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138 Part II: Physical Optics

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139

PART II

PHYSICAL ORWAVE OPTICS

In geometrical optics, we used the ray model to representthe behavior of light, which included the laws of reflectionand refraction. The only hint that a more comprehensivemodel was needed occurred when we talked about color interms of wavelength. This more inclusive model is calledwave optics, or by the older name, physical optics, wherephysical is the classical term that refers to matter or the ma-terial world. Explicitly, here in Part II, we now model lightas a wave that travels through space or matter.

Whether light behaved as a stream of particles or waveshas an interesting history. Newton argued for the particlemodel, but then experiments were performed that more eas-ily explained in terms of the wave model. Now we know thatlight is both, it has both wave-like and particle-like properties.In the chapters that follow, we describe the wave-like prop-erties, which ultimately are a part of electromagnetic theorydevised by Maxwell in latter part of the 19th century.

The first experiments that required the wave theory forexplanation occurred in the early part of the 19th century.

These experiments showed that light could produce interfer-ence patterns, regions of alternating dark and bright areas.But the vibration that composes a wave can be longitudinal,a vibration in the same direction as wave travel (like soundwaves), or transverse, a vibration perpendicular to the direc-tion of travel. Eventually, it was concluded that light behavedas a transverse wave. Then, with the advent of electromag-netic theory, the vibration was envisioned as an electric vectorand a magnetic vector.

We describe waves in terms of cosine and sine functions,which means that trigonometry is important. A big help inperforming the mathematics of these functions is provided bythe exponential function in the complex plane for it combinesin one function both the cosine and the sine properties. Al-though the mathematics is more abstract, the manipulationis usually much simpler. We start our discussion on waveoptics by developing the mathematics in the complex plane.Then we review the properties of waves, and show how todescribe waves in the complex plane.

140 Chapter 5: Complex Algebra and Harmonic Waves


141

Chapter 5—OutlineComplex Algebra and Harmonic Waves

Chapter 5: Complex Algebra and Harmonic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1435.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1435.2 Complex Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

5.2.1 Representation of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143Example 5.2.1 Rectangular to polar form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144Example 5.2.2 Polar to rectangular form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144Example 5.2.3 Several important complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

5.2.2 The negative of a complex number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1455.2.3 Addition of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1455.2.4 Subtraction of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

Example 5.2.4 Addition and subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1465.2.5 Multiplication of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1465.2.6 Division of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1475.2.7 The complex conjugate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1475.2.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

Example 5.2.5 Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148Example 5.2.6 Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148Example 5.2.7 The complex conjugate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

5.2.9 More properties of complex quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148de Moivre’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148The conjugate of a conjugate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148The complex conjugate of a product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149The real part of a sum equals the sum of the real parts . . . . . . . . . . . . . . . . . . . . . . . . . . 149The real part of a product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

Example 5.2.8 Work with Equations 5.35 and 5.36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1495.2.10 The basic idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

5.3 Harmonic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1515.3.1 Traveling harmonic waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

Travel in the +z direction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151Travel in the −z direction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152The phase θ and the phase constant φ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152The period T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153The frequency ν and the wave number σ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153The angular wave number k and angular frequency ω . . . . . . . . . . . . . . . . . . . . . . . . . . 154

5.3.2 Representation in the complex plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1545.3.3 Summary of traveling harmonic waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1555.3.4 The intensity I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1555.3.5 Traveling harmonic waves in other media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1575.3.6 The importance of the optical path difference � and the phase difference δ . . . . . . . . . . . . . . . . 157

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160Answers to Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

(x, y)

Re

Im

x

y (x, y)

Re

Im

x

y

z

(a) (b)

(x, y)

Re

Im

x

y

θ

z

|z|

143

Chapter 5

Complex Algebra and Harmonic Waves5.1 Introduction

Many of the properties of light are modeled in terms of si-nusoidal or harmonic waves. By a harmonic wave we meana sine function, a cosine function, or one of these functionswith a phase constant; that is, we use the name harmonic asa generic term. In general, these harmonic waves are trav-eling waves: their characteristics we shall describe later inthe chapter. Because complex algebra provides an efficientmeans of manipulating expressions that contain harmonicwaves, we start with a description of some of the features ofcomplex algebra.

5.2 Complex Algebra

5.2.1 Representation of complex numbersComplex algebra deals with complex numbers: they are rep-resented in rectangular form or Cartesian form as

z = x + iy (5.1)

where

x = a real number called the real part of z

= Re(z) (5.2a)

y = a real number called the imaginary part of z

= Im(z) (5.2b)

i = √−1 (5.2c)

As an aid to understand the properties of complex numbers,we represent a complex number in a plane called the complexplane (also called an Argand diagram). Two ways of display-ing a complex number in the complex plane are shown in Fig-ure 5.1. As the diagrams show, we graph the real part x alongthe horizontal axis (the Re axis), and the imaginary part yalong the vertical axis (the Im axis). In Figure 5.1 (a), werepresent the complex number as a point; in Figure 5.1 (b), asa vector-like quantity called a phasor—thus, we have a pha-sor diagram for the complex number z. Phasors have someof the properties of vectors in that they add and subtract like

Figure 5.1

the vectors used in mechanics; however, unlike vectors, pha-sors are also multiplied and divided. Just like vectors, thex and y are regarded as the components of the phasor z.The i is always inserted somewhere in the y expression; inour notation, we usually put it in front of the expression asin Equation 5.1.

From Equation 5.2c, we get the important relations

i2 = −1 (5.3a)

i = −1

i(5.3b)

and in general:

i n =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

i; n = 1, 5, 9, . . .

− 1; n = 2, 6, 10, . . .

− i; n = 3, 7, 11, . . .

1; n = 4, 8, 12, . . .

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭

(5.4)

A complex number z also has a polar form representa-tion, as we illustrate with the phasor diagram in Figure 5.2.

Figure 5.2

By inspection we read off

z = x + iy = |z| cos θ + i |z| sin θ (5.5a)

= |z|(cos θ + i sin θ) (5.5b)

where

|z| = mag(z) = the magnitude (or absolute value) of z

=√

x2 + y2 (5.6a)

θ = arg(z) = the angle the phasor makeswith the positive Re axis (or witha line parallel to the Re axis)

= tan−1( y

x

)(5.6b)

By default, we usually write θ so that −π < θ ≤ π ; however,any multiple of 2π when added or subtracted is an acceptablevalue for θ .

3.61

Re

Im

z

−2

−3 −123.7◦


We now want to develop a new relationship, which willallow us to set the sine/cosine expression of Equation 5.5bequal to a new and shorter expression that makes complexalgebra so useful for working with harmonic expressions. Westart with the power series for the exponential function ex ,where x is a real number:

ex = 1 + x + x2

2!+ x3

3!+ x4

4!+ x5

5!+ x6

6!+ x7

7!+ . . . (5.7)

This equation suggests that to define the exponential functionin the complex plane, we should replace x by the complexnumber z; thus

ez = 1 + z + z2

2!+ z3

3!+ z4

4!+ z5

5!+ z6

6!+ z7

7!+ . . . (5.8)

Now let’s suppose that z equals the complex number iθ ,where θ is an angle measured in radians. Then Equation 5.8becomes, with the help of Equation 5.4,

eiθ = 1 + iθ + (iθ)2

2!+ (iθ)3

3!+ (iθ)4

4!

+ (iθ)5

5!+ (iθ)6

6!+ (iθ)7

7!+ . . .

= 1 + iθ − θ2

2!− i

θ3

3!+ θ4

4!

+iθ5

5!− θ6

6!− i

θ7

7!+ . . .

= 1 − θ2

2!+ θ4

4!− θ6

6!+ − . . .

+i

(θ − θ3

3!+ θ5

5!− θ7

7!+ − . . .

)

= cos θ + i sin θ (5.9)

because the power series for cos θ is

cos θ = 1 − θ2

2!+ θ4

4!− θ6

6!+ − . . . (5.10a)

and the analogous one for sin θ is

sin θ = θ − θ3

3!+ θ5

5!− θ7

7!+ − . . . (5.10b)

The result embodied in Equation 5.9, namely

eiθ = cos θ + i sin θ (5.11)

is the famous equation called Euler’s identity: This equationhas played an important role in the development of mathe-matics. A related equation is quickly obtained from Euler’sidentity as follows:

e−iθ = ei(−θ) = cos(−θ) + i sin(−θ)

= cos θ − i sin θ (5.12)

Finally, with Euler’s identity in Equation 5.11, we changeEquation 5.5b to give the elegant and useful expression for acomplex number z in polar form as

z = |z| eiθ (5.13)

When adding and subtracting complex numbers, therectangular form of z in Equation 5.1 is easiest to use; whenmultiplying and dividing, the polar form in Equation 5.13 ismost convenient.

