Physical Optics - 2005
and
∂2ψ
∂
=
∂t2
∂t
Notes by Paul Walsh
pwalsht@maths.tcd.ie
∂f
±v 0
∂x
∂
= ±v 0
∂x
∂f
∂t
Since
∂f
∂ψ
=
∂t
∂t
Introduction to Optics - Light
as a wave
it follows, using eqn(2.9) that
Envision a disturbance ψ moving in the positive
x-direction with constant speed v. The specific nature of the disturbance is unimportant.
Combining these
Since the disturbance is moving, it must be a function of both position and time and can therefore be
∂2ψ
written as:
∂x2
ψ = f (x, t)
∂2ψ
∂2f
= v 2 02
2
∂t
∂x
eqns, we obtain
=
1 ∂2ψ
v 2 ∂t2
(2.11)
which is the desired one-dimensional differential
wave equation.
(2.1)
The shape of the disturbance at any instant, say
t = 0, can be found by holding time constant at Harmonic Waves
The simplest wave form is the sine or cosine curve.
that value. In this case:
These are called sinusoidal waves, simple harmonic
waves, or more succinctly as harmonic waves.
ψ(x, t)t=0 = f (x, 0) = f (x)
(2.2)
Choose at the profile the simple function
represents the shape or profile of the wave at any
ψ(x, t)|t=0 = ψx = Asin(kx) = f (x)
(2.12)
time.
where k is a positive constant known as the propagation number.
It’s necessary to introduce the constant k simply
because we cannot take the sine of a quantity that
has physical units. The sine is the ratio of two
lengths and the therefore unitless.
Accordingly, kx is properly in radians, which is not
a real physical unit. The sine varies from +1 to
-1 so that the maximum value of ψx is A. This
maximum disturbance is known as the amplitude
of the wave.
To transform eqn(2.12) into a progressive wave
travelling at speed v in the positive x-direction, we
need merely replace x by (x − vt), in which case
The Differential Wave Equation
Taking the partial derivative of ψ(x, t) with respect
to x, hold t constant. Using x0 = x ± vt we have
∂ψ
∂f ∂x0
∂f
∂x0
.
since
=
=
=1
∂x
∂x0 ∂x
∂x0
∂x
(2.8)
Holding x constant, the partial derivative with respect to time is
∂ψ
∂f ∂x0
∂f
=
.
= ±v 0
∂t
∂x0 ∂t
∂x
(2.9)
Combining eqns(2.8) and eqns(2.9) yields
∂ψ
∂ψ
= ±v 0
∂t
∂x
ψ(x, t) = ψ(x ± λ, t)
(2.10)
(2.14)
In the case of a harmonic wave, this is equivalent
The second partial derivatives of eqns(2.8) and to altering the sine function by ±2π. Therefore,
(2.9) are
sin k(x − vt) = sin k[(x ± λ) − vt] = sin [k(x − vt) ± 2π]
∂2ψ
∂2f
=
and so
∂x2
∂x02
1
|kλ| = 2π
κ=
or since both k and λ are positive numbers,
2π
λ
k=
ψ = A sin k(x ± vt)
The temporal period, τ is the amount of time it
takes for one complete wave to pass a stationary
observer.
In this case, it is the repetitive behaviour of the
wave in time that is of interest, so that
(2.16)
= sin [k(x − vt) ± 2π]
Therefore,
|kvτ | = 2π
But these are all positive numbers ; hence
(2.17)
or
v = fλ
(2.23)
ψ(x, t) = A sin (kx − ωt)
from which it follows that
f≡
ψ = A sin 2π(κx ± f t)
Phase and Phase Velocity
Examine any one of the harmonic wave functions,
such as
2π
vτ = 2π
λ
λ
v
(2.22)
x
t
±
λ τ
ψ = A sin (kx ± ωt)
(2.24)
x
±t
(2.25)
ψ = A sin 2πf
v
Eqns (2.13) and (2.24) will be encountered most
frequently.
Each wave has a single constant frequency, and is
therefore monochromatic, or even better, monoenergetic.
