Chap 2. Wave optics

2.1 POSTULATES OF WAVE OPTICS 2.2 MONOCHROMATIC WAVES

A C l R t ti d th H l h lt E ti A. Complex Representation and the Helmholtz Equation B. Elementary Waves C. Paraxial Waves

*2.3 RELATION BETWEEN WAVE OPTICS AND RAY OPTICS 2.4 SIMPLE OPTICAL COMPONENTS

A R fl ti d R f ti A. Reflection and Refraction Transmission Through Optical Components C. Graded-Index Optical Components

2.5 INTERFERENCE A. Interference of Two Waves

B M ltiple Wa e Interference B. Multiple-Wave Interference 2.6 POLYCHROMATIC LIGHT

A. Fourier Decomposition B. Light Beating

What is the wave?

Wave is the type that energy transformation

visible ()=>between 390 nm and 760 nm Ultraviole () =>10 to 390nm.Ultraviole () 10 to 390nm. Infrared () => 760 nm to 300 m.

NIR: near infrared

MIR: mid infrared

FIR: far infrared

NUV: near ultraviolet

VUV: vacuum ultraviolet

MUV: mid ultraviolet

FUV: far ultraviolet

EUV or XUV: extreme ultraviolet (soft X-rays)

Wave theory of light encompasses the ray theory

Ray optics is the limit of wave optics when the l th i i fi it i ll h twavelength is infinitesimally short

The light waves propagate through and around objects whose dimensions arearound objects whose dimensions are much greater than the wavelength,

the ray theory suffices for describing most phenomena

In wave optics, light is described by a scalar function (called

wavefunction)

obeys the wave equation.

The wave equation, constitute the postulates of the scalar wave model known as wave opticsthe scalar wave model known as wave optics.

The one-dimensional wave equation

Well derive the wave equation from Maxwells equations. Here it is in its one-dimensional form for scalar (i.e., non-vector) functions, f:

2 21f f The wave equation has the simple solution:

2 2 2

1 0v

f fx t

Light waves (actually the electric fields of light waves) will be a solution to this equation. And v will be the velocity of light.

The solution to the one-dimensional wave equation

( , ) ( v ) f x t f x twhere f (u) can be any twice-differentiable function.

How to describe the wave-function ?

( , ) ( v )f x t f x t x: position,

To displace any function f(x) to the right, just change its argument from x

t: time

to x-a, where a is a positive number.

If we let a = v t, where v is positive and i i h h di l ill

f(x)f(x-3)

f(x-2)f(x-1)

t is time, then the displacement will increase with time.

f( t) t i ht d f(x-vt) represents a rightward, or forward, propagating wave.

f (x+vt) represents a leftward orf (x+vt) represents a leftward, or backward, propagating wave.

v will be the velocity of the wave. x0 1 2 3

Proof that f (x vt) solves the wave equation

u u2 2

2 2 2

1 0v

f fx t

Write f (x vt) as f (u), where u = x vt. So and

Now use the chain rule: =>

1ux

vut

f f u f f Now, use the chain rule: >And =>

f fx u x

f f u x u

vf f

S d

t u t vt u 2 2

2f f 2 2f f So andSubstituting into the wave equation:

22 2vf f

t u 2 2

f fx u

g q2 2 2 2

22 2 2 2 2 2

1 1 v 0v v

f f f fx t u u

v vx t u u

The 1D wave equation for light waves

2 2

2 2 0E E

where E is the light electric field

Well use cosine- and sine-wave solutions:

2 2x t

( ) [ ( )] i [ ( )]k C k

electric field

( , ) cos[ ( v )] sin[ ( v )] E x t B k x t C k x t( v)kx k t

or ( , ) cos( ) sin( ) E x t B kx t C kx t ( v)kx k t

or ( , ) cos( ) sin( ) E x t B kx t C kx t 1 c

k The speed of light in vacuum,

ll ll d i 3 1010k usually called c0, is 3 x 1010cm/s.

: electric permittivity, 0: magnetic permeability ()

A simpler equation for a harmonic wave:

E(x,t) = A cos[(kx t) ]Use the trigonometric identity ():

( ) ( ) ( ) + i ( ) i ( )cos(zy) = cos(z) cos(y) + sin(z) sin(y)where z = k x t and y = to obtain:

E(x,t) = A cos(kx t) cos() + A sin(kx t) sin()which is the same result as before as long as:which is the same result as before, as long as:

A cos() = B and A sin() = CA cos() B and A sin() C( , ) cos( ) sin( ) E x t B kx t C kx t

For simplicity, well just use the forward-propagating wave.