Example 5.2.1 Rectangular to polar form.Suppose a complex number is given in rectangular form as(see Equation 5.1)

z = −2 − i3

To change to polar form we use Equations 5.6 to calculate

|z| =√

x2 + y2 =√

(−2)2 + (−3)2 = 3.61

θ = tan−1( y

x

)= tan−1

(−3

−2

)= −2.16 rad = −123.7◦

and then write z in polar form with Equation 5.13:

z = 3.61 ei(−2.16) = 3.61 ei(−123.7◦)

Note that for simplicity we do not write the rad unit (the ra-dian) beside the −2.16 value in ei(−2.16) because the rad issupplementary unit (and can be dropped when no confusionarises); thus, we treat the −2.16 as a real number with nophysical units (see Section 1.1.2). However, for clarity, weinclude the degree unit written as ◦. Also, note that whenwe substitute the values into tan−1(y/x) we include the signs(here, minus signs) so that we can determine the proper quad-rant for θ . The phasor diagram for z is shown in Figure 5.3.

Figure 5.3

Example 5.2.2 Polar to rectangular form.We start with a complex number in polar form:

z = 10 ei30◦ = 10 ei0.5236

With the help of Equation 5.5a, we calculate

x = |z| cos θ = 10 cos 30◦ = 8.66

y = |z| sin θ = 10 sin 30◦ = 5

Re

Im

z

8.66

510

30◦

Re

Im

1 z

(a)

Re

Im

−1

π

180◦or

1z

(b)

Re

Im

1

z

π/290◦

or

(a)

Re

Im

1−1

−π/2−90◦

or

z

(b)

θ − π

θ − 180◦

Re

Im

z

(a) (b)

−z

θ

Re

Im

z

−z

θ

θ + π

θ + 180◦

Re

Im

z1

z2

x1

x2y1

y2

Re

Im

z1

z2

(a) (b)

z 1+ z 2

z 1+ z 2

5.2 Complex Algebra 145

and substituting into Equation 5.1, we obtain z in rectangularor Cartesian form:

z = x + iy = 8.66 + i5

In Figure 5.4, we display the phasor diagram for z.

Figure 5.4

Example 5.2.3 Several important complex numbers.In rectangular and polar form, we have

z = 1 = ei0 (5.14a)

z = −1 = 1 eiπ = eiπ = ei180◦(5.14b)

with their phasor diagrams shown in Figure 5.5. For z = −1in Figure 5.5(b), the angle is written as π rad or 180◦; also, the−1 in the diagram represents the value of x when z is writtenin rectangular form—the 1 is the magnitude of z when writtenin polar form. It is interesting to note that when −1 = eiπ iswritten as eiπ + 1 = 0, we have a relation that connects fiveof the fundamental numbers of mathematics.

Figure 5.5

In a similar manner, we write

z = i = 1 eiπ/2 = eiπ/2 = ei90◦(5.15a)

z = −i = 1 ei(−π/2) = ei(−π/2) = ei(−90◦) (5.15b)

and show the phasor diagrams in Figure 5.6.

Figure 5.6

5.2.2 The negative of a complex numberIf a complex number z in rectangular form is

z = x + iy (5.16)

then the negative of z is defined as

−z = −x + i(−y) = −x − iy (5.17)

In polar form, we have

−z = |z| ei(θ±π) = |z| ei(θ±180◦) (5.18)

We write ±π or ±180◦ in the argument to put the angle inthe proper quadrant; that is, −π < (θ ± π) ≤ π , or in termsof degrees, −180◦ < (θ ± 180◦) ≤ 180◦ (see the statementunder Equation 5.6b). We illustrate with the phasor diagramsin Figure 5.7. The solid dots are used to mark the tails of thephasors at the origin.

Figure 5.7

5.2.3 Addition of complex numbersIf z1 and z2 are complex numbers in rectangular form, namely,

z1 = x1 + iy1 and z2 = x2 + iy2

then addition is defined as

z = z1 + z2 = (x1 + x2) + i(y1 + y2) (5.19)

and is illustrated in the phasor diagrams of Figure 5.8. In

Figure 5.8 The (b) diagram is a simplified form of (a).

Re

Im

z1

z2

Re

Im

(a) (b)

z1

−z2

z1 −

z2

z1

z2 z1 −

z2

Re

Im

z 1+ z 2z1

z2

Re

Im

Re

Im

z1

z2

z1 − z2

z2 − z1z1

z2

(b)(a)

Re

Im

z1

z2

z1z2

θ1

θ2

θ1 + θ2

|z1||z2||z1||z2|


Figure 5.8(a), we include the real part x1 and imaginarypart y1 of z1, and similarly (x2, y2) for z2, to better displayhow they determine the new complex number z1 + z2. InFigure 5.8(b), we simply draw the phasor diagram of thethree complex numbers. We note how the phasor diagramfor addition is a head-to-tail diagram just like that for vectoraddition. Also, we note by looking at z2 that the tail of aphasor does not have to begin at the origin—it is permissibleto move a phasor parallel to itself.

5.2.4 Subtraction of complex numbersWe define subtraction in terms of addition and the negativeof a complex number; thus

z = z1 − z2 = z1 + (−z2) (5.20)

The phasor diagram for subtraction is shown in Figure 5.9.The drawing in Figure 5.9(a) shows a parallelogram with thedashed phasors displaying z1+(−z2); the solid phasors showa quick way to think of z1 − z2, namely, that in subtractionthe tails of the phasors touch with the phasor that representsz1 − z2 having its head touching the head of z1. The phasordiagram in Figure 5.9(b) emphasizes this latter approach, andis the way we normally draw subtraction. As a check, welook at Figure 5.9(b) and note that the phasors z2 and z1 − z2

have the head-to-tail property indicating that these phasorsare added to give the sum z1; that is,

z2 + (z1 − z2) = z2 + z1 − z2 = z1

Figure 5.9 The (b) drawing is a shorthand form of (a).

Example 5.2.4 Addition and subtraction.Suppose

z1 = 2 + i3 and z2 = 5 + i2

thenz1 + z2 = 7 + i5

and is diagrammed in Figure 5.10.

Figure 5.10

If we continue to let

z1 = 2 + i3 and z2 = 5 + i2

then we obtain two different answers by subtraction:

z1 − z2 = 2 + i3 − (5 + i2) = −3 + i

and

z2 − z1 = 5 + i2 − (2 + i3) = 3 − i

The phasor diagrams for z1 − z2 and z2 − z1 are shown in thedrawings of Figure 5.11.

Figure 5.11

5.2.5 Multiplication of complex numbers

Complex numbers are most easily multiplied when they arein polar form, although it is possible to multiply them inrectangular form as well, as we discuss later in this section.The definition in polar form is

z = z1z2 = (|z1| eiθ1) (|z2| eiθ2

)= |z1||z2| ei(θ1+θ2) (5.21)

The phasor diagram in Figure 5.12 illustrates multiplication,where it is assumed that both |z1| and |z2| are greater thanone. Except to observe how the angles add, there is no easygeometric interpretation for multiplication like there is foraddition and subtraction.

Figure 5.12

Re

Im

z1

z2

θ1

θ2

|z1||z2|

θ1 − θ2

z1

z2

|z1||z2|

(b)

Re

Im

z

z

|z|θ

−θ

|z|

(a)

Re

Im

z

z

x

y

−y

x


If z1 and z2 are given in rectangular form, then the ex-pressions are multiplied term by term to obtain

z = z1z2 = (x1 + iy1)(x2 + iy2)

= (x1x2 − y1 y2) + i(x2 y1 + x1 y2) (5.22)

where we have used i2 = −1.We can obtain Equation 5.22 from Equation 5.21 by us-

ing Euler’s identity in Equation 5.11 and some trig identities:

|z1||z2| ei(θ1+θ2) = |z1||z2| [cos(θ1 + θ2) + i sin(θ1 + θ2)]

= |z1||z2| [cos θ1 cos θ2 − sin θ1 sin θ2]

+i |z1||z2| [sin θ1 cos θ2 + cos θ1 sin θ2]

= |z1||z2|[

x1

|z1|x2

|z2| − y1

|z1|y2

|z2|]

+i |z1||z2|[

y1

|z1|x2

|z2| + x1

|z1|y2

|z2|]

= (x1x2 − y1 y2) + i(x2 y1 + x1 y2) (5.23)

5.2.6 Division of complex numbersDivision of complex numbers is also most easily defined interms of the polar form:

z = z1

z2= |z1| eiθ1

|z2| eiθ2= |z1|

|z2| ei(θ1−θ2) (5.24)

To illustrate division, we draw the phasor diagram in Fig-ure 5.13. For convenience, this diagram is based on the onefor multiplication in Figure 5.12. Again, just like in multi-plication, there is no easy geometrical interpretation.

