100%

PATH DIFFERENCE AGAINST WHAT YOU SEE

Path difference	In words	Result on the screen
Δ = 0	The central maximum	Bright — m = 0
Δ = λ/2	Half a wavelength apart	Dark — first minimum
Δ = λ	One wavelength	Bright — m = 1
Δ = 3λ/2	One and a half wavelengths	Dark
Δ = 2λ	Two wavelengths	Bright — m = 2
Δ = (m + ½)λ	Any half-integer	Dark
Δ = mλ	Any whole number	Bright — order m

For equal sources the intensity runs from zero at a dark fringe to four times a single source's intensity at a bright fringe, and the pattern repeats every time Δ grows by one whole wavelength. Because a bright fringe becomes the next bright fringe, the pattern is periodic in Δ with period λ — that is what makes the fringe spacing uniform.

Graphs & Relationships

How the fringe spacing responds to wavelength, screen distance and slit separation; how sin θ grows with order; and how the path difference in wavelengths varies across the screen.

Fringe spacing against one variable

Wavelength λ and the spacing are directly proportional — a straight line through the origin. Yellow marker: the values currently set (D = 2 m, d = 0.5 mm).

sin θ against order — a straight line of gradient λ/d

Gradient = λ/d = 0.001. The highest order with these slits is m = 833.

Path difference in wavelengths across the screen

Every crossing of a green line is a bright fringe: the path difference is a whole number of wavelengths there. The dashed white line is the centre. The near-constant slope over this range is why the fringes are evenly spaced.

Set the values used in these graphs

Wavelength λ (nm)

Screen distance D (m)

Slit separation d (mm)

Reading the gradient of the sin θ against m graph gives λ/d, so a straight-line measurement is an alternative way to find either the wavelength or the slit separation — exactly how a diffraction grating is calibrated.

How Interference Works

Superposition, coherence, path difference and the small-angle approximation — everything you need before you trust a number from the calculators.

Superposition — waves add

Where two waves meet, their displacements add. Two crests together make a bigger crest (constructive interference); a crest meeting a trough cancels to nothing (destructive interference). Light behaves the same way: bright fringes are places where two waves always arrive in step, dark fringes where they always arrive exactly out of step.

Coherence is essential

For a steady pattern the two sources must be coherent: the same frequency, and a constant phase difference. Two separate lamps are never coherent, because they emit randomly and the pattern would shift trillions of times a second and blur out. That is why one source is split into two — by two narrow slits, a laser, or a beam splitter.

Path difference decides everything

The only thing that matters at a point on the screen is the difference in the distances travelled from the two slits: the path difference Δ. If Δ is a whole number of wavelengths the waves arrive in step (bright); if it is a half number of wavelengths they arrive in antiphase (dark).

Why Δ = d sin θ

Drop a perpendicular from the upper slit onto the ray leaving the lower slit. For a screen far away the two rays leave almost parallel, so the extra distance the lower ray travels is the slit separation d multiplied by sin θ, where θ is the angle the rays make with the straight-through direction. Hence Δ = d sin θ.

The small-angle approximation

Real patterns are observed within a few degrees of the centre, where sin θ ≈ tan θ ≈ θ (in radians). That turns Δ = d sin θ into y = mλD/d, so consecutive bright fringes are equally spaced by w = λD/d. At larger angles the exact position is y = D tan θ and the fringes slowly get further apart.

Real slits have width

Each slit is not a point, so each one also spreads its own diffraction pattern. The two-slit pattern is therefore multiplied by the single-slit envelope, which kills the fringes that fall on a diffraction minimum — the missing orders. Narrower slits widen the envelope and let more fringes show.

THE FOUR QUANTITIES IN THE DIAGRAM

▸d is the slit separation (centre to centre) and a is the width of each slit. They are different quantities and they control different things.
▸θ is the angle between the direction to the point on the screen and the straight-through direction.
▸y is the distance along the screen from the central maximum. The central maximum is the bright fringe at y = 0.
▸D is the distance from the barrier to the screen. It must be large compared with d for the far-field picture to hold — that is why D is measured in metres and d in millimetres.