Definitions

Spatial quantities:

c = = /c /

Temporal quantities:

The Phase Velocity

How fast is the wave traveling?

Velocity is a reference distancedivided by a reference time.

The phase velocity (c) is the wavelength / period:

c = In terms of the k-vector, k = 2, and the angular frequency, = 2, this is: c = / k

2.1 POSTULATES OF WAVE OPTICS

Light propagates in the form of waves. It is described by the wave equation

u ( r, t): wave functionr = (x, y, z): position function, t: time

: Laplacian operator2 2 2

2 2 2+ +x y z2

In free space , light waves travel with speed c0.x y z

In a medium of refractive index n, light waves travel with a reduced speed

The principle of superposition applies because the wave equation is linear.

if ul(r, t) and u2(r, t) represent optical wavesl( , ) 2( , ) p p u(r, t) = z+u1(r, t) + u2(r, t) also represents a possible optical wave.

At the boundary between two media the wave functionAt the boundary between two media, the wave-functionchanges in a way that depend on their refractive indexes.

Th ti i l i t l li bl t diThe wave equation is also approximately applicable to media with position-dependent refractive indices ( n = n(r)).

Provided the variation of refractive index is slow within distances of a wavelength (n ( r < )

Intensity, Power, and Energy Th i l i i I( ) The optical intensity I(r, t),

Defined as the optical power per unit area (units: watts/cm2), i l h f h f i proportional to the average of the squared wavefunction,

2 TE

The time interval T is ( , ) sin( )u x t E kx wt 2

0

2 s i n ( )E

I k x w t d tT

much longer than the time of an optical cycle but much shorter than any other time of interest (such as duration of

pulse)pulse) The duration of optical cycle

For example

1tc

t=2x10-15 (s) = 2 fs for =600 nm

Th i l P ( ) ( i ) i h i d i i The optical power P (t) (units: watts) is the integrated intensity

A : area normal to the direction of propagation of light

Th ti l ( i j l ) i h i i l f h The optical energy (units: joules) is the time integral of the optical power over the time interval

( t ) = ( r , t ) dTE P T collected in a given time interval.

2.2 MONOCHROMATIC WAVES

A monochromatic wave is represented by a wavefunction with harmonic time dependence

( r , t) = ( r )c o s[2 t + (r)] u r: amplitude wavelength=1/T: frequency (s-1) angular frequency (radius per second or s-1)g q y ( p )k=2wave numberrphase (initial phase)rphase (initial phase)

complex amplitude wave-function u(t)Complex wavefunction

A. Complex Representation and the Helmholtz EquationEquation

Complex Wavefucntion

So that

The symbol * signifies complex conjugateThe symbol signifies complex conjugateThe complex Wavefunction satisfy the wavefucntion

Complex Amplitude The complex wave function can be written in the form

Time independent factor U(r)=(r)exp[j(r)]: complex amplitude at a given position r

|U(r)|=(r): amplitude of the wavearg{U(r)}=(r) :phase{

The relation between wavefunction u(r, t) and complex amplitude is given by

The Helmholtz Equation Substituting U(r, t) = U(r)exp(j2t) into the wave equation

(2.2-4) =>Helmholtz equation

Wave number

Optical intensity

Optical IntensityF E 2 2 1From Eq.2.2-1

=>

U(r)=(r)exp[j(r)]: complex amplitudeAveraged over a time longer than an optical period, l/ causes

d i hsecond term vanish=>

Th i l i i f h i i h b l f i The optical intensity of a monochromatic wave is the absolute square of its complex amplitude.

The intensity of monchromatic wave does not vary with timeThe intensity of monchromatic wave does not vary with time.

The wavefronts

The wavefronts are the surfaces of equal phase, It is often defined by

(r) = constant =2q, q: an integer. The wavefront normal (at position r) is parallel to the gradient ( p ) p g

vector (r) The direction at which the rate of change of the phase is

maximum

Plane wave Spherical wavewave

B. Elementary WavesTh i l l i f h H l h l i i l The simplest solutions of the Helmholtz equation in a plane wave () and the spherical wave ().

