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Interference and Young's double slit

Two coherent sources reinforce where their path difference is a whole number of wavelengths and cancel where it is an extra half wavelength. Young's double slit turns that into evenly spaced fringes, and the fringe spacing gives the wavelength of the light. The sources must be coherent, or the fringes shift about and wash out.

Young's double-slit experiment

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Year 12AQA 3.3.2.1 CIE 8.3.1, 8.3.2, 8.3.3, 8.3.4 OCR A 4.4.3(a)(i), 4.4.3(c), 4.4.3(d), 4.4.3(e), 4.4.3(f), 4.4.3(g)(i)EDEXCEL 65

Builds on Diffraction and the single slit.

IN THIS TOPIC

Predict constructive or destructive interference from a path difference.
Explain what coherence means, and why a pattern you can see needs it.
Use $w = \lambda D / s$ , keeping straight which of the three lengths is the large one.
Describe the white-light pattern, and give a specific laser safety precaution.

COMMON MISCONCEPTION

Shine two lights on the same spot and you simply get a brighter spot.

Ordinary lamps have no constant phase difference, so no steady fringes form; only coherent sources hold a fixed phase relationship long enough for a stable pattern.

Superposition and path difference

When two waves occupy the same place, their displacements add. Two crests arriving together make a larger crest, constructive interference; a crest arriving with a trough makes cancellation, destructive interference. Whether a given point gets reinforcement or cancellation depends on the path difference, how much further that point sits from one source than from the other.

Watch Finding the path difference, a worked question on the InkPhysics YouTube channel

\text{constructive: path difference} = n\lambda

NOT IN THE AQA DATA BOOKLET: LEARN IT

\text{destructive: path difference} = (n + \tfrac{1}{2})\lambda

NOT IN THE AQA DATA BOOKLET: LEARN IT

Two-source interference, and why slits act as sources (animated figure) — FIG. 1Two panels. The right panel shows diffraction: plane waves meet a gap, and where the gap is much wider than the wavelength they carry on nearly straight with a slight curl at the edges, while a gap comparable to the wavelength gives near-circular wavefronts, which is what lets two narrow slits act as sources. The left panel shows interference: two coherent sources of the same frequency emit circular wavefronts, and along each amber direction the path difference from the two sources is a whole number of wavelengths, so the waves arrive in phase and the reinforcement there is steady in time.

A whole number of wavelengths of path difference delivers the two waves in phase. An extra half wavelength delivers them in antiphase. For the dark places to stay dark, the sources must be coherent, meaning the same frequency and a constant phase difference between them. Two separate lamps emit light in random, unrelated bursts, so their pattern reshuffles billions of times a second and the eye sees an even glow. A laser is the convenient source, its light coherent and effectively monochromatic. The older arrangement puts one lamp behind a single slit and splits that light into two.

Young's double slit

Illuminate two narrow slits, a distance s apart, with coherent light and view a screen a distance D away. Each slit diffracts, the two spreading waves overlap, and the screen shows fringes, bright wherever the path difference is $n\lambda$ and dark in between.

Double-slit geometry: the path difference between the two routes sets where the bright fringes fall, with the construction repeated magnified — FIG. 2The geometry. Rays from the two slits to a bright fringe differ in length by the short highlighted segment, the path difference. The fringe spacing on the screen is w.

The fringe spacing w, centre of one bright fringe to the centre of the next, comes straight from the geometry.

w = \frac{\lambda D}{s}

IN THE AQA DATA BOOKLET

A cosine-squared intensity graph beneath a strip of evenly spaced bright bands: double-slit fringes all have the same brightness and the same spacing w. — FIG. 3Adjacent bright fringes are separated by w, as are adjacent dark fringes. A dark fringe lies halfway between each pair of bright fringes, so the bright-to-dark separation is w/2.

Because w is the same between every pair of neighbours, the fringes come out evenly spaced, and, in the idealised treatment, equally bright. A longer wavelength widens the spacing, and so does a more distant screen or a closer pair of slits. Visible wavelengths are a few hundred nanometres, so with slits a millimetre apart you need D of a metre or more before w is worth measuring.

WORKED EXAMPLE

A laser of wavelength 650 nm illuminates two slits 0.50 mm apart, and fringes form on a screen 3.0 m away. Find the fringe spacing.

Convert to metres first, so $\lambda = 650 \times 10^{− 9}$ m and $s = 0.50 \times 10^{− 3}$ m.

