Nature and propagation of light

Unit 10 · Waves and Optics · 3 teaching hours · 14 min read

Nature and propagation of light

What the syllabus asks for 2
  • Huygen's principle
  • Reflection and Refraction according to wave theory

Ray optics treats light as a straight line and gets mirrors and lenses right, but it cannot say why light bends, or what happens when it meets an obstacle only a few thousandths of a millimetre across. Wave optics can. This unit replaces the ray by the wavefront, states Huygens' principle, and uses it to rebuild the two laws you already know — reflection and refraction — from scratch. Everything in the next three units (interference, diffraction, polarization) stands on the geometry you learn here.

What the examiner asks from this unit

Two things, almost every year: state Huygens' principle and explain the construction, and prove the laws of reflection or Snell's law using Huygens' construction. Both are pure marks if you can draw the diagram correctly, so practise the diagram until you can produce it from memory in ninety seconds.

10.1 Huygen's principle

A source of light sends energy out in all directions. At any instant, the points that the disturbance has just reached all started vibrating at the same moment, so they are all vibrating in step.

Wavefront

A wavefront is the locus of all points of a medium that are vibrating in the same phase. The perpendicular drawn to a wavefront, in the direction in which the wave advances, is a ray. The speed with which a wavefront advances is the speed of the wave in that medium.

Close to a point source the wavefronts are spheres; close to a long thin slit they are cylinders. Very far from any source a small patch of the sphere is indistinguishable from a flat plane, so sunlight reaching the Earth arrives as plane wavefronts. A converging lens turns an incoming plane wavefront into a spherical wavefront that shrinks onto the focus.

Type of wavefront Shape Source Amplitude falls as
Spherical sphere point source 1/r
Cylindrical cylinder linear source (slit) 1/\sqrt{r}
Plane plane very distant source constant

Statement of the principle

Christiaan Huygens (1678) gave a rule for advancing a wavefront from one instant to the next. It has two parts:

  1. Every point on a wavefront acts as a fresh source of secondary wavelets, which spread out in all forward directions with the speed of the wave in that medium.
  2. The new wavefront, after a time t, is the surface that touches all these secondary wavelets — their forward envelope or common tangent surface.

If the wave speed in the medium is v, every secondary wavelet drawn from the old wavefront is a sphere of radius

r = vt

So to advance a wavefront you take a compass, set it to vt, strike arcs from a row of points on the old wavefront, and draw the tangent.

figure
Figure 1. Huygens' construction. (a) A plane wavefront AB advances to the plane A'B'. (b) A spherical wavefront from S advances to a larger sphere. In each case the new wavefront is the forward envelope of secondary wavelets of radius ct.

Why there is no backward wave

Drawn honestly, the secondary wavelets are complete spheres, so they also have a backward envelope — a wavefront travelling back towards the source. We never observe one. Huygens simply asserted that the wavelets have zero intensity in the backward direction. Fresnel and later Kirchhoff showed the amplitude of a secondary wavelet actually varies with the angle \theta measured from the forward direction as the obliquity factor

K = \tfrac{1}{2}(1+\cos\theta)

which equals 1 straight ahead and 0 straight backwards. That is why only the forward envelope survives. For NEB purposes it is enough to say: secondary wavelets are effective only in the forward direction.

Wavelets are not rays

A secondary wavelet is a tiny spherical wave, not a ray. Drawing straight arrows from points on the wavefront instead of arcs earns no marks — the whole proof depends on the arcs and on their common tangent.

10.2 Reflection and Refraction according to wave theory

Reflection at a plane surface

Let a plane wavefront AB travel through a medium in which light has speed c and strike a plane mirror. Take the moment when the end A has just touched the mirror at A while the other end is still at B. Let the wavefront meet the mirror again at C after a further time t.

figure
Figure 2. Reflection of a plane wavefront. While B travels to C, the secondary wavelet from A grows to radius ct, and CD is the reflected wavefront. Since BC = AD, the angle of incidence equals the angle of reflection.

While B travels the distance BC, every point of the mirror between A and C has in turn started a secondary wavelet. The wavelet from A, which started earliest, has had the full time t to grow, so it is a hemisphere of radius AD = ct. The wavelet from C has just started and has zero radius. The tangent drawn from C to the wavelet centred on A is the reflected wavefront CD.

The laws of reflection

In the right-angled triangles ABC and ADC:

  • BC = ct is the distance travelled by B in the medium, and AD = ct is the radius of the wavelet from A. Hence BC = AD.
  • AC is common, and angle ABC = angle ADC = 90^\circ (a wavefront is always perpendicular to its ray).

From triangle ABC: \sin i = \dfrac{BC}{AC}, and from triangle ADC: \sin r = \dfrac{AD}{AC}.

\frac{\sin i}{\sin r} = \frac{BC}{AD} = \frac{ct}{ct} = 1 \;\Longrightarrow\; i = r

That is the second law of reflection. Also, the incident wavefront, the mirror and the reflected wavefront are all built on the same plane of the diagram, so the incident ray, the reflected ray and the normal are coplanar — the first law.