In reality though, all waves comprise a band of frequencies, and when that band is narrow, the wave
is said to be quasimonochromatic.
sin k(x − vt) = sin k[x − v(t ± τ )]
τ=
(2.13)
ψ = A sin 2π
and
kvτ = 2π
(2.21)
Using the above definitions, a number of equivalent expressions can be written for the travelling
harmonic wave:
(2.15)
ψ(x, t) = ψ(x, t ± τ )
1
λ
(2.26)
The entire argument of the sine of the phase φ of
the wave, where
(2.18)
φ = (kx − ωt)
1
τ
(2.27)
At t = x = 0
ψ(x, t)|x=0,t=0 = ψ(0, 0) = 0
(2.19)
Two other quantities are often used in the literature which is certainly a special case. More generally,
of wave motion. One is the angular temporal we can write
frequency
ψ(x, t) = A sin (kx − ωt + )
(2.28)
2π
ω≡
= 2πf
(2.20)
where is the initial phase.
τ
The phase of a disturbance such as ψ(x, t) given by
eqn (2.28) is
The wave number
2
The Superposition Principle
Suppose that the wavefunctions ψ1 and ψ2 are each
φ(x, t) = (kx − ωt + )
separate solutions of the wave equation ; it follows
that (ψ1 + ψ2 ) is also a solution. This is known
and is obviously a function of x and t. In fact, the as the Superposition Principle and it can easily be
partial derivative of φ with respect to t, holding x proven since it must be true that
constant, is the rate-of-change of phase with respect
to time, or
1 ∂ 2 ψ1
∂ 2 ψ2
1 ∂ 2 ψ2
∂ 2 ψ1
= 2
and
= 2
2
2
2
∂x
v ∂t
∂x
v ∂t2
∂φ
|
|=ω
(2.30)
Adding
∂t x
∂ 2 ψ1
∂ 2 ψ2
1 ∂ 2 ψ1
1 ∂ 2 ψ2
The rate-of-change of phase at any fixed position
+
= 2
+ 2
2
2
2
∂x
∂x
v ∂t
v ∂t2
is the angular frequency of the wave, the rate at
which a point on a rope experiencing an harmonic
wave oscillates up and down.
∂ 2 (ψ1 + ψ2 ) ∂ 2 ψ2
1 ∂ 2 (ψ1 + ψ2 )
+
=
Similarly, the rate-of-change of phase with distance,
∂x2
∂x2
v2
∂t2
holding t constant, is
which establishes that (ψ1 +ψ2 ) is indeed a solution.
∂φ
The
resulting disturbance occurring when two
|
|=k
(2.31)
∂x t
waves meet at each point in the region of overlap
is the algebraic sum of the individual constituent
These two expressions should bring to mind an waves at that location.
equation from the theory of partial derivatives.
The Complex Representation
∂φ
−
∂t
The complex
(2.35)
∂x
√ number z̄ = x + iy
= x
(2.32)
where i = −1
∂φ
∂t φ
∂x
In terms of polar coordinates (r, θ)
t
The term on the left represents the speed of propx = r cos θ
y = r sin θ
agation of the condition of constant phase.
Taking the appropriate derivatives of φ as given, for and
example, by eqn(2.29) and substituting them into
eqn(2.32), we get
z̄ = x + iy = r cos θ + ir sin θ
∂x
ω
The Euler formula
= ± = ±v
(2.23)
∂t φ
k
eiθ = cos θ + i sin θ
This is the speed at which the profile moves and
is known commonly as the phase velocity of the leads to the expression
waves.
e−iθ = cos θ − i sin θ
Any point on a harmonic wave having a fixed magnitude moves such that φ(x, t) is constant in time,
and this yields
in other words,
∂φ(x, t)
=0
∂t
− ∂ψ
∂t
±v = x
∂ψ
∂x
(2.32)
cos
eiθ + e−iθ
2
sin
eiθ − e−iθ
2i
Moreover the Euler formula allows us to write
t
3
z̄ = reiθ
The complex conjugate of z̄ is found by replacing i
with −i wherever it appears
z̄ ∗ = re−iθ
Multiplication and division are most simply expressed in polar form
z¯1 z¯2 = r1 r2 ei(θ1 −θ1 )
r1
z¯1
= e−i(θ1 −θ2 )
z¯2
r2
A harmonic wave can be written as
ψ(x, t) = Re[Aei(ωt−kx+) ]
(2.36)
Plane Waves
If you detect any errors in these notes pls cont act
me at pwalsht@maths.tcd.ie and I will try to amend
them.
Notes after this point currently being drafted.
4