11 min readPathway sheet3-mark proof with a diagram, or a 5-mark question that adds a numerical on speed, wavelength and frequency in a medium

Refraction of a plane wave: proving Snell's law

Light slows down in a denser medium, so the wavefront turns. Huygens' construction gives sin i / sin r = v₁/v₂, which is Snell's law.

Snell's lawsin i / sin r = v₁/v₂ = n₂/n₁, or n₁ sin i = n₂ sin r

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

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

In a medium of refractive index nv = c/n and λ′ = λ/n

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/n₁ and v₂ = c/n₂: sin i / sin r = n₂/n₁, so n₁ sin i = n₂ 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

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.

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. When r reaches 90°, i is the critical angle: sin i_c = n₂/n₁. Beyond it, no refracted wavefront forms, and total internal reflection takes place.

How to draw the refraction diagram

Draw a horizontal line PP′. Write medium 1 (v₁) above it and medium 2 (v₂) below it.
Draw the incident wavefront AB above the surface, touching it at A. Mark angle i between AB and PP′.
Draw incident rays at right angles to AB, reaching A and C. Write BC = v₁τ.
With A as centre, draw an arc of radius v₂τ below the surface. Make it clearly shorter than BC for a denser medium.
From C, draw a tangent CE to the arc; label it refracted wavefront. Mark angle r between CE and PP′.
Draw refracted rays at right angles to CE, and dashed normals at A and C.

Worked example: light entering glass

Problem

Light of wavelength 600 nm in air enters glass of refractive index 1.5. Find (a) its frequency, (b) its speed in glass, (c) its wavelength in glass. (c = 3 × 10⁸ m/s)

Working

Given: λ = 600 nm = 6 × 10⁻⁷ m, n = 1.5.
(a) f = c/λ = (3 × 10⁸ m/s) ÷ (6 × 10⁻⁷ m) = 5 × 10¹⁴ Hz.
(b) v = c/n = (3 × 10⁸ m/s) ÷ 1.5 = 2 × 10⁸ m/s.
(c) λ′ = λ/n = 600 nm ÷ 1.5 = 400 nm.
Check: v/λ′ = (2 × 10⁸ m/s) ÷ (4 × 10⁻⁷ m) = 5 × 10¹⁴ Hz, the same frequency as in air.

Answer

(a) 5 × 10¹⁴ Hz; (b) 2 × 10⁸ m/s; (c) 400 nm. The frequency does not change.

Worked example: angle of refraction in water

Problem

Light in air falls on water (n = 4/3) at an angle of incidence of 53°. Find the angle of refraction and the speed of light in water. (Take sin 53° = 0.8 and sin 37° = 0.6; c = 3 × 10⁸ m/s.)

Working

Snell's law: n₁ sin i = n₂ sin r, with n₁ = 1 for air.
1 × sin 53° = (4/3) × sin r.
sin r = 0.8 × 3/4 = 0.6, so r = 37°.
Speed in water: v = c/n = (3 × 10⁸ m/s) ÷ (4/3) = 2.25 × 10⁸ m/s.

Answer

r = 37°; v = 2.25 × 10⁸ m/s

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. The optical fibres that carry internet across India work by total internal reflection, which the same wavefront picture explains.

Exam angle

3 marks: “Using Huygens' principle, verify Snell's law of refraction 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 = n₂/n₁; 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 n. Frequency stays the same, because it is fixed by the source.

3 marks: “Light of wavelength 500 nm in air enters a medium of refractive index 1.25. Find its speed, wavelength and frequency in the medium.” v = (3 × 10⁸) ÷ 1.25 = 2.4 × 10⁸ m/s; λ′ = 500 ÷ 1.25 = 400 nm; f = (3 × 10⁸) ÷ (5 × 10⁻⁷) = 6 × 10¹⁴ Hz, unchanged.

Remember it

Speed and wavelength shrink by n; frequency never changes. A slower medium gives a smaller wavelet, so the ray bends towards the normal.