IB Physics HL Topic 3 — Oscillations & Waves Paper 1 & 2 θ = λ/b ~16 min read

Single-Slit Diffraction

Take away one of Young’s two slits. You’d expect a single bright rectangle on the screen — but you get a broad, dazzling band in the middle with faint ghosts either side. One slit, all on its own, interferes with itself. The light from the top of the slit cancels the light from the middle, and the whole pattern falls out of a single quantity: the slit width compared with the wavelength.

📘 What you need to know

A single slit gives a wide, bright central maximum with much dimmer side maxima
Dark fringes have zero intensity; each bright fringe is dimmer than the one inside it
The first minimum is at θ = λ/b (in radians), where b is the slit width
Further minima: θ = nλ/b, and in general b sin θ = nλ
Width of the central maximum = 2λD/b
Longer λ or a narrower slit → more diffraction → wider maxima
In a real double slit, the fringes sit inside a single-slit envelope

The single-slit pattern

Shine monochromatic light through one narrow rectangular slit onto a distant screen. Like the double slit, you get bright fringes (constructive) and dark fringes (destructive) — but the pattern looks quite different.

The central maximum is much wider and brighter than everything else — and wider than any double-slit fringe
Either side sit narrower, much dimmer maxima, fading fast as the order increases
The dark fringes have zero intensity — complete cancellation

The curve is a true intensity plot. The first side maximum reaches only 4.7% of the central peak and the second only 1.6% — the side bands in the strip below have been brightened so you can see them at all.

Why? Huygens and the wavelets

To explain the pattern we treat the slit not as one source but as a row of tiny sources. That’s Huygens’ principle:

Every point on a wavefront acts as a source of secondary wavelets
All the wavelets are coherent, so they interfere with one another

Straight ahead (θ = 0) every wavelet has travelled the same distance, so they all arrive in phase and add. That’s the central maximum.

Where does the first minimum come from?

Here is the clever pairing argument. Split the slit down the middle and pair each point in the top half with the point exactly b/2 below it. Every pair has the same path difference. If that path difference equals λ/2, then every pair cancels, so the whole slit cancels — total darkness.

The ray from the top edge travels an extra (b/2) sin θ compared with the ray from the centre. Make that extra path λ/2 and the two cancel — and so does every other pair across the slit.

Setting the path difference equal to λ/2:

Deriving the first minimum λ/2 = (b/2) sin θ λ = b sin θ

Because the screen is far away (D >> b), the angle θ is tiny, and we can use the small-angle approximation sin θ ≈ θ. That gives the equation you must know:

Angle of the first minimum θ = λ / b

Where θ is the angle of diffraction of the first minimum in radians, λ is the wavelength (m) and b is the slit width (m). Pairing up quarters, sixths and so on gives all the other minima:

All the minima θ = nλ / b (small angles) b sin θ = nλ (always), n = 1, 2, 3…

Watch the n here. For the double slit, nλ meant a bright fringe. For the single slit, nλ = b sin θ gives the dark fringes. Same-looking equation, opposite meaning. Say to yourself: “double slit, nλ is bright; single slit, nλ is dark.”

Width of the central maximum

The central maximum stretches from the first minimum on one side to the first minimum on the other. If y is the distance from the centre of the screen to the first minimum, then for small angles tan θ ≈ θ ≈ y/D, so y ≈ Dθ. The full width is 2y:

Width of the central maximum width = 2y = 2Dθ width = 2λD / b

Here D is the slit-to-screen distance. Notice the central maximum is twice as wide as every other bright fringe, which run from one minimum to the next.

Small-angle approximation: sin θ ≈ tan θ ≈ θ only works with θ in radians, and only for angles up to about 10° (π/18 rad). Beyond that, go back to b sin θ = nλ.

Changing the wavelength and the slit width

Longer λ
or narrower b

makes

Bigger θ
more diffraction

Wider, dimmer
maxima

Read θ = λ/b as a fraction. λ is on top, so red light diffracts more than blue. b is underneath, so a narrower slit diffracts more.

Both curves are true intensity plots for the same slit. The blue central maximum is narrower in exactly the ratio of the wavelengths — because θ is proportional to λ.

Change	Angle θ	Central maximum	Intensity of maxima
Increase λ (blue → red)	Bigger	Wider	Spread over more area, so dimmer
Narrow the slit (smaller b)	Bigger	Wider	Less light gets through, so dimmer
Widen the slit (bigger b)	Smaller	Narrower	Brighter
Move screen further (bigger D)	Unchanged	Wider	Dimmer

Back to the double slit: the envelope

Real slits have width. So in Young’s experiment two things happen at once:

Light from slit A interferes with light from slit B — the double-slit fringes, evenly spaced, all of equal intensity
Light from one part of a slit interferes with light from another part of the same slit — the single-slit pattern

The single-slit pattern acts as an envelope that modulates the fringes. You still get evenly spaced bright fringes, but their brightness is dialled down by the single-slit curve — which is why the outer fringes look faint, and why some vanish entirely.

