12 min readPathway sheetSection B short answer (3 marks) or long answer (5 marks) proving the law of reflection or Snell's law with Huygens' construction, often with a choice

Reflection and refraction explained with wavefronts

Huygens' wavelets prove both laws: equal speeds give i = r on reflection; a slower second medium gives sin i / sin r = v₁/v₂ on refraction.

Law of reflectioni = r; the incident ray, the reflected ray and the normal lie in one plane

Snell's lawsin i / sin r = v₁/v₂ = μ₂₁, the refractive index of medium 2 with respect to medium 1

Refractive indexμ = c/v, where c = 3 × 10⁸ m s⁻¹ is the speed of light in vacuum

On refractionThe frequency stays the same; speed and wavelength change: λ₁/λ₂ = v₁/v₂

Proof of the law of reflection

A plane wavefront AB falls on a plane reflecting surface MN at an angle i. Light travels at speed v in the medium.

1
End A of the wavefront touches the surface first. End B still has to travel a distance BC to reach the surface at C.
2
Let this take a time τ. Then BC = vτ.
3
In the same time τ, a secondary wavelet from A spreads back into the same medium. Its radius is AE = vτ.
4
From C, draw a tangent CE to this wavelet, touching it at E. CE is the reflected wavefront.
5
In triangles ABC and AEC: AC is common, BC = AE = vτ, and ∠ABC = ∠AEC = 90°.
6
So the two triangles are congruent (RHS rule). Hence ∠BAC = ∠ECA.
7
∠BAC is the angle of incidence i, and ∠ECA is the angle of reflection r.
The angle between a wavefront and the surface equals the angle between the ray and the normal, because both pairs are at 90°.
8★
So, i = r: the angle of incidence equals the angle of reflection.

★ key step

Proof of Snell's law (rarer to denser medium)

A plane wavefront AB falls at an angle i on the surface PP′. Light travels at speed v₁ in medium 1 and v₂ in medium 2, with v₂ < v₁.

1
End A of the wavefront touches the surface first. End B must travel BC in medium 1 to reach the surface at C.
2
Let this take a time τ. So BC = v₁τ.
3
In the same time, the secondary wavelet from A spreads into medium 2 with radius AE = v₂τ.
4
From C, draw a tangent CE to this wavelet. CE is the refracted wavefront, making angle r with the surface.
5
In right triangle ABC (right angle at B): sin i = BC/AC = v₁τ/AC.
6
In right triangle AEC (right angle at E): sin r = AE/AC = v₂τ/AC.
7
Divide the two: sin i / sin r = v₁/v₂.
8
Put v₁ = c/μ₁ and v₂ = c/μ₂: sin i / sin r = μ₂/μ₁ = μ₂₁, so μ₁ sin i = μ₂ sin r.
9★
So, sin i / sin r is a constant for the two media: this is Snell's law. Since v₂ < v₁, r < i and the ray bends towards the normal.

★ key step

How to draw the two constructions

Reflection: draw a horizontal line MN for the reflecting surface. Draw the incident wavefront AB slanting down so that A touches MN, and mark the angle i between AB and MN.
Draw rays at right angles to AB, reaching A and C. Write BC = vτ. With A as centre, draw an arc of radius vτ above MN.
From C, draw the tangent CE to the arc; label it reflected wavefront and mark r.
Refraction: draw a surface PP′ with medium 1 (v₁) above and medium 2 (v₂) below. Draw the incident wavefront AB touching it at A, and write BC = v₁τ.
With A as centre, draw an arc of radius v₂τ below the surface, clearly shorter than BC for a denser medium.
From C, draw the tangent CE; label it refracted wavefront and mark r. Add rays at right angles to the wavefronts and dashed normals.

Speed, wavelength and frequency on refraction

The frequency of light is fixed by the source, so it does not change on refraction. In the same time, the wavefront covers v₁τ in medium 1 and v₂τ in medium 2. If BC is one wavelength λ₁, then AE is one wavelength λ₂. So λ₁/λ₂ = v₁/v₂. In a denser medium, light is slower and its wavelength is shorter. Colour depends on frequency, so light keeps its colour inside water or glass.

Worked example: light entering glass

Problem

Light of wavelength 600 nm in air falls on glass of refractive index 1.5 at an angle of incidence of 60°. Find (a) its frequency, (b) its speed in glass, (c) its wavelength in glass, (d) the angle of refraction. (c = 3 × 10⁸ m s⁻¹)

Working

Given: λ = 600 nm = 6 × 10⁻⁷ m, μ = 1.5, i = 60°.
(a) ν = c/λ = (3 × 10⁸ m s⁻¹) ÷ (6 × 10⁻⁷ m) = 5 × 10¹⁴ Hz.
(b) v = c/μ = (3 × 10⁸ m s⁻¹) ÷ 1.5 = 2 × 10⁸ m s⁻¹.
(c) λ′ = λ/μ = 600 nm ÷ 1.5 = 400 nm. Check: v/λ′ = (2 × 10⁸) ÷ (4 × 10⁻⁷) = 5 × 10¹⁴ Hz, the same frequency.
(d) sin r = sin i / μ = 0.866 ÷ 1.5 ≈ 0.577, so r ≈ 35°.

Answer

(a) 5 × 10¹⁴ Hz; (b) 2 × 10⁸ m s⁻¹; (c) 400 nm; (d) about 35°

Into a denser medium vs into a rarer medium

Rarer to denser (v₂ < v₁)The wavelet in medium 2 is smaller than BC, so r < i. The ray bends towards the normal. Example: air to glass.

Denser to rarer (v₂ > v₁)The wavelet in medium 2 is larger than BC, so r > i. The ray bends away from the normal. Example: water to air. When r reaches 90°, i equals the critical angle C, with sin C = μ₂/μ₁. Beyond it no tangent can be drawn, no refracted wavefront forms, and total internal reflection takes place.

In real life

A coin at the bottom of a bucket of water looks raised, because light from it bends away from the normal as it leaves the water. A DTH dish on a rooftop reflects the plane wavefronts from a far-off satellite so that they converge on the receiver at its focus.

Exam angle

3 marks: “Using Huygens' principle, verify the law of reflection at a plane surface. Draw a labelled diagram.” Wavefront AB at angle i; BC = vτ; wavelet from A of radius AE = vτ; tangent CE is the reflected wavefront; triangles ABC and AEC are congruent (right angle, common hypotenuse AC, BC = AE), so i = r; diagram with normals, i and r marked.

3 marks: “Using Huygens' principle, verify Snell's law for a plane wave going from a rarer to a denser medium.” Diagram; BC = v₁τ; AE = v₂τ; tangent CE is the refracted wavefront; sin i = v₁τ/AC and sin r = v₂τ/AC; so sin i / sin r = v₁/v₂ = μ₂₁; since v₂ < v₁, the ray bends towards the normal.

2 marks: “When light passes from air into glass, which of these change: frequency, wavelength, speed? Give a reason.” Speed and wavelength both decrease by the factor μ. The frequency stays the same, because it is fixed by the source.

Remember it

Same medium, same radius: i = r. Slower medium, smaller wavelet: bend towards the normal. Frequency never changes.