Figure 5.13

To obtain a rectangular form for division, we start withz1 and z2 in rectangular form and write

z = z1

z2= x1 + iy1

x2 + iy2(5.25a)

To get z in rectangular form, we must remove the i in the de-nominator. A very useful procedure for manipulating com-plex number expressions of this type is to multiply the fractionby another fraction composed of the complex conjugate ofthe denominator; here, the fraction is (x2 − iy2)/(x2 − iy2).Thus, starting with Equation 5.25a,

z = x1 + iy1

x2 + iy2

x2 − iy2

x2 − iy2

= x1x2 + y1 y2

x22 + y2

2

+ ix2 y1 − x1 y2

x22 + y2

2

(5.26b)

an expression that is certainly more complicated than thecorresponding polar form one in Equation 5.24.

Equation 5.26b can also be obtained directly from thepolar form in Equation 5.24 following the method used toobtain Equation 5.23. We leave this derivation as a problemfor the reader.

5.2.7 The complex conjugateSuppose z is a complex number given in rectangular or polarform as

z = x + iy = |z| eiθ (5.27)

then the complex conjugate is named z and defined as

z = x − iy = |z| e−iθ (5.28)

In short, to turn a complex number into its conjugate simplyreplace i by −i ; even if the complex number is given by along expression, applying this rule works. We draw phasordiagrams to display the properties of z, z in Figure 5.14: inthe (a) diagram, we represent the phasors in rectangular form,and in (b), polar form.

Figure 5.14

One of the results we quickly obtain from Equations 5.27and 5.28 is that

zz = |z|2 (5.29)

Two more important results are

Re(z) = 1

2(z + z) (5.30a)

Im(z) = 1

2i(z − z) (5.30b)


Equations 5.30 provide an analytical method for finding thereal or imaginary part of a complex expression. We provethese equations by using the rectangular form as follows:

1

2(z + z) = 1

2(x + iy + x − iy) = x = Re(z)

1

2i(z − z) = 1

2i(x + iy − x + iy) = y = Im(z)

5.2.8 Summary

Addition and subtraction are most easily done when complexnumbers or expressions are in rectangular form; multiplica-tion and division when in polar form. In general, the complexconjugate helps to find the magnitude, the real part, and theimaginary part of a complex number or expression (see Equa-tions 5.29 and 5.30).

Example 5.2.5 Multiplication.Suppose

z1 = 10eiπ/6 and z2 = 2eiπ/18

Then Equation 5.21 says that

z = z1z2 = (10eiπ/6

) (2eiπ/18

) = 20ei2π/9

Example 5.2.6 Division.If

z1 = 4eiπ/3 and z2 = 2eiπ/15

then Equation 5.24 gives

z = z1

z2= 4eiπ/3

2eiπ/15= 2ei4π/15

Example 5.2.7 The complex conjugate.Suppose

z = 1

x + iy

and we want to find |z|2, Re(z), and Im(z). First, Equa-tion 5.29 leads to

|z|2 = zz =(

1

x + iy

) (1

x − iy

)= 1

x2 + y2

Then, Equation 5.30a gives

Re(z) = 1

2(z + z) = 1

2

(1

x + iy+ 1

x − iy

)

= 1

2

x − iy + x + iy

(x + iy)(x − iy)= x

x2 + y2

and in a similar way, Equation 5.30b yields

Im(z) = 1

2i(z − z) = 1

2i

(1

x + iy− 1

x − iy

)

= 1

2i

x − iy − x − iy

(x + iy)(x − iy)= − y

x2 + y2

5.2.9 More properties of complex quantitiesde Moivre’s theorem. We obtain an important result,

called de Moivre’s (deh Mwahv′ ris) theorem, that is help-ful in finding trig identities. We apply a rule of exponentsto obtain (

eiθ)n = einθ (5.31a)

and then use Euler’s identity in Equation 5.11 to get

(cos θ + i sin θ)n = cos nθ + i sin nθ (5.31b)

where it is Equation 5.31b that states de Moivre’s theorem.In practice, we usually apply this theorem with the left andright sides interchanged:

cos nθ + i sin nθ = (cos θ + i sin θ)n

We illustrate with n = 2, but any value of n works. Thus,

cos 2θ + i sin 2θ = (cos θ + i sin θ)2

= cos2 θ − sin2 θ + i(2 sin θ cos θ)

Equating the real and imaginary parts, we have the well-known identities

cos 2θ = cos2 θ − sin2 θ

sin 2θ = 2 sin θ cos θ

The conjugate of a conjugate. It is easy to prove that(z

) = z (5.32)

by using the rectangular form of a complex number (the polarform works just as well):(

z) = (

(x + iy)) = (x − iy) = x + iy = z

and we have proved Equation 5.32.


The complex conjugate of a product. To prove that

(z1z2) = z

1z2 (5.33)

it is easier to use the polar form; thus

(z1z2) = ((|z1|eiθ1

) (|z2|eiθ2))

= (|z1|e−iθ1) (|z2|e−iθ2

) = z1z

2

where we have found the complex conjugate by replacing iwherever it appeared by −i (see Section 5.2.7).

The real part of a sum equals the sum of the real parts.In equation form, this important statement reads

Re(z1 + z2) = Re(z1) + Re(z2) (5.34)

This equation includes subtraction because subtraction is de-fined in terms of addition (see Section 5.2.4). Using therectangular form, we first evaluate the left side

Re(z1 + z2) = Re(x1 + iy1 + x2 + iy2)

= Re[(x1 + x2) + i(y1 + y2)]

= x1 + x2

and then the right side

Re(z1) + Re(z2) = Re(x1 + iy1) + Re(x2 + iy2)

= x1 + x2

Either way we have obtained the same result proving Equa-tion 5.34. A similar equation holds for the imaginary parts.

The real part of a product. We might think that anexpression similar to Equation 5.34 should hold for a product;that is, multiplication (or division). However, the situation ismore complicated. First of all, Equation 5.30a always holds,even for a product:

Re(z1z2) = 1

2

[z1z2 + (z1z2)

]

(5.35)

This equation provides a simple way to express the calcula-tion of a product. But we shall show that

Re(z1z2) = 2 Re(z1) Re(z2) − Re(z1 z2) (5.36)

which does have some similarity with Equation 5.34; never-theless, it is definitely different. To prove Equation 5.36, westart by applying Equation 5.33 to Equation 5.35:

Re(z1z2) = 1

2

[z1z2 + (z1z2)

] = 1

2

[z1z2 + z

1z2

]

Next, we add zero twice in two different forms:

1

2

[z1z2 + z

1z2

]

= 1

2

[z1z2 + (z

1z2 − z1z2) + (z1z

2 − z1z2) + z

1z2

]Then we rearrange the terms, factor, and use Equations 5.32,5.33, and 5.30a:

1

2

[z1z2 + (z

1z2 − z1z2) + (z1z

2 − z1z2) + z

1z2

]

= 1

2

[z1z2 + z

1z2 + z1z2 + z

1z2 − z

1z2 − z1z2

]

= 1

2

[(z1 + z

1)z2 + (z1 + z1)z

2 − (z

1z2 + z1z2)

]

= 1

2

[(z1 + z

1)(z2 + z2) − (

z1z2 + (z

1z2))]

= 21

2(z1 + z

1)1

2(z2 + z

2) − 1

2

(z

1z2 + (z1z2)

)

= 2 Re(z1) Re(z2) − Re(z1z2)

and we have proved Equation 5.36. A somewhat similarequation holds for Im(z1z2).