Multiply the top two graphs together and you get the bottom one. The fringes stay equally spaced; only their heights change, following the dashed envelope. Fringes landing on an envelope minimum disappear altogether.

This picture assumes the slit width is not negligible, and that the slit separation is much bigger than the slit width (d >> b).

🧮 Using the small-angle approximation safely

Work out λ/b first. If it is well under 0.17, the small-angle approximation is fine.
Use θ = nλ/b and keep θ in radians throughout.
If λ/b is large (slit only a few wavelengths wide), go back to b sin θ = nλ and take an inverse sine.
For a distance on the screen, multiply the angle by D: y = Dθ.

WE 1

Light of wavelength 600 nm passes through a single slit of width 0.10 mm. Calculate the angle of diffraction of the first minimum, and confirm the small-angle approximation is valid.

Step 1 — convert to metres λ = 6.00 × 10⁻⁷ m, b = 1.0 × 10⁻⁴ m Step 2 — use θ = λ/b for the first minimum θ = (6.00 × 10⁻⁷) / (1.0 × 10⁻⁴) θ = 6.0 × 10⁻³ rad Step 3 — check the approximation 6.0 × 10⁻³ rad = 0.34°, far below the 10° limit. A third of a degree. Tiny angles like this are exactly why the small-angle approximation is safe in optics.

WE 2

The screen in WE 1 is placed 2.5 m from the slit. Calculate the width of the central maximum. State what happens to this width if the slit is narrowed to 0.050 mm.

Step 1 — distance from the centre to the first minimum y = Dθ = 2.5 × 6.0 × 10⁻³ = 0.015 m Step 2 — the central maximum spans both sides width = 2y = 2λD/b = 0.030 m width = 3.0 cm Step 3 — halve the slit width b is on the bottom, so halving it doubles the width → 6.0 cm. Narrower slit, wider pattern. Squeeze the light and it fights back by spreading further.

WE 3

The same 600 nm light now passes through a slit of width 1.5 µm. Calculate the angle of the first minimum, and explain why θ = λ/b should not be used here.

Step 1 — check the ratio λ/b (6.00 × 10⁻⁷) / (1.5 × 10⁻⁶) = 0.40 This is nowhere near small, so the approximation fails. Step 2 — use the exact condition b sin θ = nλ with n = 1 sin θ = 0.40 Step 3 — take the inverse sine θ = sin⁻¹(0.40) = 0.41 rad θ = 23.6° The approximation would have given 0.40 rad = 22.9°, an error of nearly 3%. Once the slit is only a few wavelengths wide, the angles stop being small.

💡 Top tips

θ = λ/b gives the first minimum, not a maximum. Say it out loud before you substitute.
Keep θ in radians. The approximation is meaningless in degrees.
The central maximum is twice the width of the other bright fringes.
For a distance on the screen use y = Dθ, then double it for the full central width.
If λ/b is bigger than about 0.17, drop the approximation and use b sin θ = nλ.
Sketch the pattern with a wide, tall centre and small side bumps — never equal-height peaks.

⚠ Common mistakes

Using nλ for a bright fringe — in single-slit diffraction b sin θ = nλ locates the dark ones
Leaving the calculator in degrees when using θ = λ/b
Forgetting the factor of 2 in the central maximum’s width
Drawing the side maxima as tall as the centre, or drawing evenly spaced equal fringes (that’s the double slit)
Confusing b (slit width) with d (slit separation)
Thinking a wider slit gives a wider pattern — it gives a narrower, brighter one

Quick recap: A single slit diffracts light into a wide central maximum flanked by faint side maxima. Huygens’ wavelets, paired across the slit, cancel when the path difference is λ/2, giving b sin θ = nλ for the minima and, for small angles, θ = λ/b. The central maximum has width 2λD/b. Longer λ or narrower b means more spreading — and in a real double slit, the fringes sit inside this pattern as an envelope.

Two slits gave sharpish fringes. One slit gave a broad blob. What if we use thousands of slits? The maxima become razor-thin and blazingly bright, separated by wide darkness — precise enough to measure the wavelength of a star’s light or to split white light into a spectrum. That’s the diffraction grating, and it’s the last page of this sub-section.