Young's Double-Slit Experiment
Two slits, one light: bright and dark stripes that measure the wavelength.
Lesson at a glanceShow the key points
- Two slits lit by the same source act as coherent sources; light from them overlaps on a screen and gives bright and dark fringes.
- Bright: δ = 2nπ, Δx = nλ. Dark: δ = (2n + 1)π, Δx = (n + ½)λ (17.5, 17.6). The central point (Δx = 0) is bright.
- Identical slits: I = 2I′(1 + cos δ) = 4I′cos²(δ/2) (17.7), between 0 and 4I′.
- For D ≫ d, Δx ≈ d sin θ ≈ yd/D; bright fringes at y = nDλ/d; dark at (n + ½)Dλ/d.
- Fringe width w = Dλ/d (17.8); measuring D, d and w gives λ. Large d makes the fringes too fine to see.
- Unequal slits: I_max/I_min = (A₁ + A₂)²/(A₁ − A₂)².
In 1801 Thomas Young let sunlight through two pinholes and saw something particles could never produce: stripes of light and dark where the two beams overlapped. Light plus light can make darkness.
Young’s pinholes
Young let sunlight through one pinhole in a card, and the light from it fall on a second card with two pinholes side by side. On a screen beyond, the light from the two holes overlapped and formed a pattern of bright and dark patches. At any point the two waves arrive with different phases and interfere. With white sunlight the pattern was coloured, but Young still estimated the wavelengths of extreme red and violet light surprisingly well.
The double-slit experiment
In the modern version, two long narrow parallel slits S₁ and S₂ replace the pinholes, and they are lit by a parallel beam of monochromatic light. A screen parallel to the slits shows a row of bright and dark stripes called fringes.
At a point P on the screen, the waves have travelled S₁P = x and S₂P = x + Δx. Their fields are
so they differ in phase by
This is exactly the problem of two waves on a string from chapter 15: the result is a wave of amplitude .
How bright?
For identical slits each giving intensity I′ alone, E₀₁ = E₀₂, and since intensity ∝ amplitude²:
Fringe width
Let the slits be d apart and the screen a distance D away, with D ≫ d. Then S₁P and S₂P are almost parallel, at a small angle θ to the centre line, and the extra path is
where y is the distance of P from the centre of the pattern. Bright fringes need yd/D = nλ:
Slits 0.20 mm apart are lit with 500 nm light. How far apart are the bright fringes on a screen 1.2 m away?
With slits 0.40 mm apart and the screen 1.5 m away, the 8th bright fringe is 15 mm from the centre. Find λ.
Why fringes are hard to see with wide slit spacing
w shrinks as d grows. If d is many thousands of wavelengths, the fringes crowd so closely that the screen looks evenly lit. This is a general rule: wave effects are hard to notice when the wavelength is small compared with the openings involved.
Unequal slits
If one slit lets through more light, the dark fringes aren’t completely dark. With amplitudes A₁ and A₂, bright fringes have amplitude A₁ + A₂ and dark ones |A₁ − A₂|, so
One slit is three times as wide as the other, and the amplitude from each slit is proportional to its width. Find I_max/I_min.
- 1Two slits lit by one source give bright fringes where Δx = nλ and dark ones where Δx = (n + ½)λ.
- 2Identical slits: I = 4I′cos²(δ/2).
- 3Δx ≈ yd/D; bright fringes at y = nDλ/d; fringe width w = Dλ/d.
- 4Unequal slits: I_max/I_min = (A₁ + A₂)²/(A₁ − A₂)².
Check yourself
Keep exploring
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