
Optional unit · institution-dependent · Interactive lesson
Single-Slit Diffraction
The slit is the whole experiment. Cut it into Huygens wavelets, add them with their phases, and the width of the opening decides where the light goes — including the fact that shrinking the opening widens the beam.
UPII-14.04 · Lesson 4 of 8 in this unit. Position is not a completion score.Optional unit · institution-dependent. Your university may teach this optics or introductory modern-physics material separately. Follow your institution’s required sequence. Optional work is not a required completion gate.
Build the model
A wide slit lets light through; a narrow one throws it sideways.
The key idea
Single-slit diffraction is the pattern formed when light passing through one aperture of width a interferes with itself, because wavelets leaving different parts of the opening travel different distances to the same point.
Example. A 0.120 mm slit lit by a 633 nm laser puts its first dark fringe at y₁ = λL/a = 13.2 mm on a screen 2.50 m away, so the bright central band is 26.4 mm across — far wider than the slit that made it.
The complete physical model and assumptions
A slit is not a source point, it is a continuum of them. Huygens's construction says every element of the wavefront filling the aperture relaunches the wave, so what reaches a distant point at angle θ is a sum of contributions whose phases run smoothly from one edge of the opening to the other. The entire spread of phase is fixed by the edge-to-edge path difference a sinθ, and the pattern depends on nothing else. When that spread reaches a whole wavelength the wavelets can be paired off — each one in the top half against its partner a/2 below, half a wavelength out of step — and the aperture extinguishes itself, so a sinθ = mλ marks the dark directions, not the bright ones. Adding the wavelets as phasors rather than pairing them returns the whole curve, I = I₀ (sin α/α)² with α = πa sinθ/λ: one tall central band of angular half-width λ/a, flanked by lobes under 5% of the peak. The content of the model is that inverse. Open the aperture and the light collapses onto the geometric shadow of the slit; close it toward a wavelength and the light floods the forward half-space. It stops working when the screen is too close for the rays to count as parallel, when a < λ so no minimum exists at all, and when the illumination is not the uniform, coherent, normally incident wave the integral assumed.
1The aperture is a continuum of sources
Huygens's construction replaces a wavefront by sources sitting on it: every element of the wave filling the slit relaunches it, and what arrives at a distant point is the sum of all of them. Label a strip of width dy lying a distance y below the top edge of a slit of width a. Light from that strip travels y sinθ further than light from the top edge to reach a screen point at angle θ, so it arrives with a phase lag 2πy sinθ/λ. The lag runs smoothly from 0 at one edge to 2πa sinθ/λ at the other, and it is worth naming half of that total spread α = πa sinθ/λ now, because every result below is written in it. None of this is new physics — it is the same superposition that gives two slits their fringes — but the sum is an integral over a continuum instead of the addition of two terms, and that one change replaces a row of equal fringes with an envelope that has a bright middle.
2A whole wavelength across the slit means darkness
Take the direction in which the edge-to-edge path difference is exactly one wavelength, a sinθ = λ, and split the slit down the middle. Every strip in the top half has a partner exactly a/2 below it, and the extra path between the members of a pair is (a/2) sinθ = λ/2. Half a wavelength: the pair cancels. The strips pair off one to one, so the aperture extinguishes itself completely. Nothing about that first λ was special. Split the slit into four equal strips and pair each with the one a/4 away, and the light dies whenever (a/4) sinθ = λ/2, that is a sinθ = 2λ. Repeating with 2m strips gives the general condition a sinθ = mλ with m = ±1, ±2, ±3 and so on. The argument needs at least two strips, so m = 0 is simply not available: on the axis every wavelet travels the same distance and they all add, which is why the centre of the pattern is its brightest point.
3Adding the wavelets: the phasor arc
Pairing locates the zeros but says nothing about what lies between them. For that, add the contributions with their phases. Each strip dy contributes an equal amplitude with a phase that advances linearly in y, so the resultant is ∫₀a cos(ωt − 2πy sinθ/λ) dy. Drawn as phasors — equal arrows, each turned a little further than the last — the sum is an arc of a circle. Its arc length is fixed at the on-axis amplitude E₀, and the resultant is the chord. An arc of length E₀ subtending an angle 2α has a chord E₀ sin α/α, and intensity goes as amplitude squared, so I(θ) = I₀ (sin α/α)² with α = πa sinθ/λ. Both earlier results fall out of it. As α → 0 the arc straightens, sin α/α → 1 and I → I₀. Whenever α = mπ the arc closes into a whole number of complete circles, the chord vanishes, and a sinθ = mλ reappears as the zeros of the curve.