It is usually easier to evaluate Re(z1z2) directly, ratherthan evaluate the right side of Equation 5.36. We give Equa-tion 5.36 for reference and to show that an equation similarto the one for addition, Equation 5.34, does not exist formultiplication; that is,

Re(z1z2) �= Re(z1) Re(z2) (5.37)

Example 5.2.8 Work with Equations 5.35 and 5.36.(a) Suppose the complex numbers are

z1 = 2 − i3 and z2 = −5 + i

Then we calculate directly

z1z2 = (2 − i3)(−5 + i) = −7 + i17

and we quickly read the real part as −7. But to show thatEquation 5.35 works, we evaluate it:

Re(z1z2) = 1

2

[z1z2 + (z1z2)

]

= 1

2[(−7 + i17) + (−7 − i17)] = −7

To use Equation 5.36, we first calculate

z1z2 = (2 + i3)(−5 + i) = −13 − i13


and then continue with Equation 5.36. By inspection andcalculation, we get

2 Re(z1) Re(z2) − Re(z1z2) = 2(2)(−5) − (−13) = −7

which agrees with what we got before.(b) We use the same complex numbers, but divide:

z1

z2= z1

(1

z2

)= (2 − i3)

(1

−5 + i

−5 − i

−5 − i

)

= (2 + i3)

(− 5

26− i

1

26

)= −1

2+ i

1

2

Then, we can read off the Re part by inspection to get −1/2,or use Equation 5.35 to obtain the same result:

Re

(z1

z2

)= 1

2

[(z1

z2

)+

(z1

z2

)]

= 1

2

[(−1

2+ i

1

2

)+

(−1

2− i

1

2

)]= −1

2

To apply Equation 5.36, we first calculate

z1

(1

z2

)= (2 + i3)

(− 5

26− i

1

26

)= − 7

26− i

17

26

and then

2 Re(z1) Re

(1

z2

)− Re

(z

11

z2

)

= 2(2)

(− 5

26

)−

(− 7

26

)= −1

2

5.2.10 The basic ideaWe frequently deal with functions of the form

E = A cos (bx + φ) (5.38)

which we can write as a complex expression, based on theinformation in the previous sections (in particular, see Euler’sidentity as given in Equation 5.11),

E = Re[Aei(bx+φ)

](5.39)

We usually want to perform mathematical operations on thecomplex expression inside the brackets, and it is these mathe-matical operations that are easier to achieve with the complexexpression than the noncomplex one. But it is often awk-ward to carry the Re notation forward in the mathematicaloperations when we simply want to manipulate the complexexpression. Thus, to remind us that we are working with a

complex expression and that in the end we need to take theRe part, we write the name in bold face:

E = Aei(bx+φ) (5.40)

Some of the quick operations we can perform with this polarnotation use a property of exponents:

E = Aei(bx+φ) = Aei(bx) eiφ = Aei(bx)

e−iφ(5.41)

However, the Re part of any of the expressions in Equa-tion 5.41 must equal the equation we started with, namelyEquation 5.38. To illustrate how this statement is true, weshall take the last expression in Equation 5.41, apply Euler’sidentity as listed in Equation 5.11, use some trig identities,multiply, and rearrange:

E = Aei(bx)

e−iφ= A

cos bx + i sin bx

cos φ − i sin φ

= Acos bx + i sin bx

cos φ − i sin φ

cos φ + i sin φ

cos φ + i sin φ

= A

[cos bx cos φ − sin bx cos φ

cos2 φ + sin2 φ

+ isin bx cos φ + cos bx sin φ

cos2 φ + sin2 φ

]

= A cos (bx + φ) + i A sin (bx + φ)

and we get the desired result of

E = Re(E) = A cos (bx + φ)

which agrees with Equation 5.38. But, we had to work hardto show this agreement; the quick and efficient ways of ma-nipulating harmonic expressions as shown in Equation 5.41is the reason for using complex notation.

However, we must be careful. Suppose we want tosquare E in Equation 5.38; that is,

E2 = A2 cos2 (bx + φ) (5.42)

Like before, we set

E = Aei(bx+φ) (5.43)

so thatE = Re (E) = A cos (bx + φ)

But thenE2 = E E = Re(E) Re(E)

which is not equal to Re(E2), or Re(E E), according to Equa-tions 5.36 and 5.37 (the boldface Es correspond to the zs).

0 z

E

E0

−E0

λ

−λ/4 λ/4 λ/2 3λ/4 5λ/4λ

z

EE0

E

z′

vt

v

P

t = 0 t = t

z

z′

5.3 Harmonic Waves 151

A cleaner method for handling this situation is to use Equa-tion 5.30a and write

E = Re(E) = 1

2

(E + E

)(5.44)

Then

E2 = (Re(E))2

= 1

4

(E + E

)2

= 1

4

[E2 + (

E)2 + 2EE

](5.45)

As a check, to show that this equation yields Equation 5.42,we substitute Equation 5.43, use Euler’s identity, use a trigidentity, and manipulate:

E2 = 1

4

[E2 + (

E)2 + 2EE

]

= 1

4

[(Aei(bx+φ)

)2 + (Ae−i(bx+φ)

)2 + 2A2]

= A2

4

[ei2(bx+φ) + e−i2(bx+φ) + 2

]

= A2

2[cos 2(bx + φ) + 1]

= A2

2

[2 cos2(bx + φ)

]= A2 cos2(bx + φ)

which does agree with Equation 5.42.

5.3 Harmonic Waves

5.3.1 Traveling harmonic wavesWe start with a very simple harmonic wave in space (that is,in a vacuum):

E = E0 cos2π

λz (5.46)

which we graph in Figure 5.15. The variable E is called the

Figure 5.15

displacement, the coefficient E0 the amplitude, and λ thespatial period, or wavelength (the distance in one cycle, orone repetitive unit).

Travel in the +z direction. Now we imagine that the wavemoves to the right with the speed v, so that in Figure 5.16 thesolid curve represents the wave position at time t = 0, andthe dashed curve the wave position at a later time t = t . If weimagine a new coordinate system, the z′ coordinate system,

Figure 5.16

to move with the dashed wave, then we see that any point Pon this wave is described by an equation in terms of z′ that isjust like the one in Equation 5.46, namely

E = E0 cos2π

λz′ (5.47)

As we show in Figure 5.16, in a time t the z′ origin (thepoint where the vertical dashed line crosses the z, z′ axis)moves a distance vt , and we read off for any point P on themoving wave

z′ = z − vt (5.48)

Substituting Equation 5.48 in Equation 5.47, we obtain anequation that describes the traveling wave in terms of z, aswell as t ,

E = E0 cos2π

λ(z − vt) (5.49)

which is an equation of a traveling harmonic wave down thepositive z axis. We observe that the traveling harmonic wavein Equation 5.49 is not only a function of z and t , but alsoof (z − vt); in fact, all waves that travel along the positivez axis, whether harmonic or not, are functions of (z − vt).

We use the cosine function to describe our harmonicwaves simply for convenience: first of all, the cosine functionis an even function; and second, the real part of a complexexpression in polar form involves the cosine. However, thedescription of harmonic waves could be done in terms of thesine function just as well.

Light is an electromagnetic field composed of electricand magnetic fields that travel together through a vacuum orother media. Because the human eye is sensitive only to theelectric field, we use E and E0 to represent the electric fieldin the preceding equations. In terms of SI units, a convenientunit of electric field is the volt/meter, or V/m; for z, λ themeter, and for t the second. However, multiples of theseunits may be preferable in some problems.

z′z

t = 0

z

EE0

E

z′

v

P

t = t

−vt

z

E0

E �z

E0 cos φ

t = 0


Travel in the −z direction. Now suppose the harmonicwave travels down the negative z axis, instead of the positivez. Then the diagram that is analogous to Figure 5.16 is shownin Figure 5.17. We insert a minus sign in front of vt because

Figure 5.17

both the speed v and time t are positive quantities, and thevt direction is in the negative direction, as we see from Fig-ure 5.17. Just as before in Figure 5.16, the dashed E axismoves with the traveling dashed wave, and the equation forthis wave is the same as Equation 5.47 (we renumber forconvenience):

E = E0 cos2π

λz′ (5.50)

Treating the dimension arrows for z′, z, and −vt like vec-tors, as we have done in previous chapters, we read off byinspection of Figure 5.17

z′ = z − (−vt) = z + vt (5.51)

We substitute Equation 5.51 into Equation 5.50, and obtainthe equation of a harmonic wave traveling along the negativez axis:

E = E0 cos2π

λ(z + vt) (5.52)

We observe the interesting feature of these travelingwave equations: those that represent travel along the posi-tive z axis are functions of (z − vt); those along the negativez axis are functions of (z + vt).

The phase θ and the phase constant φ. We do onemore thing to traveling harmonic waves to make them moregeneral: we subtract a constant φ (in radians) so that for aharmonic wave traveling in the +z direction

E = E0 cos2π

λ(z − vt) (5.53)

it becomes

E = E0 cos

[2π

λ(z − vt) − φ

](5.54)

We subtract φ because it makes its effect somewhat easier todescribe; however, φ can be positive or negative (or zero).

