Superposition and coherent sources
When two waves meet, the total disturbance at a point is the sum of the two: this is the principle of superposition. If two light waves of amplitude a meet with phase difference φ, the resultant intensity is I = 4I₀ cos²(φ/2).
- Constructive interference: φ = 0, 2π, 4π… (path difference 0, λ, 2λ…) → I = 4I₀ (bright).
- Destructive interference: φ = π, 3π… (path difference λ/2, 3λ/2…) → I = 0 (dark).
For a steady pattern, the sources must be coherent: same frequency and a constant phase difference. Two separate bulbs are not coherent — their phases change randomly millions of times a second, so the fringes shift too fast to see and the screen looks evenly lit (I = I₁ + I₂). That is why Young split light from one source into two slits. A laser gives highly coherent light.
Young's double slit experiment
Thomas Young (1801) let light from a single narrow slit fall on two close, parallel slits S₁ and S₂ (gap d). A screen was kept at a distance D (D ≫ d). He saw equally spaced bright and dark bands called fringes.
Path difference
For a point P at height y above the centre O, the extra distance from S₂ is Δ = S₂P − S₁P ≈ yd/D.
Bright fringes
Δ = nλ → yₙ = nλD/d (n = 0, 1, 2…). The centre O (n = 0) is always bright.
Dark fringes
Δ = (n + ½)λ → yₙ = (n + ½)λD/d.
Fringe width
The distance between two neighbouring bright (or dark) fringes is
β = λD/d
All fringes have the same width. Angular fringe width θ = λ/d.
What changes the fringes?
- Longer λ (red) → wider fringes; blue light gives narrower fringes.
- Screen farther (bigger D) → wider fringes.
- Slits closer (smaller d) → wider fringes.
- Whole set-up in water → λ becomes λ/n, so β becomes β/n.
- White light → central fringe white, the others coloured (violet edge inside, red outside), and only a few fringes are seen.
- Energy is not lost: it is only moved from dark places to bright places.
Single slit diffraction
Diffraction is the bending and spreading of light around the edge of a small opening or obstacle. It is noticeable when the opening is not much bigger than the wavelength.
Shine light on one narrow slit of width a. Think of the slit as many tiny Huygens sources side by side. Straight ahead, they all arrive in step: the central maximum is bright.
First dark band
At an angle where the path difference between the top and bottom edges is λ, split the slit into two halves: each point in the top half cancels a partner in the bottom half. So the first minimum is at a sin θ = λ, or θ ≈ λ/a. Minima: a sin θ = nλ.
Width of the central maximum
From −λ/a to +λ/a, so the angular width is 2λ/a and on the screen
central maximum width = 2λD/a
It is twice as wide as each side band, and the side maxima (near a sin θ ≈ (n + ½)λ) are much fainter. A narrower slit gives a wider central band.
Interference vs diffraction
| Interference | Diffraction (single slit) |
|---|---|
| Waves from two coherent sources overlap | Wavelets from different parts of one slit overlap |
| All fringes have equal width | Central maximum is twice as wide as the others |
| All bright fringes about equally bright | Brightness falls quickly away from the centre |
| Dark fringes are fully dark | Minima are dark, side maxima weak |
In a real double slit, both happen: the interference fringes sit inside the broad diffraction pattern of each slit.
Try it: see diffraction with your fingers
Hold two fingers very close together in front of one eye, leaving a hair-thin gap, and look at a bright tube light through it. You will see thin dark lines running along the gap — a diffraction pattern. Squeeze the gap narrower: the pattern spreads wider (smaller a, bigger 2λD/a). In the 3D, predict what happens to β when you double d, then move the slider to check.
Key formulas and definitions
- I = 4I₀ cos²(φ/2); φ = (2π/λ) × path difference
- Path difference Δ = yd/D
- Bright: Δ = nλ, yₙ = nλD/d
- Dark: Δ = (n + ½)λ, yₙ = (n + ½)λD/d
- Fringe width β = λD/d; angular width = λ/d
- Single slit minima: a sin θ = nλ
- Central maximum width = 2λD/a (angular 2λ/a)
Worked examples
1. In Young's experiment, λ = 600 nm, d = 0.5 mm, D = 1 m. Find the fringe width.
Step 1: β = λD/d. Step 2: = 600 × 10⁻⁹ × 1 / 0.5 × 10⁻³. Step 3: β = 1.2 × 10⁻³ m = 1.2 mm.
2. With the same set-up (β = 1.2 mm), find the distance of the 3rd bright fringe and the 2nd dark fringe from the centre.
Step 1: 3rd bright: y = 3β = 3.6 mm. Step 2: 2nd dark (n = 1): y = (1 + ½)β = 1.5 × 1.2 = 1.8 mm.
3. Fringe width is 2 mm with D = 1.2 m and d = 0.3 mm. Find the wavelength.
Step 1: λ = βd/D. Step 2: = 2 × 10⁻³ × 0.3 × 10⁻³ / 1.2. Step 3: λ = 5 × 10⁻⁷ m = 500 nm.
4. What happens to β = 1.2 mm if (a) D is doubled, (b) d is doubled, (c) the apparatus is put in water (n = 4/3)?
Step 1: β ∝ D, so (a) β = 2.4 mm. Step 2: β ∝ 1/d, so (b) β = 0.6 mm. Step 3: λ becomes λ/n, so (c) β = 1.2 × 3/4 = 0.9 mm.
5. At a point P on the screen, the path difference is 1800 nm for light of λ = 600 nm. Is P bright or dark? What if λ = 720 nm?
Step 1: 1800/600 = 3, a whole number → bright (3rd bright fringe). Step 2: 1800/720 = 2.5 → half-number → dark.
6. Two coherent waves each of intensity I₀ meet with phase difference π/2. Find the resultant intensity.
Step 1: I = 4I₀ cos²(φ/2). Step 2: = 4I₀ cos²(45°) = 4I₀ × ½. Step 3: I = 2I₀.
7. Light of λ = 500 nm falls on a single slit of width 0.2 mm. The screen is 2 m away. Find the width of the central maximum.
Step 1: width = 2λD/a. Step 2: = 2 × 500 × 10⁻⁹ × 2 / 0.2 × 10⁻³. Step 3: = 1 × 10⁻² m = 10 mm.
8. In a single slit experiment, the first minimum for red light (660 nm) is at 30°. Find the slit width.
Step 1: a sin θ = λ. Step 2: a = 660 nm / sin 30° = 660/0.5. Step 3: a = 1320 nm = 1.32 μm.
9. Two wavelengths 650 nm and 520 nm are used together in a double slit (d = 2 mm, D = 1.2 m). Find the least distance from the centre where their bright fringes coincide.
Step 1: n₁ × 650 = n₂ × 520 → n₁/n₂ = 4/5. Step 2: least y = 4 × 650 × 10⁻⁹ × 1.2 / 2 × 10⁻³. Step 3: y = 1.56 × 10⁻³ m = 1.56 mm.
Common mistakes
- Writing β = λd/D. The correct formula is β = λD/d (D on top, d below).
- Forgetting unit conversion: d in mm and λ in nm must be changed to metres.
- Saying the central maximum of a single slit has width λD/a. It is 2λD/a.
- Thinking two separate torches can make stable fringes. They are not coherent.