4How wide the band is, and what is under the lobes
The central maximum runs from the m = −1 minimum to the m = +1 minimum, so its angular half-width is θ₁ = arcsin(λ/a). On a screen at distance L, with angles small enough that sinθ ≈ tanθ, it is w = 2λL/a across — exactly twice the spacing of every fringe that follows it, which is the quickest way to recognise a single-slit pattern by eye. Beyond it the lobes are not equal. They peak where the derivative of (sin α/α)² vanishes, that is where tan α = α, at α = 4.4934, 7.7253 and 10.9041 rad, or 1.4303π, 2.4590π and 3.4709π — always a little inside the half-integer guess. Their heights are 4.72%, 1.65% and 0.83% of I₀. In practice the pattern is one bright band with a faint fringed skirt, which is why a single slit is measured from its dark fringes rather than its bright ones.
5Narrow the aperture and the beam gets wider
Slit width enters every result so far only through λ/a, so width and wavelength act inversely. With a 633 nm laser and a screen 2.00 m away, a 0.50 mm slit puts its first minimum 2.53 mm off axis, while a 0.050 mm slit — ten times narrower — puts it at 25.3 mm. Restricting light makes it spread. Push the trend both ways to find the edges of the model. For a ≫ λ the pattern shrinks onto the geometric image of the slit: a 1.0 mm slit gives θ₁ = 0.036°, which is why ray optics is safe for everyday openings. For a < λ the condition sinθ₁ = λ/a asks for a sine greater than one, so no minimum exists in any direction — a sub-wavelength slit produces no pattern at all, just a single Huygens wavelet flooding the forward half-space. Narrowing a slit past a wavelength does not give a narrower beam; it stops giving a beam.
6Far field: when parallel rays are allowed
The path difference y sinθ assumed that all the rays heading for one screen point leave the slit parallel. They do not, quite: the point sits at a finite distance L, and the exact path carries a correction of order a²/(8L) from the curvature of the wavefront across the aperture. Comparing that correction with λ gives the Fraunhofer condition, usually quoted through the Fresnel number F = a²/(λL) ≪ 1. For a = 0.10 mm at 633 nm, a²/λ = 1.6 cm, so a screen at 2.00 m has F = 0.0079 and the far field is reached with room to spare, while a screen 1 cm away, with F = 1.6, would show a quite different structured near-field pattern. A lens after the slit buys the same thing at any distance, since it brings genuinely parallel rays to one point of its focal plane. Once the condition holds, sinθₘ = mλ/a is exact, so measured sinθₘ plotted against m is a straight line through the origin of slope λ/a.
Key equations
Read the units, assumptions and limits before substituting.
Phase across the aperture
α = π a sinθ / λ
One variable carries the whole pattern: everything below is a statement about α.
a slit width (m), θ measured from the forward axis, λ wavelength (m); α in rad is half the edge-to-edge phase spreadDiffraction minima
a sinθ = mλ, m = ±1, ±2, ±3, …
The zeros are sharp, so these are the angles a slit width is measured from.
m = 0 is excluded: θ = 0 is the central maximum, not a zeroFraunhofer intensity
I(θ) = I₀ (sin α / α)²
Turns the pattern into one continuous curve instead of a list of fringes.
I₀ the on-axis intensity in W m⁻²; sin α/α → 1 as α → 0, so I(0) = I₀Width of the central maximum
w = 2λL / a
The inverse law made visible: halve the slit and the bright band doubles.
L slit-to-screen distance (m), w the full width between the first minima (m); assumes sinθ ≈ tanθSecondary maxima
tan α = α at α ≈ 1.4303π, 2.4590π, 3.4709π
Their heights are 4.72%, 1.65% and 0.83% of I₀ — the envelope falls fast.
each sits just inside the half-integer guess a sinθ = (m + ½)λFar-field (Fraunhofer) condition
F = a² / (λL) ≪ 1
Below F ≈ 0.1 the parallel-ray geometry behind a sinθ = mλ is safe to use.
F the dimensionless Fresnel number; a lens makes its own focal plane the far field at any LInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Halve the slit width and watch the central band double while the zeros march outward; then hold the width fixed and slide the wavelength to see red spread further than blue.
Live results
- FIRST MINIMUM y₁
- 12.60 mm
- CENTRAL WIDTH 2λL/a
- 25.20 mm
- ANGLE θ₁ = arcsin(λ/a)
- 0.361 °
- FRESNEL NUMBER a²/(λL)
- 0.0079
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Slit width a: 0.10 mm; Wavelength λ: 630 nm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The condition a sinθ = mλ looks exactly like the double-slit maximum condition d sinθ = mλ, and it gets read the same way — as bright directions. It gives the dark ones. The difference is what the two lengths measure. In a two-slit pattern d separates two sources, so a path difference of mλ puts those two sources back in step. In a single slit a is the distance between the aperture's edges, and the edges are only the extremes of a continuum: when the edge-to-edge difference is a full λ, the phases inside sweep once around the circle, every wavelet finds a partner a/2 away that is λ/2 out of step, and the sum is zero. The same misreading also invents an m = 0 minimum on the axis. Put α = 0 into sin α/α and the limit is 1, not 0 — θ = 0 is where every wavelet arrives in phase, the brightest point on the screen.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Sodium light of wavelength 589 nm falls normally on a single slit 0.0850 mm wide, with a screen 1.60 m beyond the slit. Find the angle of the first minimum and the width of the central bright band on the screen.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- The first minimum is m = 1 in a sinθ = mλ, so sinθ₁ = λ/a = 589 × 10⁻⁹ m ÷ 8.50 × 10⁻⁵ m = 6.93 × 10⁻³.
