Physics 9702 · Paper 2

The Double-Slit Experiment

Shine light through two very small slits. The screen does not show one bright patch. It shows a row of bright and dark stripes called .
In a dark stripe, light from the two slits meets and adds to zero. Only a wave can do that, so this experiment shows that light is a wave.

How the fringes form

Each slit acts as a source of light spreading out by diffraction. Past the slits, the two spreading beams overlap.
You already met path difference in the earlier interference pages. This is the very same idea, now with two slits.
A bright fringe is where the two paths differ by a whole number of wavelengths. The waves arrive in step and add: constructive interference.
A dark fringe is where the two paths differ by half a wavelength. (Or one and a half, two and a half, and so on.) The waves arrive out of step and cancel: destructive interference.

A row of bright fringes with dark gaps between them on a dark screen, brightest in the middle — What lands on the screen: bright fringes with dark gaps between them, brightest in the middle.

A laser shining through a double slit onto a screen — The setup: a laser lights two tiny slits, the spread-out beams overlap, and a row of bright and dark fringes appears on the screen.

Keeping the slits in step

For clear fringes the two slits must stay in step with each other — we say they are . Coherent means they keep a constant phase difference.
To fix this, put a single slit before the double slit (or just use a laser). One source now feeds both slits, so they can never drift out of step.
A laser also gives light — one single wavelength — which makes the fringes sharp and easy to see.

Two slits, each diffracting, overlap and interfere: whole wavelengths of path difference give bright fringes, half wavelengths give dark ones. A single slit (or a laser) first keeps the two slits coherent.

Your turn— tap to reveal the worked answer (9702/21/M/J/25 Q4)

State the principle of superposition, then say why bright and dark fringes appear.

Answer: where two waves meet, the resultant displacement is the sum of the two displacements. At a bright fringe the paths differ by a whole number of wavelengths, so the waves add (constructive); at a dark fringe they differ by a half wavelength, so they cancel (destructive). (9702/21/M/J/25 Q4)

Measuring the wavelength: λ = ax/D

A wavelength of light is tiny, well under a thousandth of a millimetre. No ruler can measure it directly. But the fringes are a millimetre or two apart, which you can measure — and one formula turns those ruler measurements back into the wavelength.
Measure three things: the gap between the two slits, the gap between neighbouring bright fringes, and how far the screen is from the slits.
The path each ray travels links the three. When the angle is small, the extra path to the first bright fringe is $ax/D$ . That extra path must equal one wavelength. So $ax/D = \lambda$ .

\lambda = \dfrac{ax}{D}

Symbols

$a$ = slit separation, the gap between the two slits (m)
$x$ = fringe separation, the gap between the middles of neighbouring bright (or dark) fringes (m)
$D$ = distance from the slits to the screen (m)
$\lambda$ = wavelength of the light (m)

This only works when D is much bigger than a and the angles are small. That is how a normal double slit is set up.

Both rays reach the same bright fringe, but the lower ray is one wavelength longer (amber). The small a–x–D triangle gives λ = ax / D.

Worked example

Wavelength from a double-slit pattern

Slit separation $a = 1.0$ mm. Ten fringes span 1.5 cm. The screen is $D = 2.40$ m from the slits. Find the wavelength.

Get one fringe spacing from ten fringes: $x = \dfrac{1.5 \times 10^{-2}}{10} = 1.5 \times 10^{-3}$ m.
Put everything in metres: $a = 1.0 \times 10^{-3}$ m, $x = 1.5 \times 10^{-3}$ m, $D = 2.40$ m.
Substitute: $\lambda = \dfrac{ax}{D} = \dfrac{(1.0 \times 10^{-3})(1.5 \times 10^{-3})}{2.40}$ .
This gives $\lambda = 6.25 \times 10^{-7}$ m, which rounds to $6.3 \times 10^{-7}$ m.

Answer

\lambda = 6.3 \times 10^{-7}\ \text{m} = 630\ \text{nm}

What changes the fringe spacing

Rearrange for the fringe spacing: $x = \dfrac{\lambda D}{a}$ . From this you can read off what makes the fringes wider or narrower.

Change	Fringe spacing x	Why
Move the screen back (bigger D)	wider	x is proportional to D
Slits closer together (smaller a)	wider	x is proportional to 1/a
Longer wavelength (redder light)	wider	x is proportional to λ

In the exam this is usually a ratio: change one quantity and $x$ scales by the same factor. Double $D$ , and the fringe spacing doubles.

Your turn— tap to reveal the worked answer (9702/12/M/J/25 Q31)

The screen is moved from 0.60 m to 1.50 m. The fringe spacing was 1.8 mm. What is it now?

Answer: x is proportional to D, so multiply by the ratio of distances: $x = 1.8 \times \dfrac{1.50}{0.60} = 4.5$ mm. (9702/12/M/J/25 Q31)

Using white light instead of a laser

With white light the central fringe is white — at the centre every colour has zero path difference, so they all peak together. The other fringes show colours (blue nearest the middle, red furthest, because x is proportional to λ), and only a few show before the colours overlap and mix back into white. You rarely need this for the exam; the marks are almost always on the single-wavelength formula above.

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