1.4 Superposition Principle

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1.3.3 Light 8 −1 Light moves in a vacuum at a speed of cvac =3 × 10 m s . In transparent materials it moves at a speed less than cvac by a factor n which is called the refractive index of the material:

c = cvac/n. (1.14)

Often the refractive index takes the form

2 A n ≈ 1+ 2 , (1.15) 1 − (k/kR) where k is the wavenumber and kR and A are positive constants characteristic of the material. The angular frequency of light in a transparent medium is thus ω = kc = kcvac/n. (1.16)

1.4 Superposition Principle

It is found empirically that as long as the amplitudes of waves in most media are small, two waves in the same physical location don’t interact with each other. Thus, for example, two waves moving in the opposite direction simply pass through each other without their shapes or amplitudes being changed. When collocated, the total wave displacement is just the sum of the displacements of the individual waves. This is called the superposition principle. At suﬃciently large amplitude the superposition principle often breaks down — interacting waves may scatter oﬀ of each other, lose amplitude, or change their form. Interference is a consequence of the superposition principle. When two or more waves are superimposed, the net wave displacement is just the algebraic sum of the displacements of the individual waves. Since these displacements can be positive or negative, the net displacement can either be greater or less than the individual wave displacements. The former case is called constructive interference, while the latter is called destructive interference. Let us see what happens when we superimpose two sine waves with different wavenumbers. Figure 1.5 shows the superposition of two waves with wavenumbers k1 = 4 and k2 = 5. Notice that the result is a wave with about the same wavelength as the two initial waves, but which varies in amplitude 8 CHAPTER 1. WAVES IN ONE DIMENSION

0.50 k1, k2 = 4.000000 5.000000

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Figure 1.5: Superposition (lower panel) of two sine waves (shown individually in the upper panel) with equal amplitudes and wavenumbers k1 = 4 and k2 = 5. depending on whether the two sine waves are in or out of phase. When the waves are in phase, constructive interference is occurring, while destructive interference occurs where the waves are out of phase. What happens when the wavenumbers of the two sine waves are changed? Figure 1.6 shows the result when k1 = 10 and k2 = 11. Notice that though the wavelength of the resultant wave is decreased, the locations where the amplitude is maximum have the same separation in x as in figure 1.5. If we superimpose waves with k1 = 10 and k2 = 12, as is shown in figure 1.7, we see that the x spacing of the regions of maximum amplitude has decreased by a factor of two. Thus, while the wavenumber of the resultant wave seems to be related to something like the average of the wavenumbers of the component waves, the spacing between regions of maximum wave amplitude appears to go inversely with the difference of the wavenumbers of the component waves. In other words, if k1 and k2 are close together, the amplitude maxima are far apart and vice versa. We can symbolically represent the sine waves that make up figures 1.5, 1.6, and 1.7 by a plot such as that shown in figure 1.8. The amplitudes and wavenumbers of each of the sine waves are indicated by vertical lines in this 1.4. SUPERPOSITION PRINCIPLE 9

0.50 k1, k2 = 10.000000 11.000000

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Figure 1.6: Superposition of two sine waves with equal amplitudes and wavenumbers k1 = 10 and k2 = 11.

0.50 k1, k2 = 10.000000 12.000000

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-1.00 -10.0 -5.0 0.0 5.0 sine functions vs x

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Figure 1.7: Superposition of two sine waves with equal amplitudes and wavenumbers k1 = 10 and k2 = 12. 10 CHAPTER 1. WAVES IN ONE DIMENSION

2 waves

k1 k2

k0 amplitude

∆k ∆k

wavenumber

Figure 1.8: Representation of the wavenumbers and amplitudes of two superimposed sine waves.