$w = \lambda D / s$ = (650 × 10^{− 9} × 3.0) / (0.50 × 10^{− 3}) = 3.9 × 10^{− 3} m. Fringes 3.9 mm apart are comfortably visible and can be measured with a millimetre rule.

GUIDED PRACTICE

Working backwards to the slits

Fringes of spacing 2.6 mm appear on a screen 2.4 m from a pair of slits lit with 520 nm light. Rearrange w = λD/s for the slit separation and find it.

Show the working

s = λD/w = (5.20 × 10⁻⁷ × 2.4)/(2.6 × 10⁻³) = 4.8 × 10⁻⁴ m, about half a millimetre.

Check it against the shape of the formula. A tiny λ upstairs and a small w downstairs must leave a slit separation of only a fraction of a millimetre. An answer in centimetres means a power of ten has slipped.

INDEPENDENT PRACTICE

Interference you can hear

Two loudspeakers play the same 680 Hz tone in phase; sound travels at 340 m s⁻¹. At a listening point, the paths from the speakers differ by 0.75 m. What is heard there?

Show the working

λ = 340/680 = 0.50 m, so the path difference is 0.75/0.50 = 1.5 wavelengths.

A half-integer number of wavelengths puts the waves in antiphase, so you get destructive interference and a quiet spot.

White light, and a word on safety

Replace the laser with white light and every wavelength draws its own set of fringes with its own spacing. The central fringe, where the path difference is zero for every colour, stays white. The fringes either side are spectra, with blue on the inner edge of each, because the shorter wavelength has the smaller spacing, and red on the outer edge. A few fringes out, the overlapping colours merge and the pattern fades.

Lasers make the experiment easy and also make it the one place in the optics lab where care is compulsory. Never stare into the beam or point it at anyone, keep it away from reflective surfaces, and stand behind the laser while it is on. A school laser is a Class 2 visible one, whose safety case counts on the natural blink and aversion response limiting an accidental glance to a fraction of a second; deliberately staring into the beam or a mirror-like reflection defeats that protection, which is why the rule is absolute.

ASSESSMENT FOCUS

Coherent means constant phase difference and the same frequency. Include the word constant: “in phase” is a stronger claim and is wrong as a definition, since coherent sources need only hold a fixed phase difference.
Convert nm and mm to metres before anything goes into $w = \lambda D/s$ ; mixed units here move the answer by powers of ten.
Know each symbol's place. The two small lengths are w and s, the fringe spacing and the slit separation, and D is the large one. Swapping w and s is the error to watch for, and w landing near a millimetre is the sanity check.
Double-slit fringes are evenly spaced and equally bright. Keep that phrase clear of the single-slit pattern, whose central maximum is wider and brighter than its neighbours.
Asked for a laser safety precaution, name a specific one: do not look along the beam, or clear reflective objects off the bench. “Be careful” names no precaution and describes no hazard.

CHECK YOURSELF

In a double-slit experiment, light of wavelength 630 nm falls on slits 0.45 mm apart, and the screen is 2.0 m away. Calculate the fringe spacing.

Show a hint

Put every length into metres before you substitute.

Show the answer

$\lambda = 630 \times 10^{− 9}$ m, $s = 0.45 \times 10^{− 3}$ m, D = 2.0 m.

$w = \lambda D / s$ = (630 × 10^{− 9} × 2.0) / (0.45 × 10^{− 3}) = 2.8 × 10^{− 3} m, so the bright fringes sit 2.8 mm apart.

Bright where the paths differ by nλ.

Dark where they differ by (n + ½)λ.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the interference and Young's double slit questions page.

9 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Required practical 2: interference: double slit and diffraction grating puts this topic in the lab, and the written papers ask about it.
Trigonometry and resolving vectors is the maths this lesson leans on, worked through step by step.

USED LATER IN

CHECK YOUR PROGRESS

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Predict constructive or destructive interference from a path difference.
Explain what coherence means, and why a pattern you can see needs it.
Use $w = \lambda D / s$ , keeping straight which of the three lengths is the large one.
Describe the white-light pattern, and give a specific laser safety precaution.

Open the full revision checklist to track your progress across the whole unit.

Interference and Young's double slit

Superposition and path difference

Young's double slit

White light, and a word on safety

Question workbook

Question workbook

Answer book