Young's double slit experiment and fringe width
Two narrow slits lit by one source give equally spaced bright and dark fringes. The fringe width is β = λD/d, so it grows with λ and D.
The experiment
In 1801, Thomas Young showed that light is a wave. Light from a monochromatic source falls on a narrow slit S. Two narrow slits S₁ and S₂, very close together, are placed at equal distances from S. S₁ and S₂ receive light from the same wavefront, so they act as coherent sources. The waves spreading from them overlap and interfere. On a screen at a distance D, we see alternate bright and dark bands called interference fringes. The centre O of the screen is equally far from S₁ and S₂, so the path difference there is zero and the central fringe is bright.
Results to remember (no derivation needed)
| Quantity | Expression | Meaning |
|---|---|---|
| nth bright fringe | xₙ = nλD/d | n = 0 is the central bright fringe at O |
| nth dark fringe | xₙ = (n + ½)λD/d | The first dark fringe is at λD/(2d) |
| Fringe width | β = λD/d | Distance between two neighbouring bright (or dark) fringes |
| Angular fringe width | θ = λ/d | Angle one fringe makes at the slits |
| In a liquid of refractive index μ | β′ = β/μ | The wavelength becomes λ/μ |
How the pattern changes
- β ∝ λ: red light gives wider fringes than blue light.
- β ∝ D: moving the screen away from the slits widens the fringes.
- β ∝ 1/d: bringing the slits closer together widens the fringes.
- Dipping the whole set-up in water (μ = 4/3) makes λ shorter, so β becomes 3/4 of its value in air.
- The fringes are equally wide and the pattern is symmetric about the central fringe.
- Extra reading, not in your textbook. White light: the central fringe is white, because the path difference is zero for every colour. Beside it, blue light is the first to go dark, so the nearest fringe looks reddish. After a few coloured fringes, the colours overlap and the screen looks almost white.
How to draw the Young's double slit set-up
- 1On the left, draw a source and a narrow slit S.
- 2To its right, draw a barrier with two slits S₁ and S₂, one above and one below the axis. Mark their separation d.
- 3On the far right, draw the screen. Mark the distance D from the slits to the screen.
- 4Draw arcs (wavefronts) spreading from S, and overlapping arcs from S₁ and S₂.
- 5Mark the centre O on the screen and a point P at height x. Join S₁P and S₂P.
- 6On the screen, draw alternate bright and dark bands. Label central bright fringe at O and β between two bright fringes.
Worked example: fringe width and fringe positions
In Young's experiment, the slits are 0.5 mm apart and the screen is 1.0 m away. Light of wavelength 600 nm is used. Find (a) the fringe width, (b) the distance of the third bright fringe from the centre, (c) the distance of the second dark fringe, (d) the fringe width when the set-up is dipped in water (μ = 4/3).
- Given: d = 0.5 mm = 5 × 10⁻⁴ m, D = 1.0 m, λ = 600 nm = 6 × 10⁻⁷ m.
- (a) β = λD/d = (6 × 10⁻⁷ m × 1.0 m) ÷ (5 × 10⁻⁴ m) = 1.2 × 10⁻³ m = 1.2 mm.
- (b) x₃ = 3β = 3 × 1.2 mm = 3.6 mm.
- (c) The second dark fringe has n = 1: x = (1 + ½)β = 1.5 × 1.2 mm = 1.8 mm.
- (d) In water, β′ = β/μ = 1.2 mm × 3/4 = 0.9 mm.
(a) 1.2 mm; (b) 3.6 mm; (c) 1.8 mm; (d) 0.9 mm
Worked example: two wavelengths at once
Light with two wavelengths, 600 nm and 480 nm, is used in Young's experiment with d = 0.6 mm and D = 1.2 m. Find the least distance from the central fringe where bright fringes of the two wavelengths fall on each other.
- Bright fringes coincide when n₁λ₁ = n₂λ₂.
- n₁ × 600 = n₂ × 480, so n₁/n₂ = 480/600 = 4/5.
- The smallest whole numbers are n₁ = 4 and n₂ = 5.
- x = n₁λ₁D/d = 4 × (6 × 10⁻⁷ m × 1.2 m) ÷ (6 × 10⁻⁴ m) = 4 × 1.2 × 10⁻³ m.
- x = 4.8 × 10⁻³ m = 4.8 mm.
- Check with λ₂: 5 × (4.8 × 10⁻⁷ × 1.2) ÷ (6 × 10⁻⁴) m = 5 × 0.96 mm = 4.8 mm.
4.8 mm from the central fringe
Look at a distant street lamp through the fine weave of a muslin dupatta. You see a neat pattern of bright spots, made by light waves from many tiny gaps adding and cancelling, the same idea as Young's experiment.
2 marks: “In Young's experiment, what happens to the fringe width if (a) the slit separation is halved, (b) the whole set-up is dipped in water?” (a) β = λD/d, so β doubles. (b) λ becomes λ/μ, so β becomes β/μ, that is 3/4 of its value in air.
3 marks: “In a double slit experiment, d = 0.4 mm, D = 1 m and λ = 500 nm. Find the fringe width and the distance of the fourth bright fringe from the centre.” β = λD/d = (5 × 10⁻⁷ × 1) ÷ (4 × 10⁻⁴) m = 1.25 × 10⁻³ m = 1.25 mm; x₄ = 4β = 5.0 mm.
5 marks: “Describe Young's double slit experiment with a labelled diagram. Write the expression for fringe width and state how it depends on λ, D and d.” Diagram with S, S₁, S₂, d, D and screen; slits coherent as they share one wavefront; alternate bright and dark fringes, central one bright; β = λD/d (result only); β grows with λ and D and falls as d grows; all fringes equally spaced.
β = λD/d: a bigger λ or a bigger D widens the fringes; a bigger d squeezes them.