2 The Wave Equation in Classical Optics

The Wave Equation in Classical Optics The concept of the interference of waves, developed in mechanics in the eighteenth century, was introduced into optics by Thomas Young at the beginn

The Wave Equation in Classical Optics

The concept of the interference of waves, developed in mechanics in the eighteenth century, was introduced into optics by Thomas Young at the beginning of the nineteenth century In the eighteenth century the mathematical physicists Euler, d’Alembert, and Lagrange had developed the wave equation from Newtonian mechanics and investigated its consequences, e.g., propagating and standing waves It is not always appreciated that Young’s ‘‘leap of genius’’ was to take the ideas developed in one field, mechanics, and apply them to the completely different field of optics

In addition to borrowing the idea of wave interference, Young found that it was also necessary to use another idea from mechanics He discovered that the superposition of waves was insufficient to describe the phenomenon of optical inter-ference; it, alone, did not lead to the observed interference pattern To describe the interference pattern he also borrowed the concept of energy from mechanics This concept had been developed in the eighteenth century, and the relation between the amplitude of a wave and its energy was clearly understood In short, the mechanical developments of the eighteenth century were crucial to the work of Young and to the development of optics in the first half of the nineteenth century It is difficult to imagine the rapid progress which took place in optics without these previous devel-opments In order to have a better understanding of the wave equation and how it arose in mechanics and was then applied to optics, we now derive the wave equation from Newton’s laws of motion

2.2 THE WAVE EQUATION

Consider a homogeneous string of length l fixed at both ends and under tension T0,

as shown inFig 2-1.The lateral displacements are assumed to be small compared with l The angle
between any small segment of the string and the straight line (dashed) joining the points of support are sufficiently small so that sin
is closely approximated by tan
Similarly, the tension T0 in the string is assumed to be unaltered by the small lateral displacements; the motion is restricted to the xy plane

The differential equation of motion is obtained by considering a small element

ds of the string and is shown exaggerated as the segment AB in Fig 2-1 The y component of the force acting on ds consists of F1and F2 If
1and
2are small, then

F1 ¼T0sin
1’T0tan
1¼T0 @y

@x



A

ð2-1aÞ

F2 ¼T0sin
2’T0tan
2¼T0 @y

@x



B

ð2-1bÞ

where the derivatives are partials because y depends on time t as well as on the distance x The subscripts signify that the derivatives are to be evaluated at points

Aand B, respectively Then, by Taylor’s expansion theorem,

@y

@x



A

¼@y

@x

@

@x

@y

@x

dx

2 ¼

@y

@x

@2y

@x2

dx

@y

@x



B

¼@y

@xþ

@

@x

@y

@x

dx

2 ¼

@y

@xþ

@2y

@x2

dx

in which the derivatives without subscripts are evaluated at the midpoint of ds The resultant force in the y direction is

F2F1¼T0 @

2

y

@x2

!

If  is the mass per unit length of the string, the inertial reaction (force) of the element ds is dsð@2y=@t2Þ For small displacements, ds can be written as ds ’ dx The equation of motion is then obtained by equating the inertial reaction to the applied force (2-3), so we have

@2y

@t2 ¼T0



@2y

Equation (2-4) is the wave equation in one dimension In optics y(x, t) is equated with the ‘‘optical disturbance’’ u(x, t) Also, the ratio of the tension to the density in the string T/ is found to be related to the velocity of propagation v by the equation:

v2¼T0

Figure 2-1 Derivation of the wave equation Motion of a string under tension

The form of (2-5) is easily derived by a dimensional analysis of (2-4) Equation (2-4) can then be written as

@2uðx, tÞ

@x2 ¼ 1

v2

@2uðx, tÞ

in which form it appears in optics Equation (2-6) describes the propagation of an optical disturbance u(x, t) in a direction x at a time t For a wave propagating in three dimensions it is easy to show that the wave equation is

@2uðr, tÞ

@x2 þ@2uðr, tÞ

@y2 þ@2uðr, tÞ

@z2 ¼ 1

v2

@2uðr, tÞ

where r ¼ ðx2þy2þz2Þ1=2 Equation (2-7) can be written as

r2uðr, tÞ ¼ 1

v2

@2uðr, tÞ

where r2is the Laplacian operator,

r2 @2

@x2þ

@2

@y2þ

@2

Because of the fundamental importance of the wave equation in both mechanics and optics, it has been thoroughly investigated Equation (2-7) shall now be solved in several ways Each method of solution yields useful insights

2.2.1 Plane Wave Solution

Let r(x, y, z) be a position vector of a point P in space, s(sx, sy, sz) a unit vector in a fixed direction Any solution of (2-7) of the form:

is said to represent a plane-wave solution, since at each instant of time u is constant over each of the planes,

Equation (2-11) is the vector equation of a plane; a further discussion of plane waves and (2-11) will be given later

Figure 2-2shows a Cartesian coordinate sytem Ox, Oy, Oz We now choose a new set of Cartesian axes, O, O, O, with O in the direction sr ¼  Then

