Big picture
Ray optics treats light as straight lines, which explains mirrors and lenses but not everything. When light passes through two narrow slits, it produces a pattern of bright and dark bands; when it passes through one narrow slit, it spreads out. These effects, interference and diffraction, can only be explained if light is a wave.
This chapter uses Huygens' principle to treat light as a wave, proves the laws of reflection and refraction, and then studies interference (Young's double-slit experiment) and diffraction at a single slit. Fringe width calculations are standard exam questions.
Learning goals
Explain wavefronts and Huygens' principle.
Prove the laws of reflection and refraction using Huygens' principle.
Explain coherent sources and the conditions for sustained interference.
Describe Young's double-slit experiment and calculate fringe width and fringe positions.
Explain the intensity distribution in interference.
Explain single-slit diffraction and calculate the width of the central maximum.
Distinguish interference from diffraction.
Core ideas
1. Wavefronts and Huygens' principle
A wavefront is the locus of all points that oscillate in phase. Light travels perpendicular to the wavefront (the ray is the normal to the wavefront).
Source | Wavefront |
|---|---|
Point source | Spherical |
Line source (slit) | Cylindrical |
Very distant source (the Sun) | Plane |
Huygens' principle:
Every point on a wavefront acts as a source of secondary wavelets, which spread out in all directions with the speed of the wave.
The new wavefront at a later time is the forward envelope (common tangent) of these secondary wavelets.
2. Laws of reflection and refraction using Huygens' principle
Refraction. A plane wavefront AB strikes the surface between medium 1 (speed v₁) and medium 2 (speed v₂) at angle i. While the wavelet from B travels a distance BC = v₁t in medium 1, the wavelet from A travels AE = v₂t in medium 2. The refracted wavefront CE makes an angle r with the surface.
From the geometry: sin i = BC/AC = v₁t/AC and sin r = AE/AC = v₂t/AC. Dividing:
sin i ÷ sin r = v₁ ÷ v₂ = n₂ ÷ n₁ (= constant)
This is Snell's law. If light enters a denser medium (v₂ < v₁), then r < i: light bends towards the normal.
Reflection. Using the same method, the incident and reflected wavelets travel equal distances in the same medium, giving angle of incidence = angle of reflection.
What changes in refraction: the speed and wavelength change (λ₂ = λ₁ v₂/v₁, so in a medium of refractive index n, λ = λ_vacuum/n), but the frequency stays the same, because it is set by the source.
3. Coherent sources and interference
Interference is the redistribution of light energy when two waves overlap, giving bright and dark regions.
Coherent sources emit waves of the same frequency with a constant phase difference. Two independent bulbs are not coherent, because their phase changes randomly millions of times a second. Coherent sources are obtained from a single source, as in Young's double slit.
Conditions for sustained interference: the sources must be coherent; they should have the same (or nearly the same) amplitude, for good contrast; they should be narrow and close together.
Conditions at a point (path difference Δ, wavelength λ):
Type | Path difference | Phase difference |
|---|---|---|
Constructive interference (bright) | Δ = nλ (n = 0, 1, 2 …) | 0, 2π, 4π … |
Destructive interference (dark) | Δ = (2n − 1)λ/2 (n = 1, 2, 3 …) | π, 3π, 5π … |
Intensity. If each wave has intensity I₀, the resultant intensity is
I = 4I₀ cos²(φ/2)
Maximum 4I₀ (bright) when φ = 0, 2π …; minimum 0 (dark) when φ = π, 3π …. Energy isn't destroyed; it is redistributed from dark to bright regions.
4. Young's double-slit experiment (YDSE)
Light from a single slit S falls on two narrow slits S₁ and S₂, separated by d. They act as coherent sources, and the interference pattern is observed on a screen at distance D (D much larger than d).
For a point at distance y from the centre of the screen, path difference Δ = yd/D.
