Physics 142 Wave Optics 1 Page ! 1

Wave Optics 1

For every complex problem there is one solution that is simple, neat, and wrong.— H.L. Mencken

Interference and diffraction of wavesThe essential characteristic of energy transport by waves is that waves obey the superposition principle. This means that two waves in the same spatial region can combine and interfere, rearranging the distribution of energy in space into a pattern often quite different from that of either wave alone. Light propagates as a wave, so we will analyze its interference phenomena.We distinguish two kinds of interference. First we treat “simple” interference of two harmonic waves with the same frequency traveling in the same direction in the same medium. The chief example of this occurs in the light reflected from the upper and lower surfaces of a thin transparent film.More complicated is the phenomenon called diffraction, which occurs when part of a wavefront is obstructed. The waves representing the part not obstructed produce a pattern of intensity varying with direction, resulting in a “bending” of the energy flow from its original direction. The most important examples of this involve passing light through an opening of some sort.

The mathematics of superposition: the Euler trickWe begin with a mathematical problem: How do we find the wave-function for the combined wave resulting from two or more waves?We consider two harmonic e-m waves of the same frequency and wavelength, both moving along the x-axis, but differing in phase by the angle ! . The wave-functions are the expressions describing the E-fields. We will assume the fields oscillate in the y-direction, so by E we mean the y-component of E. We have for the two waves:

!

The superposition of these waves results in another harmonic wave with the same frequency and wavelength, also moving in the x-direction and oscillating in the y-direction, so the combined wave function ! must have the general form:

!

Our problem is to find the constants ! and ! in terms of the amplitudes of the original waves and the phase difference ! . Our main interest is in the intensity of the resulting wave, which is proportional to ! ; we are less interested in ! .

To solve this problem we employ a trick based on a famous mathematical theorem:

δ

E1(x,t) = E01 cos(kx −ωt)

E2(x,t) = E02 cos(kx −ωt +δ )

E = E1 +E2E(x,t) = E0 cos(kx −ωt +φ)

E0 φδE02 φ

Physics 142 Wave Optics 1 Page ! 2

Here ! is the imaginary unit. This remarkable formula says that exponential functions and trigonometric functions are related through complex numbers.

This equation is discussed briefly in the Mathematical Notes.

Let us first review some facts about complex numbers. Any complex number z can be written in two forms, related by Euler’s theorem:

!

In the first form, x and y are real numbers; x is the real part of z [written as x = Re(z)] while y is the imaginary part of z [written as y = Im(z)]. In the other form, r is the amplitude and ! is the phase of z. From Euler’s theorem we find

.

One can display a complex number graphically by showing the real and imaginary parts in a two-dimensional plot, as shown. Each point on the diagram corresponds to a particular complex number. It can be represented by the vector giving the location of the point, as shown.This two-component vector representing z is often called a phasor by engineers. Addition of two complex numbers is accomplished by adding the phasors, using the usual rules for adding vectors.

Introductory textbooks often introduce phasors to treat superposition of oscillating functions and waves, usually without explaining that these are graphical representations of complex numbers in an abstract space. This leads to confusion when dealing with e-m waves because the E-field is itself a vector — in real space — and the phasor is not a direct representation of the field E. We will use the complex numbers directly, not as phasors, and use algebra rather than geometry.

The complex conjugate of a complex number (denoted by an asterisk) is obtained by replacing i by –i wherever it appears. Thus

!

The product of z and z* is a positive real number, and it is the square of the amplitude:!

Two more important facts about complex numbers:• Two complex numbers are equal if and only if their real parts are equal and their

imaginary parts are equal. This is two separate conditions.• The real part of the sum of two complex numbers is the sum of their individual

real parts. (The same is true of imaginary parts.)Now we return to waves, and apply some of these results. We note that we could write the E-field of, say, the first wave above in the form

!

and similarly for the other wave, and for the total wave resulting from their addition.

i = −1

z = x + iy, or z = reiθ

θx = rcosθ , y = rsinθ

z* = x − iy or z* = re− iθ

zz* = x2 + y2 = r2

E1(x,t) = Re[E01ei(kx−ωt)]

y

xθ

r

Euler’s theorem ! eiθ = cosθ + isinθ

Physics 142 Wave Optics 1 Page ! 3

Now we use the fact that the sum of real parts is the real part of the sum. We can temporarily replace the actual wave-functions by the corresponding complex exponential forms, add those, and then take the real part of the result to get the answer we want. That is the trick.In our case, we define (using a superscript c for complex)

!

