What is a wave? A wave is anything that moves.
To displace any function f ( x x) to the right, just change its argument from x to x-a, where a is a positive number. If we let a = v t , where v is positive and t is is time, then the displacement will increase with time.
f ( x x) f ( x x-2) f ( x x-1) f ( x x-3)
x - v t ) represents a rightward, So f ( x or forward, propagating wave. x + v t ) represents Similarly, f ( x represen ts a leftward, or backward, propagating wave, where v is the velocity of the wave.
0
1
2
3
x
The one'dimensional wave e(uation The one'dimensional wave e(uation for scalar $i.e., non'vector& functions, f ) 2
∂ f ∂ x 2
2
1 ∂ f − 2 2 = 0 v ∂t
where v will be the velocity of the wave. The wave e(uation has the simple solution)
f ( x, t ) = f ( x ± vt )
The one'dimensional wave e(uation The one'dimensional wave e(uation for scalar $i.e., non'vector& functions, f ) 2
∂ f ∂ x 2
2
1 ∂ f − 2 2 = 0 v ∂t
where v will be the velocity of the wave. The wave e(uation has the simple solution)
f ( x, t ) = f ( x ± vt )
x ± vt ) solves the wave e(uation *roof that f ( x x ± vt ) as f (u), where u = x ± vt . So ∂u Write f ( x = 1 and ∂ x
+ow, use the chain rule)
∂ f ∂ x
So
=
∂f ∂ u
∂ f
∂u ∂x
∂t
=
∂u =±v ∂t
∂f ∂u ∂u ∂t
2 ∂f ⇒ ∂ 2 f ∂ f ∂f ⇒ ∂ 2 f ∂ 2 f and ∂ f 2 ∂ f = =v =±v = 2 2 2 ∂ x ∂u ∂t ∂u ∂t ∂u 2 ∂ x ∂u
Substituting into the wave e(uation)
∂ 2 f 1 ∂ 2 f − 2 2 2 v ∂t ∂ x
∂ 2 f 1 2 ∂ 2 f = − 2 v 2 ∂u v ∂u 2
= 0
What about a harmonic wave? E = E 0 cos k ( x − ct ) ! # wave amplitude $related to the energy carried by the wave&. k =
2π λ
= 2 πν˜ # angular wavenumber
$- # wavelength ν # wavenumber # /0-& 1lternatively)
E = E 0 cos(kx − ω t ) Where ω = kc = 2πc/λ = 2π f = angular
What about a harmonic wave? E = E 0 cos k ( x − ct ); φ = k ( x − ct ) The argument of the cosine function represents the phase of the wave, ϕ, or the fraction of a complete cycle of the wave.
In'phase waves
3ine of e(ual phase # wavefront # contours of ma4imum field
2ut'of'phase
The *hase 5elocity 6ow fast is the wave traveling? 5elocity is a reference distance divided by a reference time.
The phase velocity is the wavelength 0 period) v # λ 0 τ Since f = 1/τ )
v # λ f
In terms of k, k = 2π 0 λ , and the angular fre(uency, ω = 2π 0 τ , this is)
v # ω 0 k
The 7roup 5elocity
This is the velocity at which the overall shape of the wave8s amplitudes, or the wave 9envelope8, propagates. $# signal velocity &
:ispersion) phase0group velocity depends on fre(uency
;lack dot moves at phase velocity.
:ispersion) phase0group velocity depends on fre(uency
;lack dot moves at group velocity.
+ormal dispersion of visible light
Shorter $blue& wavelengths refracted more than long $red& wavelengths.
>omple4 numbers Consider a point, P = ( x,y), on a 2D Cartesian grid.
3et the 4'coordinate be the real part and the y'coordinate the imaginary part of a comple4 number. So, instead of using an ordered pair, $ x, y&, we write) P = x + i y = A cos(ϕ) + i A sin(ϕ)
where i = √(-1)
!uler8s ormula
5oted the 9"ost ;eautiful "athematical ormula !ver8 in /ABB
3inks the trigonometric functions and the comple4 e4ponential function
exp(iϕ ) = cos(ϕ ) + i sin(ϕ ) so the point, P = A cos(ϕ) + i A sin(ϕ), can also be written) P = A exp(iϕ ) = A eiφ
where A # 1mplitude
Waves using comple4 numbers E = E 0 cos k ( x − ct ); φ = k ( x − ct ) The argument of the cosine function represents the phase of the wave, ϕ, or the fraction of a complete cycle of the wave. Csing comple4 numbers, we can write the harmonic wave e(uation as)
E = E 0 e
ik ( x − ct )
= E 0 e
i( kx −ω t )
i.e., E = E 0 cos(ϕ) + i E 0 sin(ϕ), where the 9real8 part of the e4pression actually represents the wave. We also need to specify the displacement ! at 4 # and t # , i.e., the 9initial8 displacement.
1mplitude and 1bsolute phase E ( x,t ) = A cos[(k x –
t)–
]
A # 1mplitude
θ # 1bsolute phase $or initial, constant phase& at x # , t #
kx
Waves using comple4 numbers So the electric field of an !" wave can be written) E ( x,t ) = E 0 cos(kx – ω t – θ )
Since exp(iϕ ) = cos(ϕ ) + i sin(ϕ ), E ( x,t ) can also be written)
E ( x,t )
= Re { E 0 expi(kx – ω t – θ )! "
Waves using comple4 amplitudes We can let the amplitude be comple4)
E ( x,t ) = E 0 expi(kx − ω t − θ )! E ( x,t ) = [ E 0 exp(−iθ )]expi(kx − ω t )!