@=@x ¼ ð@=@xÞ@=@, etc., so

and we can write

@

@x¼sx

@

@

@

@y¼sy

@

@

@

@z¼sz

@

Since s2xþs2yþs2z ¼1, we easily find that

r2u ¼@

2

u

so that (2-8) becomes

@2u

@21

v2

@2u

Thus, the transformation (2-12) reduces the three-dimensional wave equation to a one-dimensional wave equation Next, we set

and substitute (2-15) into (2-14) to find

@2u

The solution of (2-16) is

as a simple differentiation quickly shows Thus, the general solution of (2-14) is

where u1and u2are arbitrary functions The argument of u is unchanged when (, t)

is replaced by ( þ v, t þ ), where  is an arbitrary time Thus, u1( þ v) represents a disturbance which is propagated with a velocity v in the negative  direction Similarly, u2(  v) represents a disturbance which is propagated with a velocity v

in the positive  direction

2.2.2 Spherical Waves

Next, we consider solutions representing spherical waves, i.e.,

Figure 2-2 Propagation of plane waves

where r ¼ rj j ¼ ðx þy þz Þ Using the relations

@

@x¼

@r

@x

@

@r¼

x r

@x

one finds after a straightforward calculation that

r2ðuÞ ¼1

r

@2ðruÞ

The wave equation (2-8) then becomes

@2ðruÞ

@r2 1

v2

@2ðruÞ

Following (2-14) the solution of (2-22) is

uðr, tÞ ¼u1ðr  vtÞ

u2ðr þ vtÞ

where u1and u2are, again, arbitrary functions The first term in (2-23) represents a spherical wave diverging from the origin, and the second term is a spherical wave converging toward the origin; the velocity of propagation being v in both cases 2.2.3 Fourier Transform Method

The method for solving the wave equation requires a considerable amount of insight and experience It would be desirable to have a formal method for solving partial differential equations of this type This can be done by the use of Fourier transforms Let us again consider the one-dimensional wave equation:

@2uð, tÞ

@2 ¼ 1

v2

@2uð, tÞ

The Fourier transform pair for u(, t) is defined in the time domain, t, to be uð, tÞ ¼ 1

2

Z1

1

and

uð, !Þ ¼

Z1

1

We can then write

@2uð, tÞ

@2 ¼ 1

2

Z1

1

@2uð, !Þei!t

@2 d!

@2uð, tÞ

@t2 ¼ 1

2

Z1

1

so (2-24) is transformed to

@2uð, !Þ

@2 ¼!2uð, !Þ

Equation (2-27) is recognized immediately as the equation of a harmonic oscillator whose solution is

where k ¼ !=v We note that the ‘‘constants’’ of integration, A(!) and B(!), must be written as functions of ! because the partial differentiation in (2-24) is with respect to

 The reader can easily check that (2-28) is the correct solution by differentiating it according to (2-27) The solution of (2-24) can then be found by substituting u(, !)

in (2-28) into the Fourier transform u(, t) in (2-25a)

uð, tÞ ¼ 1

2

Z1

1

or

¼ 1 2

Z1

1

Að!Þei!ðtþ=vÞd! þ 1

2

Z1

1

Bð!Þei!ðt=vÞd! ð2-30Þ

From the definition of the Fourier transform, Eq (2-25), we then see that

uð, tÞ ¼ u1 t þ

v

þu2 t 

v

ð2-31Þ

which is equivalent to the solution (2-18)

Fourier transforms are used throughout physics and provide a powerful method for solving partial differential equations Finally, the Fourier transform pair shows that the simplest sinusoidal solution of the wave equation is

where A and B are constants The reader can easily check that (2-32) is the solution

of the wave equation (2-24)

2.2.4 Mathematical Representation of the Harmonic Oscillator

Equation Before we end the discussion of the wave equation, it is also useful to discuss, further, the harmonic oscillator equation From mechanics the differential equation of the harmonic oscillator motion is

md

2x

or

d2x

dt2 ¼ k

mx ¼ !

2

where m is the mass of the oscillator, k is the force constant of the spring, and

!0¼2f is the angular frequency where f is the frequency in cycles per second

Equation (2-33b) can be solved by multiplying both sides of the equation by dx/dt ¼

v(v ¼ velocity):

vdv

dt ¼ !

2xdx

or

Integrating both sides of (2-34b) yields

v2

2 ¼ 

!2

2 x

where A2is the constant of integration Solving for v, we have

v ¼dx

dt ¼ ðA

which can be written as

dx

The solution of (2-36) is well known from integral calculus and is

where a and  are constants of integration Equation (2-37) can be rewritten in another form by using the trigonometric expansion:

sinð!0t þ Þ ¼sinð!0tÞcos  þ cosð!0tÞsin  ð2-38Þ so

where

Another form for (2-39) is to express cos !0tand sin !0tin terms of exponents; that is, cos !0t ¼e

i! 0 tþei!0 t

sin !0t ¼e

i! 0 tei!0 t

Substituting (2-41a) and (2-41b) into (2-39) and grouping terms leads to

where

C ¼A  iB

A þ iB

and C and D are complex constants Thus, we see that the solution of the harmonic oscillator can be written in terms of purely real quantities or complex quantities The form of (2-35a) is of particular interest The differential equation (2-33a) clearly describes the amplitude motion of the harmonic oscillator Let us retain the original form of (2-33a) and multiply through by dx/dt ¼ v, so we can write

mvdv

dt¼ kx

dx

We now integrate both sides of (2-43), and we are led to

mv2

kx2

where C is a constant of integration Thus, by merely carrying out a formal integra-tion we are led to a new form for describing the mointegra-tion of the harmonic oscillator