The argument of the cosine function is called the phase θ ;that is, in equation form

θ = phase = 2π

λ(z − vt) − φ (5.55)

where φ is the initial phase or the phase constant. To see theeffect of a positive phase constant, we look at waves whent = 0. Then, when φ is zero, Equation 5.54 becomes

E = E0 cos2π

λz (5.56)

as shown by the solid curve in Figure 5.18. The wave peakat the origin occurs when the phase θ is zero; thus

θ = 2π

λz1 = 0 or z1 = 0 (5.57)

Now we allow φ to have a positive value, and continue to lett = 0. Equation 5.54 reads in this case

E = E0 cos

(2π

λz − φ

)(5.58)

The peak that was at the origin is still described by θ = 0, sowe have

θ = 2π

λz2 − φ = 0 or z2 = λ

2πφ (5.59)

From Equations 5.57 and 5.59, we find

�z = z2 − z1 = λ

2πφ (5.60)

that is, the entire harmonic wave is shifted to the right by theamount �z, as shown by the dashed curve in Figure 5.18.We also see that if we set z = 0 in Equation 5.58, we have

E = E0 cos (−φ) = E0 cos φ (5.61)

where the last step follows because the cosine is an evenfunction. The value given by E0 cos φ is the point where thedashed curve crosses the E axis, as indicated in Figure 5.18.

Figure 5.18

z

E0

E

t = 0t = t1

t = t2

z = z1 = λz = 0

λ

1

2

0

v

E

E0

T

z = z1 = λ

t0 t1 t2

1

2

0

t

E

E0z = z1 = λ

t

z = z2


Thus, the effect of the phase constant φ is to shift the entireharmonic wave along the z axis. Summarizing, for fixed t ,a positive φ shifts the wave in the positive z direction; anegative φ shifts the wave in the negative z direction.

It is important to note that to follow the movement ofthe wave peak that was at the origin (look at the solid curvein Figure 5.18), we set the phase θ = 0—see Equations 5.57and 5.59. By choosing other values for z and t , we couldobtain another value for θ , and thereby follow the movementof another point on the wave.

The period T . So far we have graphed a traveling har-monic wave in terms of z at particular values of t . Now wewant to do the opposite: graph the wave in terms of t atparticular values of z. First, we imagine a harmonic wavetraveling in the +z direction with a speed v: in Figure 5.19,we show the waves at the times of t = 0, t1, and t2 to helpwith the visualization. To obtain a graph of the wave as a

Figure 5.19

function of t at a particular value of z, we choose a plane atz = z1 = λ. Then, we imagine that we set up a device thatmeasures the E value of the wave with time t as it passes theplane at z = z1 = λ, as indicated in Figure 5.19—we showthe E values we measure at times t = 0, t1, and t2 by thepoints 0, 1, 2. The graph we obtain as a function of time isthen shown in Figure 5.20. In this diagram, we denote thetime period (the time in one cycle), or simply the period, bythe symbol T . The period T plays the same role in time asthe wavelength λ does in space (see Figure 5.15).

Figure 5.20

If we choose a plane somewhat to the right of z = z1 = λ

in Figure 5.19, say at z = z2, then we obtain another graphof E versus t , shown by the dashed curve in Figure 5.21.

Figure 5.21

Now we return to Figure 5.19, and concentrate on thesolid curve. We imagine it to move to the right, and considerhow long it will take for that curve to move through thedistance of λ. We quickly see that the time would be justthe period T that is shown in Figure 5.20. Recalling thattime = distance/speed, we have

T = λ

v(5.62)

With Equation 5.62, we can rewrite the traveling harmonicwave equation of Equation 5.54 as follows:

E = E0 cos

[2π

λ(z − vt) − φ

]

= E0 cos

[2π

λ

(z − λ

Tt

)− φ

]

= E0 cos

[2π

(z

λ− t

T

)− φ

](5.63)

This form of the harmonic wave equation emphasizes thesimilar roles that λ and T have relative to space and time.

The frequency ν and the wave number σ . To helpunderstand frequency, we look at Figure 5.20, and observe

period = Ts

cycle

The frequency ν is defined as the reciprocal of the period T ;thus,

ν = frequency = 1

T

cycles

s(5.64)

The units, cycles/s, indicate that frequency refers to the num-ber of cycles in a second. The cycle unit is a tag unit; that is,it is not a real unit like the meter or the second. A tag unitis used to help understanding, and can be erased whenever itis not needed; for that reason, the frequency unit is writtensimply as s−1, or the SI unit is used: hertz (Hz).

t = t0

Re

Im

E

zE0 cos θ

θ0θ

λ

P0

P

E0 cos θ0

E0

E0

z


In space, the analog of the frequency ν is called thelinear wave number, or more simply the wave number σ .Following the format we just used to define frequency, welook at Figure 5.15, and observe that

spatial period = wavelength = λm

cycle

The wave number σ is defined as the reciprocal of the wave-length, so that

σ = wave number = 1

λ

cycles

m(5.65)

Here, the units of cycles/m are interpreted as the number ofcycles in a meter. Again, the cycle is a tag unit, and the unitfor σ is often written as m−1 and read as reciprocal meters.No special name is assigned to this unit.

With the definitions in Equations 5.64 and 5.65, we canrewrite Equation 5.63 for a traveling harmonic wave as

E = E0 cos

[2π

(z

λ− t

T

)− φ

]

= E0 cos [2π (σ z − νt) − φ] (5.66)

We make two more definitions to simplify Equation 5.66.

The angular wave number k and angular frequency ω.The final set of definitions we make streamline the travelingharmonic wave equation yet more. We define the angularwave number k as

k = 2π

λ= 2πσ (5.67)

and the angular frequency ω as

ω = 2π

T= 2πν (5.68)

Multiplying the 2π into the parentheses in Equation 5.66,and then substituting Equations 5.67 and 5.68, we get

E = E0 cos (kz − ωt − φ) (5.69)

which is the simplest form of the traveling harmonic waveequation; it is the one usually used in theoretical work. Theunits for k are the rad/m and for ω the rad/s.

5.3.2 Representation in the complex planeAs a complex expression in polar form, we write Equa-tion 5.69 as

E = E0 ei(kz−ωt−φ) (5.70)

where we write E in bold face to remind us that we have acomplex expression from which we must take the Re part toobtain Equation 5.69, as discussed in Section 5.2.10.

Next, we want to show how the phasor representationof Equation 5.70 can be interpreted graphically to give theharmonic wave E as a function of z at a fixed value of t , sayat t = t0. The result is shown in Figure 5.22; we now explainhow this diagram is constructed.

Figure 5.22

In terms of the phase θ , Equation 5.70 becomes

E = E0 eiθ (5.71)

whereθ = kz − ωt − φ (5.72)

At the particular time t = t0, we write Equation 5.72 as

θ = kz − ωt0 − φ = kz + θ0 (5.73)

whereθ0 = −ωt0 − φ = −(ωt0 + φ) (5.74)

Equation 5.71 indicates that the phasor magnitude is alwaysE0, so we draw a circle of this radius in Figure 5.22. To makeit easier to represent the cosine, we rotate the complex plane90 deg counterclockwise to make the Re axis vertical. Atz = 0, Equation 5.73 gives θ = θ0, where by Equation 5.74,θ0 < 0 for ωt0 + φ > 0. Thus, we draw θ0 as a negative anglein Figure 5.22. By inspection, we see that the projection ofthis phasor on the Re axis is E0 cos θ0, as noted in the diagram.Continuing to look at the diagram, a dashed line is drawnfrom the head of this phasor to the P0 point on the E versusz graph, the beginning point on this graph at z = 0. Nowlet z increase from 0, so that θ also increases according toEquation 5.73 making the phasor rotate counterclockwise inthe complex plane. We show one of the later locations of thisphasor in Figure 5.22 with its vertical projection on the Reaxis of E0 cos θ ; the dashed line from the head of this phasorlocates the point P on the harmonic wave graph. In thismanner, the rotating phasor builds up the E versus z graph inFigure 5.22. Note how the phasor making a complete rotationof 2π rad moves the point P through the wavelength λ (thespatial period) on the wave.

Representing a harmonic wave in polar form (see Equa-tion 5.70) in the complex plane is a helpful way to visualizethe wave. In Figure 5.22, we represented the case for z varieswith a fixed time t . A similar procedure can be described toget the E versus t graph at a fixed value of z.

wave number k = 2π

λ

= 2πσ

speed v = λ

T= λν = 1

σ T= ω

k

space time

wavelength λ

wave number σ = 1

λ

angularfrequency ω = 2π

T

= 2πν

period T

frequency ν = 1

T

angular

y

x

z

t = t1

z = z1z = z2

E0E

wavefronts

v


5.3.3 Summary of traveling harmonic waves

A traveling harmonic wave has a very simple form whengraphed, but it is more difficult to describe analytically be-cause of the several different ways to do it, as we have seenin the previous sections. In its simplest form, we write

E = E0 cos (kz − ωt − φ) (5.75)

which is the real part of the complex expression

E = E0 ei(kz−ωt−φ) (5.76)

In Section 5.3.1, we learned that k and ω could be expressedin terms of other quantities, each meaningful in its own wayin a physical sense. To try to summarize these different waysof viewing a traveling harmonic wave, we might say it is asingle entity like a coin with two sides: one side a spaceview, the other side a time view—each view with its own setof parameters. We summarize these two views in the tableof Figure 5.23. In the last row of this table, we write thespeed v of wave travel including several forms of what v

equals, starting with Equation 5.62, which we have solvedfor v. We note that each of the expressions for v contain twosymbols, one from the space column, the other from the timecolumn: in this manner, the speed v combines the two viewsof the traveling harmonic wave.