Our actual wave-functions are the real parts of these. We add the first two and set the result equal to the third. Dividing out a common factor ! , we then find

!

This equation between complex numbers is two equations, one for the real parts and one for the imaginary parts. These two equations allow us to solve for ! and ! .

The advantage of this method is that algebra of exponentials is easier than algebra of sines and cosines.

To get the intensity of the resulting wave we need only ! . We simply multiply each side of the above equation by its complex conjugate and use the fact that ! :

!

From Euler’s theorem we see that the ( ) in the last term is ! , so we have

.

This equation allows us to relate the intensity of the combined wave to the intensities of the original waves and the phase difference ! . We use the fact that for any e-m wave ! , where K is a constant.

If I represents the average intensity over a cycle, then ! .

We multiply every term of the above equation by K. Denoting the intensities of the original waves by ! and ! , and calling the intensity of the combined wave I, we find:

This is the general formula for interference of two harmonic waves of the same wavelength and frequency, moving in the same direction but out of phase by ! .

This result is valid for interference of any kind of waves, including sound.

E1c = E01e

i(kx−ωt)

E2c = E02e

i(kx−ωt+δ )

Ec = E0ei(kx−ωt+φ )

ei(kx−ωt)

E0eiφ = E01 +E02e

iδ

E0 φ

E02

eiφ ⋅e− iφ = 1

E02 = (E01 +E02e

iδ ) ⋅(E01 +E02e− iδ )

= E012 +E02

2 +E01E02 (eiδ + e− iδ )

2cosδ

E02 = E01

2 +E022 + 2E01E02 cosδ

δI = KE0

2

K = 12 cε0

I1 I2

δ

Interference of two waves ! I = I1 + I2 + 2 I1I2 cosδ

Physics 142 Wave Optics 1 Page ! 4

The maximum intensity (constructive interference) occurs when ! (i.e. ! is a multiple of ! ). Minimum intensity (destructive interference) occurs when ! (i.e., when ! is an odd multiple of ! ). We find from the above formula:

!

In most of the cases we will treat, the two waves have the same amplitude (therefore the same intensities alone), in which case the resulting wave has intensity

!

This case occurs in most of our examples and problems, so this is a very useful formula.

For this case, destructive interference gives zero intensity, while constructive interference gives four times the intensity of one wave alone.

The method outlined here will be generalized later to many waves, in the treatment of diffraction.

Interference in thin filmsAn important application of these formulas is the case of light reflected from the two surfaces of a thin transparent film. Shown in cross section is an example, where the film is the shaded region.Light is incident from above, coming from a medium with refractive index ! . Wave 1 is the wave reflected at point a. The transmitted wave enters the film, which has index ! . Part of it is reflected at point b, and part of this wave emerges back into the original medium as wave 2. We consider the interference between waves 1 and 2.

We are interested in the case of normal incidence — the angles in the drawing are exaggerated for clarity. At normal incidence the reflectivity R is usually quite small. It is shown in the assignments that when R is small waves 1 and 2 have approximately the same amplitude and other waves resulting from more reflections within the film have much smaller amplitudes and can be neglected.

The phase difference between these waves results from two causes:• The difference in path followed by the two waves before they are brought

together in a detector (such as the eye of an observer). If a wave travels a distance ! in as medium with wave number k, its phase increases by ! .