Bright: Δ = mλ for m = 0, ±1, ±2 … The two waves have travelled a whole number of wavelengths more (or less) and arrive exactly in step.

Dark: Δ = (m + ½)λ. The waves arrive half a cycle out of step, so a crest from one slit always meets a trough from the other and the resultant is zero for equal sources.

Fringe spacing: w = λD/d. Widening the screen distance or shortening the wavelength spreads the fringes apart; separating the slits squeezes them together.

Interference is the strongest experimental evidence that light is a wave: two beams of light can be arranged to give darkness, which no particle model explains. Young's experiment (1801) measured wavelengths of visible light for the first time, and the same idea now underpins interferometers, optical flats, anti-reflective coatings and the gravitational-wave detectors that measure displacements smaller than a proton.

Slit Width & the Diffraction Envelope

Each slit diffracts as well as interferes. The single-slit envelope is what removes whole orders from a real pattern and limits how many fringes you can see.

Set the slit geometry

Slit separation d (mm)

Slit width a (mm)

Ratio d / aDecides everything below

Bright fringes in the central diffraction maximum

Missing orders m = p · d / a

-10, -5, 5, 10

HOW WIDER SLITS CHANGE THE PATTERN

Slit width	d / a	Fringes in central maximum	First missing orders
a = d / 2 = 0.25 mm	2	3	-40, -38
a = d / 3 = 0.1667 mm	3	5	-60, -57
a = d / 4 = 0.125 mm	4	7	-80, -76
a = d / 5 = 0.1 mm	5	9	-100, -95
a = d / 8 = 0.0625 mm	8	15	-160, -152
a = d / 10 = 0.05 mm	10	19	-200, -190

▸Each single slit produces its own diffraction pattern, zero at sin θ = pλ/a for p = 1, 2, 3 …
▸Two-slit fringes disappear where an interference maximum falls on a diffraction minimum. That happens when mλ/d = pλ/a, so the missing orders are m = p·d/a.
▸If d/a = 5, orders 5, 10, 15 … are missing and only 2(5) − 1 = 9 bright fringes sit inside the central diffraction maximum.
▸Making the slits narrower increases a's denominator, widens the envelope, and shows more fringes — but each fringe becomes dimmer, because less light gets through.

In the virtual laboratory you can see this happening: turn the slit width up and the outer fringes fade and then vanish one by one, while the fringe spacing never changes. The spacing is a property of d; the brightness envelope is a property of a.

Worked AICE-Style Examples

Six questions worked in full: state the relationship, substitute with the units converted, evaluate, then comment on the physics.

Fringe spacing from a known wavelength

WE1

A helium-neon laser of wavelength 632.8 nm illuminates two slits 0.250 mm apart. The screen is 1.80 m away. Find the fringe spacing.

WHAT WE KNOW

λ = 632.8 nm = 632.8 × 10⁻⁹ m
d = 0.250 mm = 0.250 × 10⁻³ m
D = 1.80 m
w = ?

WORKING

1w = λD / d
2w = (632.8 × 10⁻⁹ m × 1.80 m) / (0.250 × 10⁻³ m)
3w = (1.139 × 10⁻⁶ m²) / (2.50 × 10⁻⁴ m)
4w = 4.556 × 10⁻³ m
5w = 4.56 mm

The bright fringes are 4.56 mm apart.

Convert both the nanometres and the millimetres to metres before dividing. The spacing is uniform, so it describes every adjacent pair of bright fringes.

Measuring the wavelength from the pattern

WE2

The distance between the centres of the first and the eleventh bright fringe is measured as 20.0 mm. The slits are 0.500 mm apart and the screen is 2.00 m from the slits. Calculate the wavelength of the light.

WHAT WE KNOW

Distance across 10 spacings = 20.0 mm
d = 0.500 mm = 5.00 × 10⁻⁴ m
D = 2.00 m
λ = ?