Figure 5.23 The parameters of a traveling harmonic wave.

As we have mentioned before, Equation 5.75 (or Equa-tion 5.76) describes a harmonic wave traveling in the +zdirection. This wave represents a light wave in terms of theelectric field E (there is also a magnetic field H that we ignorehere for simplicity, since the eye only responds to the electricfield), and it is traveling in the simplest medium that exists,the space of a vacuum. To represent position in this medium,we use an x, y, z coordinate system: then, at every x, y, zpoint in this system we associate an E value given by Equa-tion 5.75. Thus, we imagine harmonic waves traveling inthe +z direction everywhere in this space. To give a picture,

we choose a particular value of t , say t = t1, and drawthe diagram shown in Figure 5.24. In this way, we makea snapshot of a portion of the harmonic waves traveling inspace. We also show two surfaces called wavefronts, whichare surfaces of constant phase θ . For example, when z = z1,we look at Equation 5.75 (or Equation 5.76), and see that onthis surface

θ = θ1 = kz1 − ωt1 − φ

and on the surface z = z2, the phase θ has another constantvalue. In general, any surface of constant phase is calleda wavefront. Since in this particular case, the surfaces areplanes, harmonic waves of this type are called plane waves,and represent the simplest of light waves. In this special case,we note that the E value is also constant on these surfaces,but in general, that is not true.

We note that all the E values in Figure 5.24 are perpen-dicular to the z axis, the direction of travel; that is, light wavesare transverse waves. Although plane waves are simplewaves, they are very important because complicated wavescan be constructed by adding waves of this type with themethods of Fourier series and integrals.

5.3.4 The intensity I

All traveling waves carry energy in the direction of travel.Sound waves have this property; electromagnetic waves alsohave this property. Visible light waves are electromagneticwaves having wavelengths in a vacuum of 400 to 700 nm (seeFigure 0.1); therefore, visible light waves carry energy in thedirection of travel. Solving the equation v = λν in the tableof Figure 5.23 for ν, we calculate,

ν = v

λ= 3 × 108 m/s{

400 nm

700 nm

} ={

750 × 1012 Hz

430 × 1012 Hz(5.77)

where v = c = 3 × 108 m/s for the speed of light in a vac-uum. We note the high frequencies of visible light waves.

Figure 5.24 Plane waves.

z

Av


We represent the energy carried by the light waves withthe symbol U . Now suppose we have a detector of smallsurface area A that is able to measure U , and is placed in thepath of the light waves oriented perpendicular to the direction

of wave travel, as shown in Figure 5.25. If the detector mea-sures an energy dU passing through A in a time dt , then wedefine the instantaneous intensity Iinst as

Iinst = 1

A

dU

dt(5.78)

where the SI units for Iinst are joule/second . meter2 or, theequivalent form, watt/meter2; in symbol form: J/s . m2 orW/m2. In electromagnetic theory, it can be shown that Equa-tion 5.78 leads to

Iinst = C E2 (5.79)

where C is a constant with SI units of watt/volt2, or W/V2.The value of this constant is not needed in our work. Equa-tion 5.75 gives E for a traveling harmonic wave, which werewrite for convenience:

E = E0 cos (kz − ωt − φ) (5.80)

Substituting Equation 5.80 into Equation 5.79 gives the in-stantaneous intensity for a traveling harmonic wave, namely

Iinst = C E20 cos2 (kz − ωt − φ) (5.81)

Because cos(−θ) = cos θ , and using Equation 5.68 to writeω = 2π/T , we express Equation 5.81 in a better form for ournext task, determining the time average of Iinst:

Iinst = C E20 cos2 (kz − ωt − φ)

= C E20 cos2 (−kz + ωt + φ)

= C E20 cos2

(2π

Tt + α

)(5.82)

where

α = −kz + φ (5.83)

is independent of the time t . We now observe that the highfrequencies for visible light given in Equation 5.77 make

the period T = 1/ν (see Equation 5.64) range from 1.33 fsto 2.33 fs, where fs = femtosecond = 10−15 s. These ex-tremely small periods make the time variation of Iinst so rapidthat practical detectors (such as the eye, photographic film,photocell, and all other detectors that we can construct) donot respond to Iinst. Instead, these detectors respond to thetime average (or mean) with respect to time of Iinst, whichwe shall simply call the intensity I .

In calculus, the average or mean of a function f (t) overan interval from t1 to t2 is defined to be

fav = 1

t2 − t1

∫ t2

t1

f (t) dt (5.84)

We use this definition to calculate the time average of Iinst overa time interval from t1 = 0 to t2 = τ with the time τ >> Tto correspond to practical detectors. Then, using the trig iden-tity 2 cos2 A = 1 + cos 2A, and substituting Equation 5.82for Iinst, we have

I = (Iinst)av = 1

τ

∫ τ>>T

0Iinst dt

= 1

τ

∫ τ

0C E2

0 cos2

(2π

Tt + α

)dt

= C E20

2τ

∫ τ

0dt + C E2

0

2τ

∫ τ

0cos

(4π

Tt + 2α

)dt

= C E20

2+ T

τ

C E20

8π

[sin

(4π

Tτ + 2α

)− sin 2α

]

≈ C

2E2

0

= C E20 = C E E (5.85)

where we have used τ >> T to remove the expression in-volving the sin terms within the brackets, replaced C/2 by Cfor simplicity (both C/2 and C are constants whose valueswe never need, so it is a type of shorthand to use the same Cfor both constants), and in the last step we represent the E ofEquation 5.80 in its complex form, namely,

E = E0 ei(kz−ωt−φ) (5.86)

The I = C E E expression in Equation 5.85 provides a veryadvantageous way of quickly calculating the time averageintensity I , or for short, the intensity I . To see how E E

quickly gives the amplitude squared E20 , you might want to

review Section 5.2.7 and Equation 5.29, where |z| represents(or equals) E0. Even when E is not given simply by Equa-tion 5.86, but is a sum of several harmonic waves with thesame frequency (but with the other parameters different), thecalculation of E E gives the correct square of the amplitude.The SI units for I are the same as for Iinst, namely, J/s . m2

or W/m2.

Figure 5.25

vacuumn = 1

mediumn > 1

c

λ

λnv

interface

Pz1

z2

O1

O2


5.3.5 Traveling harmonic waves in other mediaSo far, we have assumed that the harmonic waves travel withspeed c in a vacuum, a homogeneous and isotropic medium ofindex of refraction n = 1. The space and time parameters (seeFigure 5.23) of a traveling harmonic wave in a vacuum arewritten without a subscript. Now suppose a harmonic waveenters another homogeneous, isotropic medium of index ofrefraction n > 1; then the speed of wave travel slows tov < c.Electromagnetic theory shows that the time parameters donot change; that is, T , ν, and ω remain the same. Withdefinition of n, as given in Equation 1.1, and with the help ofthe relationships in Figure 5.23, we obtain

n = speed in a vacuum

speed in medium= c

v= λν

λnν= λ

λn(5.87)

which says that

λn = λ

n(5.88)

Thus, the wavelength shortens when the harmonic wave trav-els into a medium with a greater index of refraction, as weillustrate in Figure 5.26, where the interface is the surface ofthe medium, for example, glass. The other space parametersof σn and kn in the medium would also have different valuesfrom those in the vacuum.

Figure 5.26

We now want to look in greater detail at harmonic wavesthat travel through one or more different media.