figure. The regions of large wave amplitude are called wave packets. Wave packets will play a central role in what is to follow, so it is important that we acquire a good understanding of them. The wave packets produced by only two sine waves are not well separated along the x-axis. However, if we superimpose many waves, we can produce an isolated wave packet. For example, figure 1.9 shows the results of superimposing 20 sine waves with wavenumbers k = 0.4m, m = 1, 2,..., 20, where the amplitudes of the waves are largest for wavenumbers near k = 4. In particular, we assume that the amplitude 2 2 of each sine wave is proportional to exp[−(k − k0) /∆k ], where k0 = 4 and ∆k = 1. The amplitudes of each of the sine waves making up the wave packet in figure 1.9 are shown schematically in figure 1.10. The quantity ∆k controls the distribution of the sine waves being superimposed — only those waves with a wavenumber k within approximately ∆k of the central wavenumber k0 of the wave packet, i. e., for 3 ≤ k ≤ 5 in this case, contribute significantly to the sum. If ∆k is changed to 2, so that wavenumbers in the range 2 ≤ k ≤ 6 contribute significantly, the wavepacket becomes narrower, as is shown in figures 1.11 and 1.12. ∆k is called the wavenumber spread of the wave packet, and it evidently plays a role similar to the difference in wavenumbers in the superposition of two sine waves — the larger the wavenumber spread, the smaller the physical size of the wave packet. Furthermore, the wavenumber of the oscillations within the wave packet is given approximately by the central wavenumber. 1.4. SUPERPOSITION PRINCIPLE 11

k0 = 4.000000; dk = 1.000000; nwaves = 20

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Figure 1.9: Superposition of twenty sine waves with k0 = 4 and ∆k = 1.

∆k amplitude

0 4 8 12 16 20 wavenumber

Figure 1.10: Representation of the wavenumbers and amplitudes of 20 superimposed sine waves with k0 = 4 and ∆k = 1. 12 CHAPTER 1. WAVES IN ONE DIMENSION

k0 = 4.000000; dk = 2.000000; nwaves = 20

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-5.0

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Figure 1.11: Superposition of twenty sine waves with k0 = 4 and ∆k = 2.

∆k amplitude

0 4 8 12 16 20 wavenumber

Figure 1.12: Representation of the wavenumbers and amplitudes of 20 superimposed sine waves with k0 = 4 and ∆k = 2. 1.5. BEATS 13

We can better understand how wave packets work by mathematically analyzing the simple case of the superposition of two sine waves. Let us define k0 = (k1 + k2)/2 where k1 and k2 are the wavenumbers of the component waves. Furthermore let us set ∆k =(k2 − k1)/2. The quantities k0 and ∆k are graphically illustrated in figure 1.8. We can write k1 = k0 − ∆k and k2 = k0 + ∆k and use the trigonometric identity sin(a + b) = sin(a) cos(b)+ cos(a) sin(b) to find

sin(k1x) + sin(k2x) = sin[(k0 − ∆k)x] + sin[(k0 + ∆k)x] = sin(k0x) cos(∆kx) − cos(k0x) sin(∆kx)+ sin(k0x) cos(∆kx) + cos(k0x) sin(∆kx) = 2 sin(k0x) cos(∆kx). (1.17)

The sine factor on the bottom line of the above equation produces the oscillations within the wave packet, and as speculated earlier, this oscillation has a wavenumber k0 equal to the average of the wavenumbers of the component waves. The cosine factor modulates this wave with a spacing between regions of maximum amplitude of ∆x = π/∆k. (1.18) Thus, as we observed in the earlier examples, the length of the wave packet ∆x is inversely related to the spread of the wavenumbers ∆k (which in this case is just the diﬀerence between the two wavenumbers) of the component waves. This relationship is central to the uncertainty principle of quantum mechanics.

1.5 Beats

Suppose two sound waves of diﬀerent frequency impinge on your ear at the same time. The displacement perceived by your ear is the superposition of these two waves, with time dependence

A(t) = sin(ω1t) + sin(ω2t) = 2 sin(ω0t) cos(∆ωt), (1.19) where we now have ω0 = (ω1 + ω2)/2 and ∆ω = (ω2 − ω1)/2. What you actually hear is a tone with angular frequency ω0 which fades in and out with period

Tbeat = π/∆ω =2π/(ω2 − ω1)=1/(f2 − f1). (1.20) 14 CHAPTER 1. WAVES IN ONE DIMENSION

The beat frequency is simply

fbeat =1/Tbeat = f2 − f1. (1.21)

Note how beats are the time analog of wave packets — the mathematics are the same except that frequency replaces wavenumber and time replaces space.

1.6 Interferometers

An interferometer is a device which splits a beam of light into two sub- beams, shifts the phase of one sub-beam with respect to the other, and then superimposes the sub-beams so that they interfere constructively or destructively, depending on the magnitude of the phase shift between them. In this section we study the Michelson interferometer and interferometric eﬀects in thin ﬁlms.

