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Interference and Diffraction of Light

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Senior 5 · Physics

Interference and Diffraction of Light

4 lessons

Lesson 1 of 4

Light Waves and the Conditions for Interference

25 min

Colours from nothing coloured

A soap bubble is made of colourless soapy water, yet it shimmers with bands of colour. The wing of a blue butterfly contains no blue pigment. The underside of a CD shows a full rainbow. In each case the colour comes from light waves adding together and cancelling: interference. Interference is the strongest evidence that light is a wave, and it makes possible measurements of astonishing precision, from the wavelength of light itself to the detection of gravitational waves. This unit builds the wave model of light that you need to explain it. In Unit 7 you saw light behave as particles; here you see the other half of the story.

Light as an electromagnetic wave

Maxwell showed in 1865 that a changing electric field produces a magnetic field and a changing magnetic field produces an electric field, so the two can sustain each other and travel through empty space as an electromagnetic wave, at $c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} = 3.00 \times 10^{8}\,\mathrm{m\,s^{-1}}$ . Light is such a wave.

The electric field $\vec{E}$ and the magnetic field $\vec{B}$ are perpendicular to each other and to the direction of travel: the wave is transverse.
$\vec{E}$ and $\vec{B}$ oscillate in phase, and their sizes are linked by $E = c B$ .
Wavelength and frequency are related by $c = f\lambda$ . Visible light runs from about 400 nm (violet, $7.5 \times 10^{14}$ Hz) to about 700 nm (red, $4.3 \times 10^{14}$ Hz).
The intensity (power per unit area) is proportional to the square of the amplitude: $I = \frac{1}{2}c\varepsilon_0E_0^2$ .

It is the electric field that acts on the electrons in our eyes and in detectors, so in optics "the light wave" usually means the oscillating $\vec{E}$ .

Worked example 1: the fields in sunlight

Bright sunlight at midday delivers about 1000 W m⁻². Find the amplitudes of its electric and magnetic fields ( $\varepsilon_0 = 8.85 \times 10^{-12}\,\mathrm{F\,m^{-1}}$ ).

Solution

$E_0 = \sqrt{\frac{2I}{c\varepsilon_0}} = \sqrt{\frac{2(1000)}{(3.00 \times 10^{8})(8.85 \times 10^{-12})}} = 8.7 \times 10^{2}\,\mathrm{V\,m^{-1}}$ .

$B_0 = \frac{E_0}{c} = \frac{868}{3.00 \times 10^{8}} = 2.9 \times 10^{-6}$ T, about a tenth of the Earth's magnetic field.

Huygens' principle and superposition

In 1678 Christiaan Huygens proposed a way to predict how a wavefront moves: every point on a wavefront acts as a source of secondary wavelets, spreading out at the speed of the wave; the new wavefront is the surface that touches all the wavelets. Huygens' principle explains reflection, refraction and, as you will see, the spreading of light through narrow openings.

When two or more waves meet, the principle of superposition applies: the resultant displacement (here, the electric field) at any point is the vector sum of the displacements of the individual waves. Where crest meets crest, the waves reinforce (constructive interference); where crest meets trough, they cancel (destructive interference).

Path difference and phase difference

Two sources $S_{1}$ and $S_{2}$ , oscillating in phase, send waves to a point P. The waves travel distances $S_{1} P$ and $S_{2} P$ . Their path difference $\Delta = S_2P - S_1P$ decides what happens at P, because every wavelength of extra path shifts the phase by $2\pi$ :

\phi = \frac{2\pi}{\lambda}\Delta

Constructive (bright): $\Delta = n\lambda$ , phase difference $0, 2\pi, 4\pi, \ldots$ , where $n = 0, \pm 1, \pm 2, \ldots$
Destructive (dark): $\Delta = \left(n + \tfrac{1}{2}\right)\lambda$ , phase difference $\pi, 3\pi, \ldots$

Definition

Interference is the superposition of two or more waves to give a resultant whose amplitude varies from point to point. The path difference $\Delta$ is the difference in the distances two waves have travelled to a point; the phase difference is $\phi = \frac{2\pi\Delta}{\lambda}$ . Example: for $\lambda = 600$ nm, a path difference of 1.5 μm is $2.5\lambda$ , so the waves arrive half a cycle out of step and cancel.

The conditions for observable interference

Switch on two torches and point them at the same wall: you see no fringes, just a uniformly brighter patch. Light from an ordinary lamp is emitted by countless atoms, each producing a short wave train lasting roughly $10^{-8}$ s, with a random phase. The interference pattern of two such lamps exists, but it shifts at random about $10^{8}$ times a second and averages to a uniform brightness. For a steady, observable pattern the sources must be:

Coherent: they must have a constant phase difference. This also requires them to have the same frequency (the same wavelength). In practice coherent sources are made by splitting the light from one source into two parts, as with two slits lit by the same lamp or laser, so that every random change of phase happens in both at once.
Of similar amplitude, so that the cancellation at the dark fringes is nearly complete and the contrast is good.
Of the same polarisation (their electric fields oscillating in the same plane), or at least not at right angles, since perpendicular fields cannot cancel.
Close together, with a path difference smaller than the length of the wave trains (the coherence length), so that the parts that meet came from the same wave train.

Definition

Two sources are coherent if they have the same frequency and a constant phase difference. Example: two slits lit by the same laser beam are coherent; two separate car headlights are not.

Worked example 2: bright or dark?