WORKING

110 spacings span 20.0 mm, so one spacing w = 20.0 mm / 10 = 2.00 mm = 2.00 × 10⁻³ m
2w = λD / d, so λ = wd / D
3λ = (2.00 × 10⁻³ m × 5.00 × 10⁻⁴ m) / 2.00 m
4λ = 1.00 × 10⁻⁶ m² / 2.00 m
5λ = 5.00 × 10⁻⁷ m = 500 nm

The wavelength is 500 nm — blue-green light.

Counting matters: the first to the eleventh fringe is ten spacings, not eleven. Measuring across many fringes and dividing reduces the percentage uncertainty in the measurement.

Angle and position of a chosen order

WE3

Light of wavelength 600 nm falls on slits 0.800 mm apart, and the screen is 2.50 m away. Find the angle of the second-order maximum and its distance from the centre.

WHAT WE KNOW

λ = 600 nm = 6.00 × 10⁻⁷ m
d = 0.800 mm = 8.00 × 10⁻⁴ m
D = 2.50 m
m = 2

WORKING

1sin θ = mλ / d
2sin θ = (2 × 6.00 × 10⁻⁷ m) / (8.00 × 10⁻⁴ m) = 1.50 × 10⁻³
3θ = arcsin(1.50 × 10⁻³) = 0.0859° (1.50 × 10⁻³ rad)
4y = D tan θ = 2.50 m × tan(0.0859°)
5y = 3.75 × 10⁻³ m = 3.75 mm

The second-order maximum lies 3.75 mm from the centre, at 0.0859°.

Check the small-angle route too: y = mλD/d = (2 × 6.00 × 10⁻⁷ × 2.50) / (8.00 × 10⁻⁴) = 3.75 mm. At this angle the two agree to every digit you would write down.

Missing orders and fringes in the central maximum

WE4

A double slit has a separation of 0.400 mm and each slit is 0.100 mm wide. State which orders are missing and how many bright fringes appear inside the central diffraction maximum.

WHAT WE KNOW

d = 0.400 mm
a = 0.100 mm
d / a = 4

WORKING

1Interference maxima occur where d sin θ = mλ; diffraction minima where a sin θ = pλ
2Fringes are missing when both hold at once: m = p (d / a) = 4p
3So the missing orders are m = ±4, ±8, ±12 …
4The central diffraction maximum spans the first diffraction minima at sin θ = ±λ/a
5Inside it, |m| < d / a = 4, so m = 0, ±1, ±2, ±3 → 2(4) − 1 = 7 fringes

Orders 4, 8, 12 … are missing, and 7 bright fringes sit inside the central maximum.

The seventh-order fringe is not simply missing in the middle of a smooth pattern: it falls exactly on a diffraction minimum, so it is absent altogether while its neighbours are still visible.

Finding an unknown slit separation

WE5

Sodium light of wavelength 589 nm produces fringes 1.20 mm apart on a screen 1.50 m from the slits. Calculate the slit separation.

WHAT WE KNOW

λ = 589 nm = 5.89 × 10⁻⁷ m
w = 1.20 mm = 1.20 × 10⁻³ m
D = 1.50 m
d = ?

WORKING

1w = λD / d, so d = λD / w
2d = (5.89 × 10⁻⁷ m × 1.50 m) / (1.20 × 10⁻³ m)
3d = 8.835 × 10⁻⁷ m² / 1.20 × 10⁻³ m
4d = 7.36 × 10⁻⁴ m = 0.736 mm

The slits are 0.736 mm apart.

Notice the inverse relationship: a smaller spacing gives wider fringes. Doubling the slit separation would halve the fringe spacing.

Intensity away from a bright fringe

WE6

Two coherent sources of equal intensity interfere. Find the intensity at a point where the path difference is one third of a wavelength, as a fraction of the intensity at a bright fringe.

WHAT WE KNOW

Δ = λ / 3
I_peak = 4I₀ at a bright fringe
Equal source intensities

WORKING

1Path difference in wavelengths: Δ / λ = 0.3333
2Phase difference: φ = 360° × 0.3333 = 120°
3I = I_peak cos²(φ / 2) = I_peak cos²(60°)
4cos 60° = 0.500, so cos² 60° = 0.250
5I = 0.250 I_peak = 25.0% of the maximum

The intensity is one quarter of the peak intensity of a bright fringe.