• Possible phase changes upon reflection at points a and b on the surfaces.We deal first with path difference. When light travels from one transparent medium to another with a different index of refraction, the frequency of the wave does not change. But because the wave speed changes, the wavelength ! will be different, and so will the value of ! . The ratio of k in the medium to that in the vacuum is

!

cosδ = +1 δ2π cosδ = −1δ π

Constructive interference: I = Imax = I1 + I2⎡⎣

⎤⎦

2

Destructive interference: I = Imin = I1 − I2⎡⎣

⎤⎦

2

Waves of equal amplitude: I = 2I1(1+ cosδ )

n1

n2

Δx δ path = kΔx

λ = v/ fk = 2π /λ

kkvac

= cv= n

n1

n3

n2a

b

1 2

t

Physics 142 Wave Optics 1 Page ! 5

As wave 2 travels the extra distance 2t in medium 2, the phase of its wave-function increases by ! .

We will always use ! to represent the vacuum wavelength, which is essentially the same as the wavelength in air. The wavelength in a medium with refractive index n is ! .

Now we consider phase changes on the reflections at a and b.For normal incidence there are simple rules (for any kind of wave, including light) about phase changes on reflection:

• If the waves have greater speed in the incident medium, then the reflected wave undergoes a phase change of π upon reflection.

• If the waves have lower speed in the incident medium, the reflected wave undergoes no phase change upon reflection.

In terms of light, if waves are incident from medium 1 onto medium 2:

Call ! the reflection phase change at a, and ! that at b . These numbers are either 0 or ! , according to the rule given above. The net phase difference when the two waves come back together is thus

.

The absolute value is used so the overall reflection phase change is never negative.

This formula gives the value of ! to be put into the general intensity equations given earlier. It is the general formula for all thin film cases.Case 1: ! and ! . (Example: a thin film of a fluid with air on both sides.) Then ! and � . This gives

.

As the film thickness shrinks to zero (the fluid film is about to break) ! , so there is destructive interference in the reflected light.Case 2: ! and ! . (Example: A thin film of a fluid on a glass plate, with air above.) Then ! and the reflection phase changes cancel, giving

.

As t goes to zero (maybe the fluid evaporates) ! and there is constructive interference for all wavelengths. If ! , then ! and there is destructive interference for that wavelength.

δ path = k2 ⋅2t = 4πn2t/λλ

λ/n

δ a δ bπ

δ = 4πn2t/λ + δ b −δ a

δ

n1 < n2 n2 > n3δ a = π δ b = 0

δ = 4πn2t/λ +π

δ → π

n1 < n2 n2 < n3δ a = δ b = π

δ = 4πn2t/λ

δ → 0n2t = λ/4 δ = π

Phase changes on reflection

1. If ! , the reflected wave changes phase by ! .

2. If ! , there is no phase change in the reflected wave.

n1 < n2π

n1 > n2

Physics 142 Wave Optics 1 Page ! 6

Case 3: ! and ! . (Example: a film of air between two glass surfaces.) Then ! and ! . This gives

! ,

the same as in Case 1. Where the thickness of the film is zero, there is destructive interference. The case shown in the photograph illustrates this.

Shown is the pattern called Newton’s rings, formed by the thin film of air between a flat glass plate and a slightly convex glass lens resting on it. The reflected monochromatic green light is viewed from above. The bright circles arise from constructive interference, the dark ones from destructive interference. (The glass surfaces are not perfect, so the rings are not perfect circles.) The region at the center, where the two surfaces touch, is dark.

In order for the interference effects to be easily observable the film must be no more than a few wavelengths thick. The reason for this is that no actual light source emits only a single wavelength; there is always a spread of wavelengths around the average. If the film is thick, so that the ratio ! is large, then a small change in ! will result in a large enough shift in ! to change destructive interference into constructive, or vice versa. These opposite patterns for nearby wavelengths overlap and blur each other’s effects, so no overall pattern is visible. This is why we do not observe interference patterns in light reflected from the two surfaces of ordinary window glass.

This is analyzed in one of the assignments.

n1 > n2 n2 < n3δ a = 0 δ b = π

δ = 4πn2t/λ +π

t/λ λδ