5.3.6 The importance of the optical path difference� and the phase difference δ

Suppose that two harmonic waves of the same angular fre-quency ω travel in a homogeneous, isotropic medium of in-dex of refraction n. One wave travels along the z1 axis, theother along the z2 axis, as shown in Figure 5.27. Based onthe description of harmonic waves that we gave in previoussections, we write the equations of these waves as

E1 = E01 cos (knz1 − ωt − φ1) (5.89a)

E2 = E02 cos (knz2 − ωt − φ2) (5.89b)

where the amplitudes E01, E02 are not necessarily the same,and where Figure 5.23 and Equation 5.88 gives

kn = 2π

λn= 2π

λn (5.90)

The origin O1 marks the point where z1 = 0, and O2 is theorigin for the z2 axis, where z2 = 0. The respective phasesof E1 and E2 are

θ1 = knz1 − ωt − φ1 and θ2 = knz2 − ωt − φ2

At the origins, z1 and z2 equal zero, so then the respectivephases at the origins become

θ1 = −ωt − φ1 and θ2 = −ωt − φ2 (5.91)

We wish to make these waves in phase at their respectiveorigins; that is, if we imagine that we could see these wavesat O1 and O2, then we would see them both going through

Figure 5.27 Two harmonic waves traveling in a homo-geneous, isotropic medium of index of refraction n.

their maxima, minima, and zeros at the same time—that is,the respective phase equations must be the same. Lookingat Equations 5.91, we see that we satisfy this requirement bymaking φ1 = φ2 = φ. Then we have

θ1 = −ωt − φ and θ2 = −ωt − φ (5.92)

Now we can say that E1 and E2 are in phase with each otherat O1 and O2; that is

E1 = E01 cos (−ωt − φ) (5.93a)

E2 = E02 cos (−ωt − φ) (5.93b)

at their respective origins. Because the cosine function is aneven function, we could remove the minus signs in Equa-tions 5.93, but for what we want to do next, it is simplerto leave them in place. With the phase constants the same,Equations 5.89 become

E1 = E01 cos (knz1 − ωt − φ) (5.94a)

and

E2 = E02 cos (knz2 − ωt − φ) (5.94b)

E

t

E1

E2

E

tt ′

t = 0 t ′ = 0

t0t

t ′

E E1

E2


Inspecting Figure 5.27, we see that the axes cross at thepoint P; at this point our harmonic waves in Equations 5.94become

E1 = E01 cos (knz1P − ωt − φ) (5.95a)

E2 = E02 cos (knz2P − ωt − φ) (5.95b)

where z1P is the distance from O1 to P , and in like manner,z2P is the distance from O2 to P .

By inspection of Equations 5.95, we see that the phasesof the two harmonic waves at the point P are

θ1P = knz1P − ωt − φ (5.96a)

θ2P = knz2P − ωt − φ (5.96b)

The phase difference δ is defined as the difference betweenthese phases, so with Equation 5.90 we get

δ = θ2P − θ1P

= (knz2P − ωt − φ) − (knz1P − ωt − φ)

= kn (z2P − z1P) = 2π

λn(z2P − z1P)

= 2π

(λ/n)(z2P − z1P) = 2π

λ(nz2P − nz1P)

= 2π

λ(�2P − �1P) = 2π

λ� (5.97)

where� = �2P − �1P = nz2P − nz1P (5.98)

is called the optical path difference. We frequently want toknow the phase difference δ when we know the optical pathdifference �, so Equations 5.97 and 5.98 are important equa-tions. The distance �1P = nz1P is called the optical path (orthe optical path length) from O1 to P , and �2P = nz2P isgiven a similar explanation. The distance z1P is called thegeometrical path (or the geometrical path length) from O1

to P; similarly for z2P .The phase difference δ helps to simplify work with har-

monic waves of the type shown in Equations 5.95, whichwe emphasize, describe the behavior of the harmonic waveswith time t at the point P . If we imagine an observer at thepoint P to graph these harmonic waves as functions of time t ,the graphs obtained might look like those in Figure 5.28.

Figure 5.28

Because we are looking at these waves as functions of time t ,it helps if we write the arguments (the phases) in Equa-tions 5.95 with the ωt first. We remember that the cosinefunction is an even function; that is, cos(−θ) = cos θ . There-fore, we multiply the arguments through by −1, rearrangeslightly, and obtain the equivalent equations

E1 = E01 cos (ωt − knz1P + φ) (5.99a)

E2 = E02 cos (ωt − knz2P + φ) (5.99b)

We graph these equations again in Figure 5.29, but also in-dicate that we shift the time origin by a time of t0 to a new t ′

axis origin at the E1 peak so that we have

E1 = E01 cos(ωt ′) (5.100)

which is a much simpler expression than the one in Equa-tion 5.99a. To determine t0, we start by observing in the

Figure 5.29

diagram of Figure 5.29 that arbitrary points on the waves aredescribed in terms of either t or t ′ where

t = t ′ + t0 (5.101)

We then substitute Equation 5.101 into Equation 5.99a, andset it equal to the desired Equation 5.100 to get

E01 cos[ω(t ′ + t0) − knz1P + φ] = E01 cos(ωt ′)

ω(t ′ + t0) − knz1P + φ = ωt ′

which when solved for t0 gives

t0 = knz1P − φ

ω(5.102)

Next we substitute Equation 5.101 into Equation 5.99b, re-place the t0 by Equation 5.102, and rearrange:

E2 = E02 cos[ω(t ′ + t0) − knz2P + φ]

= E02 cos[ωt ′ − kn(z2P − z1P)]

z3z1

n1 n2 n3

P1 P2

z2

�P1 P2

z P3 P4

P3 P4vacuum


Substituting Equation 5.90 for kn , and then with the help ofEquations 5.98 and 5.97, we finally obtain

E2 = E02 cos

[ωt ′ − 2π

λ(nz2P − nz1P)

]

= E02 cos

[ωt ′ − 2π

λ�

]

= E02 cos(ωt ′ − δ) (5.103)

We have now achieved an important result. Wheneverwe want to describe two harmonic waves of the form in Equa-tions 5.95 or 5.99, we can write them very simply in terms ofthe phase difference δ as

E1 = E01 cos(ωt ′) (5.104a)

E2 = E02 cos(ωt ′ − δ) (5.104b)

Because there is no t0 in the equations, and because we nor-mally do not need to know it, we drop the prime and writethe equations yet more simply as

E1 = E01 cos(ωt) (5.105a)

E2 = E02 cos(ωt − δ) (5.105b)

or in the complex plane as

E1 = E01ei(ωt) (5.106a)

E2 = E02ei(ωt−δ) (5.106b)

where δ has all the information we need to state how the twoharmonic waves differ from each other. Even when harmonicwaves travel through media that are more complicated thanthose shown in Figure 5.27, the above equations still work:for example, the waves might travel through several differentmedia on the way to P , or one wave might travel through oneset of media, and the other through another set.

The optical path has an interesting interpretation thatwe illustrate with an example. Suppose a harmonic wavetravels from P1 to P2 through several different homogeneous,isotropic media, as shown in Figure 5.30. Then the opticalpath is

�P1 P2 = n1z1 + n2z2 + n3z3 (5.107)

This equation has an important interpretation when we usethe definition of the index of refraction that we gave at the

beginning of Chapter 1: applying this definition we write

n1 = c

v1n2 = c

v2n3 = c

v3(5.108)

and, since time equals distance divided by speed,

t1 = z1

v1t2 = z2

v2t3 = z3

v3(5.109)

Figure 5.30

To obtain another expression for �P1 P2 in Equation 5.107, wefirst substitute Equations 5.108 into Equation 5.107, and sec-ond, successively solve each of equations in Equation 5.109for z1, z2, z3 before substituting into Equation 5.107; we get

�P1 P2 = n1z1 + n2z2 + n3z3

=(

c

v1

)(v1t1) +

(c

v2

)(v2t2) +

(c

v3

)(v3t3)

= ct1 + ct2 + ct3

= c(t1 + t2 + t3) = cttot = zP3 P4 (5.110)

Inspecting this equation, we observe that the time ttot is thetime for the light to travel through the media from points P1

to P2. When we multiply this total time by the speed of lightin a vacuum, we get the distance zP3 P4 (see Figure 5.30). Thatis, Equation 5.110 says that the optical path in several media(or one medium) is equal to the geometric path traveled bylight in a vacuum—the times to travel the separate paths arethe same.

In Figure 5.30, the angles of incidence and refraction arezero, but Equation 5.110 holds for nonzero angles as well.Even when there is a continuous change in the index of re-fraction for a medium, an equation similar to Equation 5.110can be obtained.


PROBLEMS

Note: Angles with no units are in radians. For the meaningof default angle, see the note under Equation 5.6b.

5.1 The following expressions are complex numbers in rect-angular form:

(a) 5 + i6 (b) − 5 + i3

(c) 2 − i2√

3 (d) −√

6 − i√

2

Write each of these expressions in polar form with thedefault angle in both radians and degrees, and draw aphasor diagram.

5.2 The following expressions are complex numbers writtenin polar form (where there is no degree unit, the angleis in radians):

(a) 6 ei10◦(b) 4 ei125◦

(c) 2 ei(−3π/8) (d) 14.7 ei(−135◦)

Write each of these expressions in rectangular form anddraw a phasor diagram.