Refraction at a plane surface

Now let the same plane wavefront AB fall on the flat surface separating medium 1 (speed v_1) from a denser medium 2 (speed v_2 < v_1). Again A touches the surface first, and B reaches C after time t, so BC = v_1 t. In that same time the wavelet that started at A has spread into medium 2 and has grown only to radius AD = v_2 t, which is smaller. The tangent CD from C to this wavelet is the refracted wavefront, and it is tilted towards the normal.

figure
Figure 3. Refraction of a plane wavefront. In time t the wavefront covers BC = v_1t in the rarer medium while the wavelet from A grows to only AD = v_2t in the denser medium, so the refracted wavefront CD bends towards the normal.

Snell's law from Huygens' construction

From the right-angled triangle ABC, with the angle at A equal to i:

\sin i = \frac{BC}{AC} = \frac{v_1 t}{AC}

From the right-angled triangle ADC, with the angle at C equal to r:

\sin r = \frac{AD}{AC} = \frac{v_2 t}{AC}

Dividing, the common factors AC and t cancel:

\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = {}_1n_2

The right-hand side depends only on the two media, not on i. So the ratio \sin i/\sin r is a constant — Snell's law.

Writing the absolute refractive index of a medium as n = c/v, we get v_1 = c/n_1 and v_2 = c/n_2, so

\frac{\sin i}{\sin r} = \frac{n_2}{n_1} \;\Longrightarrow\; n_1\sin i = n_2\sin r

Frequency, wavelength and optical path

The secondary wavelets in medium 2 are started by the vibrations arriving from medium 1, so they must vibrate at the same rate. Frequency does not change on refraction — it is fixed by the source. Since v = f\lambda and v falls,

\frac{\lambda_1}{\lambda_2} = \frac{v_1}{v_2} = \frac{n_2}{n_1}, \qquad \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{n}

Light slows down and its wavelength shrinks, but its colour (frequency) is unchanged. A path of geometrical length x inside a medium of index n holds the same number of waves as a length nx in vacuum, so we define the optical path

\Delta_{\text{opt}} = n\,x

This single idea is used constantly in the next two units: whenever light passes through glass or water, replace the real distance by the optical path.

Prediction Newton's corpuscular theory Huygens' wave theory
Nature of light stream of tiny particles disturbance spreading as wavefronts
Speed in water compared with air greater smaller
Reflection explained explained
Refraction explained, wrong speed explained, correct speed
Interference, diffraction, polarization not explained explained
Experimental verdict disproved by Foucault (1850) confirmed

Foucault's rotating-mirror measurement in 1850 showed that light travels more slowly in water than in air. That single result killed the corpuscular theory and established the wave picture used in Units 11–13.

Worked example 10.1

Problem. Light strikes a glass block of refractive index 1.5 at an angle of incidence of 60^\circ. Take c = 3\times10^{8}\ \text{m s}^{-1} and a vacuum wavelength of 600\ \text{nm}. Find (a) the angle of refraction, (b) the speed of light in the glass, (c) the wavelength in the glass.

Solution.

(a) \sin r = \dfrac{\sin 60^\circ}{1.5} = \dfrac{0.8660}{1.5} = 0.5774, so r = 35.3^\circ.

(b) v = \dfrac{c}{n} = \dfrac{3\times10^{8}}{1.5} = 2.0\times10^{8}\ \text{m s}^{-1}.

(c) \lambda_g = \dfrac{\lambda_0}{n} = \dfrac{600}{1.5} = 400\ \text{nm}.

The frequency stays at f = c/\lambda_0 = 5.0\times10^{14}\ \text{Hz} in both media.

Worked example 10.2

Problem. A beam of frequency 5.0\times10^{14}\ \text{Hz} passes from air into a liquid in which its wavelength is 375\ \text{nm}. Find the refractive index of the liquid and the speed of light in it.

Solution. Frequency is unchanged, so in the liquid

v = f\lambda_m = (5.0\times10^{14})(375\times10^{-9}) = 1.875\times10^{8}\ \text{m s}^{-1}

n = \frac{c}{v} = \frac{3\times10^{8}}{1.875\times10^{8}} = 1.6

Check with wavelengths: \lambda_0 = c/f = 600\ \text{nm}, and \lambda_0/\lambda_m = 600/375 = 1.6. The two routes agree.

Worked example 10.3

Problem. A glass plate of thickness 2.0\ \text{mm} and refractive index 1.5 is placed in a beam of light of vacuum wavelength 500\ \text{nm}. Find (a) the optical path through the plate, (b) the number of waves contained in the plate, (c) the extra time the light takes compared with travelling the same 2.0\ \text{mm} in vacuum.

Solution.

(a) \Delta_{\text{opt}} = nx = 1.5 \times 2.0\times10^{-3} = 3.0\times10^{-3}\ \text{m}.