1.6.1 The Michelson Interferometer

The American physicist Albert Michelson invented the optical interferometer illustrated in figure 1.13. The incoming beam is split into two beams by the half-silvered mirror. Each sub-beam reflects off of another mirror which returns it to the half-silvered mirror, where the two sub-beams recombine as shown. One of the reflecting mirrors is movable by a sensitive micrometer device, allowing the path length of the corresponding sub-beam, and hence the phase relationship between the two sub-beams, to be altered. As figure 1.13 shows, the difference in path length between the two sub-beams is 2x because the horizontal sub-beam traverses the path twice. Thus, constructive interference occurs when this path difference is an integral number of wavelengths, i. e.,

2x = mλ, m =0, ±1, ±2,... (Michelson interferometer) (1.22) where λ is the wavelength of the light and m is an integer. Note that m is the number of wavelengths that ﬁts evenly into the distance 2x. 1.6. INTERFEROMETERS 15

Michelson interferometer

movable mirror

half-silvered mirror

d + x

fixed mirror

Figure 1.13: Sketch of a Michelson interferometer.

incident wave fronts d A

n = 1 n > 1

Figure 1.14: Plane light wave normally incident on a transparent thin film of thickness d and index of refraction n > 1. Partial reflection occurs at the front surface of the film, resulting in beam A, and at the rear surface, resulting in beam B. Much of the wave passes completely through the film, as with C. 16 CHAPTER 1. WAVES IN ONE DIMENSION 1.7 Thin Films

One of the most revealing examples of interference occurs when light interacts with a thin film of transparent material such as a soap bubble. Figure 1.14 shows how a plane wave normally incident on the film is partially reflected by the front and rear surfaces. The waves reflected off the front and rear surfaces of the film interfere with each other. The interference can be either constructive or destructive depending on the phase difference between the two reflected waves. If the wavelength of the incoming wave is λ, one would naively expect constructive interference to occur between the A and B beams if 2d were an integral multiple of λ. Two factors complicate this picture. First, the wavelength inside the film is not λ, but λ/n, where n is the index of refraction of the film. Constructive interference would then occur if 2d = mλ/n. Second, it turns out that an additional phase shift of half a wavelength occurs upon reflection when the wave is incident on material with a higher index of refraction than the medium in which the incident beam is immersed. This phase shift doesn’t occur when light is reflected from a region with lower index of refraction than felt by the incident beam. Thus beam B doesn’t acquire any additional phase shift upon reflection. As a consequence, constructive interference actually occurs when

2d =(m +1/2)λ/n, m =0, 1, 2,... (constructive interference) (1.23) while destructive interference results when

2d = mλ/n, m =0, 1, 2,... (destructive interference). (1.24)

When we look at a soap bubble, we see bands of colors reflected back from a light source. What is the origin of these bands? Light from ordinary sources is generally a mixture of wavelengths ranging from roughly λ =4.5 × 10−7 m (violet light) to λ = 6.5 × 10−7 m (red light). In between violet and red we also have blue, green, and yellow light, in that order. Because of the different wavelengths associated with different colors, it is clear that for a mixed light source we will have some colors interfering constructively while others interfere destructively. Those undergoing constructive interference will be visible in reflection, while those undergoing destructive interference will not. 1.8. MATH TUTORIAL — DERIVATIVES 17

y y = y(x) B

∆ y tangent to A curve at point A ∆x

slope = ∆y /∆x

Figure 1.15: Estimation of the derivative, which is the slope of the tangent line. When point B approaches point A, the slope of the line AB approaches the slope of the tangent to the curve at point A.

Another factor enters as well. If the light is not normally incident on the film, the difference in the distances traveled between beams reflected off of the front and rear faces of the film will not be just twice the thickness of the film. To understand this case quantitatively, we need the concept of refraction, which will be developed later in the context of geometrical optics. However, it should be clear that different wavelengths will undergo constructive interference for different angles of incidence of the incoming light. Different portions of the thin film will in general be viewed at different angles, and will therefore exhibit different colors under reflection, resulting in the colorful patterns normally seen in soap bubbles.

1.8 Math Tutorial — Derivatives

This section provides a quick introduction to the idea of the derivative. Often we are interested in the slope of a line tangent to a function y(x) at some value of x. This slope is called the derivative and is denoted dy/dx. Since a tangent line to the function can be deﬁned at any point x, the derivative