Intensity depends on the square of the cosine — the pattern is not linear between a bright and a dark fringe. The average intensity across the whole pattern is 2I₀, half the peak value.

Common Student Mistakes

The errors that cost marks most often in this topic, and the habit that prevents each one.

Mixing nanometres, millimetres and metres

w = λD/d only works if every length is in the same unit — convert to metres first. λ = 600 nm is 600 × 10⁻⁹ m, and d = 0.50 mm is 0.50 × 10⁻³ m.

Write the conversion on its own line before substituting, exactly as the worked examples do.

Swapping d and a

d is the separation between the centres of the two slits; a is the width of each slit. Fringe spacing depends on d only. The slit width a controls the diffraction envelope that decides which fringes are visible.

Label the diagram with both before you substitute anything.

Confusing fringe spacing with fringe position

If w = 2.4 mm, the first bright fringe is 2.4 mm from the centre and the second is 4.8 mm — not 2.4 mm. Spacing is the gap between neighbours; position is measured from the centre.

Use y = mw for a position and w for a spacing, and check which one the question asks for.

Counting fringes wrongly

The distance from the first to the sixth bright fringe spans five spacings, not six. Dividing by the higher number makes the wavelength come out too small.

Number of spacings = (final order − initial order). For the first to the sixth, that is 5.

Using the wrong condition

Bright fringes need Δ = mλ; dark fringes need Δ = (m + ½)λ. Reversing them gives a pattern that is exactly half a fringe out of step and answers that are wrong by a factor of two in the fractional part.

Ask what the two waves are doing where they meet: adding (whole λ) or cancelling (half λ).

Mixing degrees and radians

The small-angle approximation sin θ ≈ tan θ ≈ θ only holds when θ is in radians. Using degrees in the approximation gives answers that are wrong by a factor of about 57.

Either convert to radians, or use sin θ = mλ/d and y = D tan θ exactly — both are always safe.

AICE Practice Challenges

Thirty original questions in three levels, with immediate feedback that names the misconception and a full explanation behind every answer.

Direct substitution into w = λD/d and the bright/dark conditions.

ANSWERED

0 / 10

CORRECT

SCORE

QUESTION 1 OF 10press 1–4 · Enter · S

Light of wavelength 600 nm passes through slits 0.500 mm apart, and the screen is 2.00 m away. What is the fringe spacing?

Conventions used throughout: wavelengths in nanometres, slit separations and slit widths in millimetres, slit-to-screen distances in metres and screen positions in millimetres, converting to SI internally. θ is measured from the straight-through direction, path difference is Δ = d sin θ, bright fringes need Δ = mλ and dark fringes need Δ = (m + ½)λ. Angles and positions are exact (θ = arctan(y/D)); the small-angle result mλD/d is shown alongside for comparison. Sources are treated as coherent, equally bright and ideal, and diffraction is described by the single-slit envelope sinc²(πa sin θ/λ).

Virtual Double-Slit Laboratory

Virtual Double-Slit Laboratory

The Pattern on the Screen

Path Difference Geometry

Fringe Spacing Calculator

Interference & Phase Calculator

Bright & Dark Conditions

Graphs & Relationships

How Interference Works

Superposition — waves add

Coherence is essential

Path difference decides everything

Why Δ = d sin θ

The small-angle approximation

Real slits have width

Slit Width & the Diffraction Envelope

Worked AICE-Style Examples

Fringe spacing from a known wavelength

Measuring the wavelength from the pattern

Angle and position of a chosen order

Missing orders and fringes in the central maximum

Finding an unknown slit separation

Intensity away from a bright fringe

Common Student Mistakes

Mixing nanometres, millimetres and metres

Swapping d and a

Confusing fringe spacing with fringe position

Counting fringes wrongly

Using the wrong condition

Mixing degrees and radians

AICE Practice Challenges

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