(b) The number of waves is the optical path divided by the vacuum wavelength:

N = \frac{nx}{\lambda_0} = \frac{3.0\times10^{-3}}{500\times10^{-9}} = 6000

(c) Time in glass = nx/c, time in vacuum = x/c, so the delay is

\Delta t = \frac{(n-1)x}{c} = \frac{0.5 \times 2.0\times10^{-3}}{3\times10^{8}} = 3.3\times10^{-12}\ \text{s}

Chapter summary

  • A wavefront is the locus of points vibrating in the same phase; a ray is the normal to the wavefront. Wavefronts may be spherical, cylindrical or plane.
  • Huygens' principle: every point of a wavefront is a source of secondary wavelets of radius vt; the new wavefront is their forward envelope. Backward wavelets are suppressed by the obliquity factor K = \frac{1}{2}(1+\cos\theta).
  • Reflection: BC = AD = ct gives \sin i/\sin r = 1, so i = r, and the incident ray, reflected ray and normal are coplanar.
  • Refraction: BC = v_1t, AD = v_2t give \dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2} = \dfrac{n_2}{n_1}, i.e. n_1\sin i = n_2\sin r.
  • On refraction the frequency is unchanged; the speed and wavelength both fall by the factor n: v = c/n and \lambda_m = \lambda_0/n.
  • Optical path = n \times geometrical path; it is the vacuum distance holding the same number of waves.
  • The wave theory predicts v_{\text{water}} < v_{\text{air}}, confirmed by Foucault in 1850; the corpuscular theory predicted the opposite and failed.

Practice questions

Group A — Multiple choice (1 mark each)

  1. A wavefront is a surface on which all points have the same [1]aamplitudebphasecwavelengthdintensity
  2. When light travels from air into glass, the quantity that does not change is [1]aspeedbwavelengthcfrequencyddirection
  3. Secondary wavelets in Huygens' construction travel with [1]athe speed of the sourcebthe speed of light in that mediumcinfinite speeddhalf the wave speed
  4. According to the wave theory of light, the speed of light in water is [1]agreater than in airbless than in aircequal to that in airdzero
  5. The optical path of a ray covering 4\ \text{cm} in a medium of index 1.5 is [1]a2.67 cmb4 cmc6 cmd9 cm
Answers to Group A

1. (b) — a wavefront is by definition a surface of constant phase.

2. (c) — frequency is set by the source and is not altered by the medium.

3. (b) — each wavelet expands at the wave speed v = c/n of that medium.

4. (b) — the wave theory needs v_2 < v_1 to bend light towards the normal; Foucault confirmed it.

5. (c) — optical path = nx = 1.5\times4 = 6\ \text{cm}.

Group B — Short answer (5 marks each)

  1. Define a wavefront and state Huygens' principle. Using it, show how a plane wavefront and a spherical wavefront advance. [5]
  2. Using Huygens' construction, prove the laws of reflection of light at a plane surface. [5]
  3. Light of wavelength 600\ \text{nm} in vacuum enters a medium of refractive index 1.25. Find its speed, frequency and wavelength in the medium. [5]
  4. Explain why the frequency of light does not change on refraction while its wavelength does. Hence define optical path. [5]
Answers to Group B

1. Outline: wavefront = locus of points in the same phase; each point emits secondary wavelets of radius vt; new wavefront = forward envelope. A plane wavefront gives a plane envelope, a spherical one a larger sphere (Figure 1).

3. v = c/n = 3\times10^{8}/1.25 = 2.4\times10^{8}\ \text{m s}^{-1}. Frequency is unchanged: f = c/\lambda_0 = 3\times10^{8}/600\times10^{-9} = 5.0\times10^{14}\ \text{Hz}. Wavelength: \lambda_m = v/f = 2.4\times10^{8}/5.0\times10^{14} = 4.8\times10^{-7}\ \text{m} = 480\ \text{nm} (equivalently 600/1.25).

4. Outline: the vibrations in medium 2 are driven by those arriving from medium 1, so they repeat at the same rate — f is fixed. Since v = f\lambda and v falls to c/n, the wavelength falls to \lambda_0/n. Optical path = n \times geometrical path is the equivalent vacuum distance containing the same number of waves.

Group C — Long answer (8 marks each)

  1. (a) State Huygens' principle and explain the meaning of a secondary wavelet. [3](b) Using Huygens' construction, derive Snell's law of refraction and hence show that the refractive index of a medium equals c/v. [5]
  2. A ray of light is incident at 45^\circ on a parallel-sided glass slab of refractive index 1.5 and thickness 6.0\ \text{cm}. Find(a) the angle of refraction,(b) the speed of light inside the slab,(c) the time taken to cross the slab, and(d) the lateral displacement of the emergent ray. [8]
Answer to Group C question 2

(a) \sin r = \sin 45^\circ/1.5 = 0.7071/1.5 = 0.4714, so r = 28.1^\circ.

(b) v = c/n = 3\times10^{8}/1.5 = 2.0\times10^{8}\ \text{m s}^{-1}.

(c) Path length inside the slab = t/\cos r = 6.0/0.8819 = 6.80\ \text{cm}, so

\text{time} = \frac{6.80\times10^{-2}}{2.0\times10^{8}} = 3.4\times10^{-10}\ \text{s}

(d) Lateral displacement

d = \frac{t\,\sin(i-r)}{\cos r} = \frac{6.0 \times \sin 16.9^\circ}{0.8819} = \frac{6.0 \times 0.2902}{0.8819} = 1.97\ \text{cm}

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