At the beginning of the eighteenth century the meaning of (2-44) was not clear Only slowly did physicists come to realize that (2-44) describes the motion of the harmonic oscillator in a completely new way, namely the description of motion in terms of energy The terms mv2/2 and kx2/2 correspond to the kinetic energy and the potential energy for the harmonic oscillator, respectively Thus, early on in the development of physics a connection was made between the amplitude and energy for oscillatory motion The energy of the wave could be obtained by merely squaring the amplitude This point is introduced because of its bearing on Young’s inter-ference experiment, specifically, and on optics, generally The fact that a relation exists between the amplitude of the harmonic oscillator and its energy was taken directly over from mechanics into optics and was critical for Young’s interference experiment In optics, however, the energy would become known as the intensity 2.2.5 A Note on the Equation of a Plane

The equation of a plane was stated in (2-11) to be

We can show that (2-11) does indeed describe a plane by referring to Fig 2-2

Inspecting the figure, we see that r is a vector with its origin at the origin of the coordinates, so,

and i, j, and k are unit vectors Similarly, from Fig 2-2 we see that

Suppose we now have a vector r0along s and the plane is perpendicular to s Then

OPis the vector r  r0and is perpendicular to s Hence, the equation of the plane is

or

where  ¼ s  r0 is a constant Thus, the name plane-wave solutions arises from the fact that the wave front is characterized by a plane of infinite extent

2.3 YOUNG’S INTERFERENCE EXPERIMENT

In the previous section we saw that the developments in mechanics in the eighteenth century led to the mathematical formulation of the wave equation and the concept

of energy

Around the year 1800, Thomas Young performed a simple, but remarkable, optical experiment known as the two-pinhole interference experiment He showed that this experiment could be understood in terms of waves; the experiment gave the first clear-cut support for the wave theory of light In order to understand the pattern that he observed, he adopted the ideas developed in mechanics and applied them

to optics, an extremely novel and radical approach Until the advent of Young’s work, very little progress had been made in optics since the researches of Newton (the corpuscular theory of light) and Huygens (the wave theory of light) The simple fact was that by the year 1800, aside from Snell’s law of refraction and the few things learned about polarization, there was no theoretical basis on which to proceed Young’s work provided the first critical step in the development and acceptance of the wave theory of light

The experiment carried out by Young is shown in Fig 2-3 A source of light, ,

is placed behind two pinholes s1and s2, which are equidistant from The pinholes then act as secondary monochromatic sources that are in phase, and the beams from them are superposed on the screen  at an arbitrary point P Remarkably, when the screen is then observed, one does not see a uniform distribution of light Instead, a distinct pattern consisting of bright bands alternating with dark bands is observed

In order to explain this behavior, Young assumed that each of the pinholes, s1and s2, emitted waves of the form:

where pinholes s1and s2are in the source plane A, and are distances l1and l2from a point P(x, y) in the plane of observation  The pattern is observed on the plane Oxy normal to the perpendicular bisector of s1s2 and with the x axis parallel to s1s2 The separation of the pinholes is d, and a is the distance between the line joining the

Figure 2-3 Young’s interference experiment

pinholes and the plane of observation  For the point P(x, y) on the screen,Fig 2-3

shows that

l21 ¼a2þy2þ x d

2

ð2-50aÞ

l22 ¼a2þy2þ x þd

2

ð2-50bÞ Thus,

Equation (2-51) can be written as

Now if x and y are small compared to a, then l1þl2’2a Thus,

l2l1¼
l ¼xd

At this point we now return to the wave theory The secondary sources s1and

s2are assumed to be equal, so u01¼u02 ¼u0 In addition, the assumption is made that the optical disturbances u1and u2can be superposed at P(x, y) (the principle of coherent superposition), so

uðtÞ ¼ u1þu2

A serious problem now arises While (2-54) certainly describes an interference behav-ior, the parameter of time enters in the term !t In the experiment the observed pattern does not vary over time, so the time factor cannot enter the final result This suggests that we average the amplitude u(t) over the time of observation T The time average of u(t) written as u(t), is then defined to be

uðtÞ

¼ lim

T!1

RT

0 uðtÞ dt

RT

¼ lim

T!1

1 T

ZT

0

Substituting (2-54) into (2-55) yields

uðtÞ

¼ lim

T!1

u0 T

ZT

0

½sinð!t  kl1Þþsinð!t  kl2Þdt ð2-56Þ

Using the trigonometric identity:

sinð!t  kl Þ ¼ sinð!tÞ cosðkl Þ  cosð!tÞ sinðkl Þ ð2-57Þ and averaging over one cycle in (2-56) yields