Where the constant stuff is separated from the rapidly changing stuff .
The resulting Dcomple4 amplitudeE) is constant in this case $as ! and # are constant&, which implies that the medium in which the wave is propagating is nonabsorbing. E exp(−iθ )
[
0
What happens to the wave amplitude upon interaction with matter?
]
>omple4 numbers simplify wavesF 1dding waves of the same fre(uency, but different initial phase, yields a wave of the same fre(uency. This isnGt so obvious using trigonometric functions, but itGs easy with comple4 e4ponentials)
Etot ( x, t ) = E1 exp i (kx − ω t ) + E2 exp i (kx − ω t ) + E3 exp i (kx − ω t ) %
%
%
= ( E1 + E2 + E3 ) exp i( kx − ω t )
%
where all initial phases are lumped into E 1, E 2, and E 3.
5ector fields 6owever, light is a H: vector field. 1 H: vector field assigns a H: vector $i.e., an arrow having both direction and length& to each point in H: space. 1 light wave has both electric and magnetic H: vector fields)
The H: wave e(uation for the electric field and its solution 1 light wave can propagate in any direction in space. So we must allow the space derivative to be H:)
r2
∂ 2 E ∇ E − µε 2 = 0 ∂t
whose solution is) E ( x, y, z , t ) = E 0 exp(− k $⋅ x )exp(i(k % ⋅ x − ω t )! Where E 0 is a constant, comple4 vector 1nd k = k % + ik $ is a comple4 wave vector the length of this vector is inversely proportional to the wavelength of the wave. Its magnitude is the angular wavenumber, k # JK0-.
x = ( x, y, z )
is a position vector
The H: wave e(uation for the electric field and its solution E ( x, y, z , t ) = E 0 exp(− k $⋅ x )exp(i( k % ⋅ x − ω t )! The vector kis% normal to planes of constant phase $and hence indicates the direction of propagation of wave crests& The vector is normal to planes of constant amplitude. +ote that these are not necessarilykparallel.
$
The amplitude of the wave at location
is now)
exp( − $ ⋅ ) E k x 0 is Lero, then the medium is nonabsorbing, since the amplitude is
So if constant .
k$
x
!" propagation in homogeneous materials The speed of an !" wave in free space is given by)
c=
1 µ 0ε 0
=
ω k
ε # permittivity of free space, µ0 # magnetic permeability of free space To describe !" propagation in other media, two properties of the medium are important, its electric permittivity & and magnetic permeability '. These are also comple4 parameters.
ε # ε$/M χ& M i σ/ω # complex permittivity σ # electric conductivity χ # electric susceptibility $to polariLation under the influence of an e4ternal field& +ote that & and ' also depend on fre(uency $ ω&
!" propagation in homogeneous materials In a non'vacuum, the wave must still satisfy "a4well8s !(uations) v=
1 µε
=
ω k
We can now define the complex index of refraction, +, as the ratio of the wave velocity in free space to the velocity in the medium) N =
µε µ 0ε 0
=
c v
or
N = n r + i ni
If the imaginary part of + is Lero, the material is nonabsorbing, and v is the phase velocity of the wave in the medium. or most physical media, + = / $i.e., the speed of light is reduced relative to a vacuum&.
!" propagation in homogeneous materials
k′ = k ′′ =
ω nr c
ω ni c
= =
2 π λ 2 πν ni
c
These are the so'called 9dispersion relations8 relating wavelength, fre(uency, velocity and refractive inde4.
1bsorption of !" radiation
1 2 F = cε 0 E 2
1bsorption of !" radiation
The scalar amplitude of an !" wave at location x is) E 0 exp(− k$⋅ x) rom the e4pression for the flu4 density we have) r
2
F F [exp( k $ )]
r
F exp( 2 k $ )
1bsorption of !" radiation
+ow substitute the e4pression for k ′′ ) 1nd we have) F = F 0 exp(−
πν ni
k ′′ = x)
ω ni c
=
2 πν ni
c
1bsorption coefficient and skin depth
F = F 0 exp(−
πν ni c
x) = F 0 e
− β a x
Where Na is known as the asorption coefficient!
πν ni c
=
π ni
λ
The (uantity /0N gives the distance re(uired for the wave8s energy to be
1bsorption coefficient and skin depth Within a certain material, and !" wave with - # / Rm is attenuated to /Q of its original intensity after propagating / cm. :etermine the imaginary part of the refractive inde4 n i.
n =1+
e
N k
∑ 2ε ! ω − ω + iγ ω 2
0
k
k
2
k
e # charge on an electron &0 # electric constant * # mass of an electron # number of charges $oscillators& of type k per unit volume ω # angular fre(uency of the !" radiation ω # resonant fre(uency of an electron bound in an atom # 9damping coefficient8 for oscillator k $oscillation cannot be permanent& What is the refractive inde4 of visible light in air? What happens as the fre(uency of !" radiation increases at constant ω ? What happens if the resonant fre(uency is in the visible range?