5.3 Suppose z1 = −5 + i5 and z2 = 2 + i2√

3, calculate(a) z1 + z2, (b) z1 − z2, (c) z1z2, (d) z1/z2. Express theresults in rectangular form.

5.4 Suppose z1 = 6 ei(−120◦) and z2 = 4 ei(−45◦), determine(a) z1 + z2, (b) z1 − z2, (c) z1z2, (d) z1/z2. Express theresults in polar form with the angle in default form inboth radians and degrees.

5.5 Suppose z = −5 + i6. Determine (a) the magnitudeusing Equation 5.29 (that is, calculate

√zz ), (b) the

real part using Equation 5.30a, and (c) the imaginarypart using Equation 5.30b.

5.6 Suppose the complex number is given in polar formz = 10 ei(−35◦). Determine (a) the magnitude usingEquation 5.29 (that is, calculate

√zz ), (b) the real part

using Equation 5.30a, and (c) the imaginary part usingEquation 5.30b.

5.7 Suppose the complex number is

z = 2 + i4

3 + i5

Determine (a) the magnitude using Equation 5.29 (thatis, calculate

√zz ), (b) the Re part using Equation 5.30a,

and (c) the Im part using Equation 5.30b.

5.8 Show the derivation of Equation 5.23 in more detail;some steps are missing.

5.9 Derive Equation 5.26b starting with Equation 5.24.

5.10 Prove Equation 5.29.

5.11 Given that z1 = 5 + i6 and z2 = 8 + i3, follow Exam-ple 5.2.8 as a guide and calculate Re(z1z2) (a) directly,(b) using Equation 5.35, and (c) using Equation 5.36.(d) Then find Re(z1) Re(z2). Is Equation 5.37 obeyed?

5.12 Using the derivation of Equation 5.36 as a guide, showthat

Im(z1z2) = 2 Im(z1) Re(z2) + Im(z1 z2)

5.13 Suppose the equation of a traveling harmonic wave is

E =√

3 cos

(3

2z − 4

5πt + 0.01

)

With Equation 5.75 and Figure 5.23 as a guide, deter-mine the (a) amplitude, (b) phase, (c) angular wave num-ber, (d) angular frequency, (e) phase constant, (f) wavenumber, (g) wavelength, (h) frequency, (i) period, and(j) speed. (k) Is the direction of wave travel in the +zdirection or the −z direction? Assuming that the lengthunit is the m, the time unit is the s, and the E unit is theV/m include units with your answers where appropriate.

5.14 Write the traveling harmonic wave in Problem 5.13 asa complex expression in polar form.

5.15 Take the polar-form expression of Problem 5.14 andcalculate the intensity I using Equation 5.85. Includeappropriate units with your answer.

5.16 Rederive Equation 5.85 putting in all the steps.

5.17 Suppose a traveling harmonic wave is given by the sumof two traveling harmonic waves in polar form:

E = 4ei(2z−3t−0.1) + 5ei(6z−3t−0.2)

Calculate the intensity I at z = 7 m using Equation 5.85.

5.18 Given that

E = E0(1 + eiδ

)eiωt

determine the intensity I using Equation 5.85.

0.0566 m

P1 P2

�P1 P2

z P3 P4

P3 P4vacuum

0.0849 m 0.0707 m

n1 = 1 n2 = 1.50 n3 = 1.33

Chapter 5: Problems 161

5.19 Using the definition of the average of a function givenby Equation 5.84, determine the average of the function

f (t) = sin ωt

(where ω is a constant) over the interval from t1 = 0 tot2 = π/ω.

5.20 Under the same conditions as Problem 5.19, calculatethe average of the function

f (t) = cos ωt

5.21 Two harmonic waves start out in phase from points O1

and O2, respectively, as in Figure 5.27. The waves travelin water, for which n = 1.33, to a common point Psuch that the geometric paths are zO1 P = 10.29 nm andzO2 P = 12.42 nm. Calculate (a) the optical path differ-ence �, and (b) the phase difference δ in both radiansand degrees for light of wavelength λ = 632.8 nm.

5.22 Using a diagram similar to the one in Figure 5.27, twoharmonic waves start out in phase from points O1 andO2 and travel to the point P . However, on the way tothe point P , the waves travel through several differentmedia such that

nO1 P1 = 1, n P1 P2 = 1.33, n P2 P = 1

nO2 P3 = 1, n P3 P4 = 1.50, n P4 P = 1

and

zO1 P1 = 20.5 mm, zP1 P2 = 25.5 mm, zP2 P = 19.3 mm

zO2 P3 = 24.5 mm, zP3 P4 = 26.0 mm, zP4 P = 23.4 mm

Calculate (a) the optical path difference �, and (b) thephase difference δ in both radians and degrees for lightof wavelength λ = 546.1 nm.

5.23 In a diagram which is like Figure 5.27, imagine that twowaves travel in glass of n = 1.518. If λ = 632.8 nm ina vacuum, calculate (a) the corresponding angular wavenumber k in a vacuum, (b) the wavelength λn in themedium, and (c) the corresponding kn . The two wavesare in phase at O1 and O2, and travel to the point Psuch that zO1 P = 100.1 nm and zO2 P = 110.2 nm. De-termine (d) the optical path difference �, and (e) thephase difference δ in rad and deg.

5.24 Suppose a harmonic wave of light travels through severalmedia from point P1 to P2, as shown in Figure 5.31. Forthese two points, calculate (a) the geometrical path zP1 P2 ,and (b) the optical path �P1 P2 . Using c = 3 × 108 m/sfor the speed of light in a vacuum, determine for each ofthe three media, (c) the speeds of wave travel v1, v2, v3,and (d) the times t1, t2, t3. Then find (e) the total time ttot

for the wave to travel from P1 to P2, and (f) the distancethat light travels in a vacuum in this time of ttot; that is,calculate zP3 P4 = cttot. Is �P1 P2 equal to zP3 P4 ?

Figure 5.31

162

Chapter 5: Answers to Problems

5.1 (a) 7.81 ei0.876, 7.81 ei50.2◦; (b) 5.83 ei2.60, 5.83 ei149.0◦

;(c) 4 ei(−1.05), 4 ei(−60◦); (d) 2.83 ei(−2.62), 2.83 ei(−150◦)

5.2 (a) 5.91 + i1.04; (b) −2.29 + i3.28; (c) 0.765 − i1.85;(d) −10.4 − i10.4

5.3 (a) −3 + i8.46; (b) −7 + i1.54; (c) −27.3 − i7.32;(d) 0.458 + i1.71

5.4 (a) 8.03 ei(−1.59), 8.03 ei(−91.2◦);(b) 6.29 ei(−2.76), 6.29 ei(−157.9◦);(c) 24 ei(−2.88), 24 ei(−165◦);(d) 1.5 ei(−1.31), 1.5 ei(−75◦)

5.5 (a) 7.81; (b) −5; (c) 6

5.6 (a) 10; (b) 8.19; (c) −5.74

5.7 (a) 0.767; (b) 13/17; (c) 1/17

5.11 (a) 22; (b) 22; (c) 22; (d) 40

5.13 (a)√

3 V/m = 1.73 V/m; (b)3

2z − 4

5πt + 0.01;

(c)3

2rad/m = 1.5 rad/m; (d)

4

5πrad/s = 0.255 rad/s;

(e) −0.01 rad; (f)3

4πm−1 = 0.239 m−1;

(g)4π

3m = 4.19 m; (h)

2

5π2Hz = 0.0405 Hz;

(i)5π2

2s = 24.7 s; (j)

8

15πm/s = 0.170 m/s

5.14√

3 ei( 32 z− 4

5πt+0.01)

5.15 3C W/m2

5.17 3.77C W/m2

5.18 4C E20 cos2(δ/2)

5.19 2/π

5.20 0

5.21 (a) 2.83 nm; (b) 0.0281 rad = 1.61 deg

5.22 (a) 13.2 nm; (b) 0.152 rad = 8.69 deg

5.23 (a) 0.009923 rad/nm; (b) 416.9 nm;

(c) 0.01507 rad/nm; (d) 15.33 nm;

(e) 0.1522 rad = 8.722 deg

5.24 (a) 0.212 m; (b) 0.278 m;

(c) 3.00 × 108 m/s, 2.00 × 108 m/s, 2.26 × 108 m/s;

(d) 0.189 ns, 0.425 ns, 0.313 ns;

(e) 0.927 ns; (f) 0.278 m

138 part ii: physical opticslearnoptics.com/optics chapters/opticschap5.pdf · 2009. 2. 8. · 139...

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