Two coherent sources emit in phase with $\lambda = 600$ nm. Point P is 12.0000 μm from $S_{1}$ and 13.5000 μm from $S_{2}$ ; point Q is 12.0000 μm from $S_{1}$ and 14.4000 μm from $S_{2}$ . Is each point bright or dark?

Solution

P: $\Delta = 1.5000\,\mu\mathrm{m} = \frac{1500}{600}\lambda = 2.5\lambda$ : an odd number of half wavelengths, so destructive: dark.

Q: $\Delta = 2.4000\,\mu\mathrm{m} = 4.0\lambda$ : a whole number of wavelengths, so constructive: bright.

Worked example 3: sources out of phase

Two loudspeakers driven by the same amplifier emit 680 Hz sound (speed 340 m s⁻¹), but one is wired the wrong way round, so the sources are half a cycle out of phase. A listener stands equidistant from them. What does she hear, and where should she move to hear a loud sound?

Solution

$\lambda = \frac{340}{680} = 0.50$ m. At the equidistant point the path difference is zero, but the sources start half a cycle apart, so the waves cancel: a quiet spot. Loud sound now needs a path difference that adds another half cycle: $\Delta = \left(n + \tfrac{1}{2}\right)\lambda$ , e.g. 0.25 m. Swapping the wires restores the usual rule. The same reasoning applies to light: the conditions for bright and dark swap if the sources are in antiphase.

Warning

Coherent does not mean "in phase". Coherent sources need a constant phase difference, which may be zero, half a cycle or anything else. And interference does not destroy energy: the energy missing from the dark fringes appears in the bright ones.

Tip

Count the half-wavelengths. Divide the path difference by $\frac{\lambda}{2}$ : an even answer means bright, an odd answer means dark (for sources in phase).

Coherence and Path Difference - A Level Physics, Physics Online. Watch how a constant phase difference and the path difference together decide where waves reinforce and where they cancel.

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Practice

State three properties of electromagnetic waves.
Find the frequency of green light of wavelength 530 nm.
A laser beam has an electric-field amplitude of 600 V m⁻¹. Find its magnetic-field amplitude and its intensity.
What is meant by coherent sources? Why do two separate lamps not produce an interference pattern?
Two coherent sources in phase emit light of wavelength 500 nm. Find the phase difference, in radians, at a point where the path difference is 1.25 μm, and state whether the point is bright or dark.
Use Huygens' principle to explain why a plane wavefront stays plane as it travels through a uniform medium.
Explain why the contrast of a fringe pattern is poor if one of the two sources is much brighter than the other.
A student claims that in destructive interference light energy is destroyed. Evaluate this claim.

Show answersHide answers

Any three of: transverse; $\vec{E}$ and $\vec{B}$ perpendicular to each other and to the direction of travel, and in phase; travel at $c$ in a vacuum; carry energy with intensity $\propto E_0^2$ ; need no medium.
$f = \frac{3.00 \times 10^{8}}{530 \times 10^{-9}} = 5.7 \times 10^{14}$ Hz.
$B_0 = \frac{600}{3.00 \times 10^{8}} = 2.0 \times 10^{-6}$ T; $I = \frac{1}{2}(3.00 \times 10^{8})(8.85 \times 10^{-12})(600)^2 = 4.8 \times 10^{2}\,\mathrm{W\,m^{-2}}$ .
Sources with the same frequency and a constant phase difference. Separate lamps emit short wave trains with random phases, so their phase difference changes about $10^{8}$ times a second and the pattern averages out to uniform brightness.
$\frac{1.25 \times 10^{-6}}{500 \times 10^{-9}} = 2.5$ wavelengths; $\phi = 2\pi \times 2.5 = 5\pi$ rad (15.7 rad), an odd multiple of $π \pi$ : dark.
Every point on the plane wavefront sends out wavelets of equal radius $c\,\Delta t$ ; the surface touching all of them (their envelope) is a plane parallel to the first.
With unequal amplitudes the waves cannot cancel completely: the dark fringes have amplitude $|A_1 - A_2| \ne 0$ , so they are not dark, while the bright fringes have $A_{1} + A_{2}$ . The difference between bright and dark is reduced.
Incorrect. Energy is redistributed, not destroyed: dark fringes receive less than the sum of the two intensities and bright fringes receive more (up to four times one source's intensity). Averaged over the pattern, the intensity equals the sum of the two separate intensities.

Key points

Light is a transverse electromagnetic wave: $\vec{E} \perp \vec{B} \perp$ direction of travel; $E = c B$ ; $c = f\lambda$ .
Huygens: every point on a wavefront is a source of wavelets. Superposition: displacements add.
Phase difference $\phi = \frac{2\pi\Delta}{\lambda}$ ; in-phase sources give bright at $\Delta = n\lambda$ , dark at $\left(n + \frac{1}{2}\right)\lambda$ .
Observable interference needs coherent sources (same frequency, constant phase difference), similar amplitudes and the same polarisation.

Sources and credits

Photo: A study of the propagation, refraction, reflection, interference and diffraction of ripple waves (1914), Internet Archive Book Images, Wikimedia Commons, no known copyright restrictions.
Video: Coherence and Path Difference - A Level Physics, Physics Online, embedded from YouTube.

Written by Ganzaa. The two diagrams (the interference curves are calculated), the worked examples and the practice exercises are by Ganzaa.

1Colours from nothing coloured

2Light as an electromagnetic wave

3Huygens' principle and superposition

4Path difference and phase difference

5The conditions for observable interference

Colours from nothing coloured

Light as an electromagnetic wave

Huygens' principle and superposition

Path difference and phase difference

The conditions for observable interference