Interference and Young’s experiment

Young’s double slit

Narrated video. Transcript:

In 1801, Thomas Young made two coherent sources of light out of one. Light from a source S falls on two narrow pinholes, S₁ and S₂, very close together in an opaque screen.

If the phase of the light from S changes suddenly, the change reaches S₁ and S₂ in exactly the same way. So the phase difference between S₁ and S₂ stays fixed: they behave as coherent sources.

The spherical waves from S₁ and S₂ overlap. On a screen beyond, they produce bright and dark bands, called interference fringes.

Let the slits be a distance d apart, and the screen a distance D away. Take a point P on the screen, at a distance x from the centre O.

By Pythagoras, S₂P² − S₁P² = [D² + (x + d/2)²] − [D² + (x − d/2)²] = 2xd.

So S₂P − S₁P = 2xd/(S₂P + S₁P). When x and d are much smaller than D, S₂P + S₁P ≈ 2D, and the path difference is simply xd/D.

P is bright when the path difference is a whole number of wavelengths: xd/D = nλ. So the bright fringes are at x = nλD/d, with n = 0, ±1, ±2, and so on.

P is dark when xd/D = (n + ½)λ, so the dark fringes are at x = (n + ½)λD/d. Bright and dark fringes alternate, equally spaced.

The distance between two neighbouring bright fringes, or two neighbouring dark ones, is the fringe width: β = λD/d.

Example: slits 1 mm apart, a screen 1 m away, and blue-green light of wavelength 500 nm. β = λD/d = (5 × 10⁻⁷ × 1)/10⁻³ = 5 × 10⁻⁴ m, just half a millimetre.

Changing the fringes

Narrated video. Transcript:

How do the fringes change when we change the set-up? Start from β = λD/d. The angular separation of the fringes is θ = β/D = λ/d.

Move the screen away from the slits. D increases, so the fringes get wider apart. But their angular separation, λ/d, stays the same.

Use light of a shorter wavelength. β gets smaller: blue light gives narrower fringes than red.

Move the slits further apart. d increases, so β = λD/d decreases, and the fringes crowd together.

The source slit matters too. Let s be its width and S its distance from the double slit. Fringes are seen only if s/S < λ/d. Bring the source closer, or widen it, and the fringes become less sharp, and finally disappear.

Now use white light. Every colour has a bright fringe at the centre, where the path difference is zero. So the central fringe is white.

Move outwards. Where the path difference is λ_b/2, about 200 nm, blue light is dark, so the fringe looks red. Further on, where it is λ_b, blue is bright and red is dark. So the fringe closest to the white centre is red, and the farthest appears blue.

After a few fringes, the patterns of all the colours overlap, and no clear fringes are seen.

Example: slits 0.28 mm apart, screen 1.4 m away. The fourth bright fringe is 1.2 cm from the central one. Since x₄ = 4λD/d, λ = x₄d/4D = (1.2 × 10⁻² × 0.28 × 10⁻³)/(4 × 1.4) = 6 × 10⁻⁷ m, or 600 nm.

Try it: Young’s double-slit lab

Interactive simulation. Things to try:

• The strip is the screen, the curve beside it the intensity. Drag P along the screen: bright where S₂P − S₁P = nλ, dark where it is (n + ½)λ.
• Read β, then double D. Then put D back and double d. Predict each new β from β = λD/d before you look.
• Slide λ from violet to red: the fringes widen. Tick “White light” to see a white centre with coloured fringes.
• Tick “Fill with water”: λ in water is λ/1.33, so the fringes move closer together.

Fringe width

In Young’s experiment the slits are 0.5 mm apart and the screen is 1 m away. Light of wavelength 600 nm is used. What is the fringe width?

1. 0.6 mm
2. 1.2 mm
3. 0.12 mm
4. 1.2 cm

Under water

A Young’s double-slit apparatus giving fringes of width β in air is immersed completely in water (n = 1.33). What happens to the fringes?

1. They disappear
2. They widen to 1.33β
3. They narrow to β/1.33
4. They do not change

Summary

• Path difference at P: S₂P − S₁P = xd/D
• Bright fringes: x = nλD/d
• Fringe width: β = λD/d
• Angular fringe width: θ = λ/d

• Young made two coherent sources from one: both slits are lit by the same source, so their phase difference stays fixed.
• Dark fringes lie halfway between, at x = (n + ½)λD/d. All fringes are equally spaced; β increases with λ and D and decreases with d.
• For fringes to be seen, the source slit must satisfy s/S < λ/d.
• With white light the central fringe is white; the nearest fringes on either side are red, the farther ones blue, and then the colours wash out.