Quantity | Formula |
|---|---|
Position of the nth bright fringe | yn = nλD/d |
Position of the nth dark fringe | yn = (2n − 1)λD/2d |
Fringe width (distance between consecutive bright or dark fringes) | β = λD/d |
All fringes have equal width; the central fringe (n = 0) is bright.
Worked example. λ = 600 nm, d = 0.2 mm, D = 1 m.
β = λD/d = (600 × 10^-9 × 1) / (0.2 × 10^-3) = 3 × 10^-3 m = 3 mm
Third bright fringe from the centre: y = 3β = 9 mm
Second dark fringe: y = (2 × 2 - 1) λD/2d = 1.5 β = 4.5 mmEffects of changes on fringe width (β = λD/d):
Change | Effect on β |
|---|---|
Increase D (move screen away) | Increases |
Increase d (slits further apart) | Decreases |
Use longer wavelength (red instead of blue) | Increases |
Immerse the apparatus in water (n = 4/3) | λ becomes λ/n, so β decreases to β/n (3 mm becomes 2.25 mm) |
Use white light | Central fringe is white; other fringes are coloured, with violet nearest the centre |
5. Diffraction at a single slit
Diffraction is the bending of light around the edges of an obstacle or aperture into the region of geometrical shadow. It is noticeable when the size of the slit is comparable to the wavelength.
When parallel light of wavelength λ passes through a single slit of width a:
There is a broad, bright central maximum.
On either side are minima at angles given by a sin θ = nλ (n = 1, 2, 3 …), or θ ≈ nλ/a for small angles.
Secondary maxima lie roughly midway between the minima, at a sin θ ≈ (n + ½)λ, with intensity falling rapidly.
Angular width of the central maximum = 2λ/a. Linear width on a screen at distance D = 2λD/a.
Example. a = 0.1 mm, λ = 500 nm, D = 1 m.
First minimum: θ = λ/a = (500 × 10^-9) / (0.1 × 10^-3) = 5 × 10^-3 rad
Width of central maximum = 2λD/a = 2 × 5 × 10^-3 × 1 = 10^-2 m = 10 mmA narrower slit gives a wider central maximum: the smaller the opening, the more light spreads.
Explanation. Divide the slit into two halves. For a minimum at a sin θ = λ, each point in the upper half has a partner in the lower half whose path differs by λ/2, so their waves cancel in pairs.
6. Interference and diffraction compared
Basis | Interference | Diffraction |
|---|---|---|
Cause | Superposition of waves from two coherent sources | Superposition of wavelets from different parts of the same wavefront |
Fringe width | All fringes have equal width | Central maximum is twice as wide as the others |
Intensity | All bright fringes have equal intensity | Intensity falls rapidly for secondary maxima |
Dark fringes | Perfectly dark | Not perfectly dark |
A reliable problem-solving routine
Convert all lengths to metres (nm = 10⁻⁹ m, mm = 10⁻³ m).
For YDSE, use β = λD/d; nth bright at nβ, nth dark at (2n − 1)β/2.
In a medium, replace λ by λ/n.
For single-slit diffraction, use a sin θ = nλ for minima and width 2λD/a for the central maximum.
For intensity, use I = 4I₀ cos²(φ/2).
Connections that matter
Ray optics treats the same light with rays; Huygens' principle derives its laws.
Electromagnetic waves explains that light is an EM wave.
Dual nature of radiation adds the particle side of light.
Common mistakes
Saying frequency changes on refraction. Only speed and wavelength change.
Writing β = λd/D. It is λD/d.
Using the double-slit formula for single-slit minima. For a single slit, a sin θ = nλ gives minima, not maxima.
Thinking energy is lost in dark fringes. It is redistributed.
Assuming two bulbs can produce sustained interference. They are not coherent.
Revision checklist
I can explain wavefronts and Huygens' principle.
I can prove Snell's law using wavefronts.
I can explain coherent sources and conditions for interference.
I can calculate fringe width and fringe positions in YDSE, including changes.
I can explain single-slit diffraction and calculate the width of the central maximum.
I can distinguish interference from diffraction.