IB Physics HL Topic 3 — Oscillations & Waves Paper 1 & 2 s = λD/d ~13 min read

Young’s Double Slit

In 1801 Thomas Young shone light through two thin slits and got something that made no sense if light were made of particles: not two bright lines, but a whole row of them, evenly spaced, with darkness in between. Light plus light was making darkness. This is the experiment that settled the argument — and the beautiful part is that the fringe spacing lets you measure the wavelength of light with nothing but a ruler.

📘 What you need to know

The experiment

The apparatus is deceptively simple. A monochromatic source (one wavelength) shines on a single slit, which diffracts the light. That single slit then illuminates two narrow slits, A and B, side by side. Because both slits are lit by the same wavefront, whatever the phase does at A it does at B: the two slits act as coherent sources.

Light diffracts again at each slit, the two beams spread out and overlap, and on a distant screen they superpose — giving a pattern of evenly spaced bright and dark fringes.

Young’s double-slit apparatus laser single slit double slit A B beams overlap screen fringes
The single slit guarantees the two slits are coherent. Each slit diffracts, the beams overlap, and the screen records where they add (bright) and where they cancel (dark).
Why the single slit? Because you need the two slits to be coherent, and a lamp’s light is a jumble of random phases. Take one wavefront, split it in two, and the phase difference between the halves is locked forever. A laser is already coherent and monochromatic, which is why modern setups often skip the single slit entirely.

Why fringes appear

Everything on this page follows from the interference conditions you already know. Pick any point on the screen and ask how much further the light from B travelled than the light from A. That’s the path difference.

Bright fringe (maximum) path difference =
Dark fringe (minimum) path difference = (n + ½)λ

Here n = 0, 1, 2, 3… is called the order of the fringe. Right in the middle of the screen the two paths are identical, so the path difference is zero — that’s the central maximum, n = 0. Step sideways and the path difference grows steadily, passing through ½λ (dark), λ (bright, n = 1), 1½λ (dark), and so on.

Because the path difference increases evenly as you move across a distant screen, the bright fringes come out evenly spaced, each one of essentially the same width and brightness.

The double-slit equation

The geometry turns that pattern into a formula worth memorising:

The double-slit equation s = λD / d
The three distances: d, D and s d s Dcentral maximum, n = 0 first order, n = 1 slits screen
Keep the three letters straight: d is between the slits, s is between the fringes, and D is the long gap in between. In a real setup D is metres while d is a fraction of a millimetre — this diagram is nowhere near to scale.
Longer λ
redder light
or bigger D
or smaller d
Bigger s
wider fringes

What makes the fringes wider?

Read the equation like a sentence. λ and D are on top, so increasing either spreads the fringes out. d is underneath, so bringing the slits closer together also spreads them out.

ChangeEffect on sWhy
Increase λ (blue → red)Fringes get widerλ is on the top
Move screen further away (bigger D)Fringes get widerD is on the top
Move slits closer (smaller d)Fringes get widerd is on the bottom
Move slits further apartFringes get narrowerbigger d, same top line
Same slits, same screen — different colourred light, λ = 650 nm sblue light, λ = 450 nm s smaller λ → smaller s
The blue fringe spacing really is 450/650 = 0.69 times the red spacing here — the two patterns are drawn to that exact ratio, because s is proportional to λ.

📐 Measuring the wavelength of light

  1. Don’t measure one fringe. Measure across many — say from the 1st bright fringe to the 11th.
  2. Count the gaps, not the fringes. From the 1st to the 11th is 10 fringe separations.
  3. Divide the total distance by the number of gaps to get s.
  4. Rearrange s = λD/d to give λ = sd/D, and check the answer lands between 400 and 700 nm.
WE 1

Laser light of wavelength 600 nm falls on a double slit with a slit separation of 0.30 mm. The screen is 2.0 m away. Calculate the fringe separation on the screen.

Step 1 — convert everything to metres λ = 6.00 × 10⁻⁷ m, d = 3.0 × 10⁻⁴ m, D = 2.0 m Step 2 — write the double-slit equation s = λD / d Step 3 — substitute s = (6.00 × 10⁻⁷ × 2.0) / (3.0 × 10⁻⁴) s = 4.0 × 10⁻³ m = 4.0 mm A few millimetres — easily measurable with a ruler. That’s the whole point of putting the screen metres away.
WE 2

A student measures 27 mm from the centre of the 1st bright fringe to the centre of the 10th. The slits are 0.40 mm apart and the screen is 1.8 m away. Calculate the wavelength of the light.

Step 1 — count the gaps between the fringes From the 1st to the 10th there are 9 gaps, not 10. s = 27 / 9 = 3.0 mm = 3.0 × 10⁻³ m Step 2 — rearrange s = λD/d for λ λ = s d / D Step 3 — substitute λ = (3.0 × 10⁻³ × 4.0 × 10⁻⁴) / 1.8 λ = 6.7 × 10⁻⁷ m = 670 nm Red light. Dividing by 10 instead of 9 gives 600 nm — wrong, and it’s the single most common slip in this experiment.
WE 3

In WE 1 the red laser (600 nm) gave fringes 4.0 mm apart. The laser is swapped for a blue one of wavelength 450 nm, with the slits and screen unchanged. Calculate the new fringe separation.

Step 1 — D and d are unchanged, so s is proportional to λ s ∝ λ Step 2 — scale the old answer by the ratio of wavelengths s = 4.0 × (450 / 600) s = 3.0 mm Shorter wavelength, tighter fringes. No need to redo the whole calculation — spotting the proportionality saves you a minute in the exam.
Why the fringes are so useful: rearranged, λ = sd/D. Every quantity on the right is something you can measure in a school lab with a ruler and a travelling microscope — and out drops the wavelength of light, a few hundred billionths of a metre.

💡 Top tips

⚠ Common mistakes

Quick recap: Coherent, monochromatic light diffracts at two slits and the beams interfere on a distant screen. Bright fringes where the path difference is , dark where it is (n + ½)λ, with n = 0 the central maximum. The evenly spaced fringes obey s = λD/d, so fringes widen with longer λ, bigger D or smaller d.
We’ve been quietly pretending the slits are infinitely thin. They aren’t. Each slit has a real width, and each one produces its own single-slit diffraction pattern — a broad, bright central blob that fades at the edges. The double-slit fringes are actually sitting inside that envelope, which is why the outer fringes look dimmer than the middle ones. That’s Single-Slit Diffraction, next.

Fringe calculations not sticking?

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