- θ₁ = arcsin(6.93 × 10⁻³) = 0.397°, small enough that sinθ and tanθ agree to better than one part in 10⁴.
- Convert angle to position on the screen: y₁ = L tanθ₁ ≈ Lλ/a = 1.60 m × 6.93 × 10⁻³ = 1.109 × 10⁻² m.
- The band runs from the minimum on one side to the minimum on the other, so w = 2y₁ = 2.22 × 10⁻² m.
Answer. θ₁ = 0.397°, and the central maximum is 22.2 mm wide, with y₁ = 11.1 mm either side of the axis.
MediumExample 2
A 0.150 mm slit is lit at normal incidence by 546 nm mercury light, the screen 2.00 m away. (a) Locate the first minimum. (b) Find the intensity at y = 10.0 mm as a fraction of I₀. (c) The first secondary maximum peaks at α = 4.4934 rad; how far from the axis is its centre?
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- (a) y₁ = λL/a = (546 × 10⁻⁹ m × 2.00 m) ÷ 1.50 × 10⁻⁴ m = 7.28 × 10⁻³ m = 7.28 mm, so the point at 10.0 mm lies beyond the first zero.
- (b) At y = 10.0 mm, sinθ ≈ y/L = 5.00 × 10⁻³, so α = πa sinθ/λ = π × 1.50 × 10⁻⁴ × 5.00 × 10⁻³ ÷ 5.46 × 10⁻⁷ = 4.315 rad, which is 1.374π.
- I/I₀ = (sin α / α)² = (−0.9222 ÷ 4.315)² = (−0.2137)² = 0.0457, about 4.6% of the central intensity.
- (c) The lobe peaks where tan α = α: sinθ = αλ/(πa) = 4.4934 × 5.46 × 10⁻⁷ ÷ (π × 1.50 × 10⁻⁴) = 5.206 × 10⁻³, so y = L sinθ = 10.41 mm.
- The half-integer guess a sinθ = 1.5λ would have put that peak at 10.92 mm; the true lobe is pulled inward because the 1/α² envelope is still falling across it.
Answer. y₁ = 7.28 mm; I/I₀ = 0.046 at y = 10.0 mm; the first secondary maximum is centred at y = 10.4 mm.
HardExample 3
A student records the first four minima of a single-slit pattern made with a 632.8 nm laser on a screen 2.40 m away: y = 15.1, 30.4, 45.3 and 60.6 mm. Find the slit width from the whole data set, test whether the far-field condition behind a sinθ = mλ was met, and say how close the screen could have come before it failed.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- In the far field at small angles yₘ = mλL/a, a straight line through the origin of slope λL/a. Fit all four points, not one: slope = Σ m yₘ ÷ Σ m² = 454.2 mm ÷ 30 = 15.14 mm per order.
- a = λL ÷ slope = (632.8 × 10⁻⁹ m × 2.40 m) ÷ 1.514 × 10⁻² m = 1.003 × 10⁻⁴ m.
- Check the small-angle step the linear fit assumed: sinθ₄ = 4λ/a = 2.523 × 10⁻² and tanθ₄ = 2.524 × 10⁻², a difference of 0.03%, far below the reading precision.
- Far-field test: F = a²/(λL) = (1.003 × 10⁻⁴ m)² ÷ (632.8 × 10⁻⁹ m × 2.40 m) = 6.6 × 10⁻³, far below the F ≈ 0.1 limit, so the parallel-ray geometry holds.
- F climbs to that limit at L = 10a²/λ = 10 × (1.003 × 10⁻⁴ m)² ÷ (632.8 × 10⁻⁹ m) = 0.159 m, so the screen would have had to come inside about 16 cm of the slit before the far-field description failed.
Answer. a = 1.00 × 10⁻⁴ m (0.100 mm); F = 6.6 × 10⁻³, and the far-field description only fails within roughly 16 cm of the slit.
Quick check
Select an answer and explain it before checking.
Review & continue
Single-slit diffraction is the pattern formed when light passing through one aperture of width a interferes with itself, because wavelets leaving different parts of the opening travel different distances to the same point.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.