Th Pl WThe Plane Wave The complex amplitude of plane wave

A( ) C l lA(r): Complex envelopek=(kx, ky, kz): wavevector ( )

h b

2 2 2 2x y zk k k k

The wavefront obey

2 arg{ } k r k x k y k z q Awhere q is integer

2 arg{ } x y zk r k x k y k z q A

The parallel plane perpendicular the wavevector k (plane wave)

These consecutive planes are separated by a distance (wavelength )

2k

(wavelength )

The wavefuction can be expressed as (z axis: the direction of the wavevector k)the wavevector k)

Space period : Space period :

Time period :/Time period :/

c : phase velocity (c = c0/n, c0 :velocity in vacuum )

/ / /c c n n As a monochromatic wave propagates through media of

0 0/ / /c c n n p p g g

different refractive indices Its frequency remains the same, Its velocity, wavelength, and wavenumber are altered:

The Spherical Wave

The complex amplitude of spherical wave

r: distance from the origingA: constant

Intensity (I(r)=|A0|2/r2 ):y ( ( ) | 0| ) inversely proportional to the square of the distance

The wavefronts fit the condition: kr = 2q or r = q q: integer

..These are concentric spheres separated by a radial distance =2/k that advance

di ll t th h l itradially at the phase velocity c

Fresnel Approximation of the Spherical Wave

A spherical wave at points r = (x, y, z) (from origin at r = 0) sufficiently close to the z axis but far from the origin

2 2( )( ) Using the Taylor-series expansion:

2 22 2

2

( ), 1 x yx y zz

The Fresnel approximation of a spherical wave (r @ z can be substituted for the magnitude):

The complex amplitude in (2.2-17)may be viewed as a plane wave (A0 exp(-jkz)) modulated by the factor

(l/z) exp[ -jk (x2 + y2)/2z]( ) p[ j ( 2 y2) ]

Paraboloidal wave : The phase factor ( k (x2 + y2)/2z ) serves to 2 2bend the planar wavefronts of the plane wave into paraboloidalsurfaces Pl h b l Plane wave: when z becomes very large.

The second condition of Fresnel approximationkz4/8

C. Paraxial Waves

A wave is said to be paraxial if its wavefront normals are paraxial rays.

The complex amplitude of the a paraxial wave

The complex envelope A is slowly varying function ofThe complex envelope A is slowly varying function of position (A = A(r))

The variation of the following function with the position zThe variation of the following function with the position zmust be slowly within the distance

a) complex envelope A(r)b) derivative of A(r)

the wave approximately maintain its underlying plane-wave nature.

The wave function of a paraxial wave:

The amplitude |A(0,0, z)| and phase (arg{A(x, y, z)}) vary slowly with z. Since the change of the phase is small within the distance of a

wavelength=> The planar wavefronts (kz = 2q) of the carrier plane wave bendThe planar wavefronts (kz 2q) of the carrier plane wave bend

only slightly Normals are paraxial rays.

(a) The magnitude of a paraxial wave as a function of the axial distance z. ( ) g p(b) The wavefronts and wavefront normals of a paraxial wave

The Paraxial Helmholtz EquationS b i i (2 2 20)Substitution (2.2-20)

into (2.2-7)

In paraxial approximation (A

A A A kA=

Helmholtz equation

(Eq.1)

Let

The following derivatives are necessary: g y

(Eq.2)

(Eq 3)(Eq.3)

Substitution (Eq 2 and Eq 3 into Eq 1)Substitution (Eq.2 and Eq.3 into Eq.1)

(Eq.4)

Paraxial Helmholtz equation: Slowly varying envelope approximation of the Helmholtz equationapproximation of the Helmholtz equation

T L l i tTransverse Laplacian operator2 2 2 2 2/ / T x y

The solution of the paraxial Helmholtz equationi ( i l i i f h h i l )

T y

Paraboloidal wave (paraxial approximation of the spherical wave) The complex envelope A(r)= (A/z) exp[ -jk (x2 + y2)/2z]

(Excise 2 2 2)(Excise 2.2-2)

Gaussian beam (will be discussed in Chap.3)

2.4 SIMPLE OPTICAL COMPONENTSA Reflection and RefractionA. Reflection and RefractionReflection from a Planar MirrorA plane wave of wavevector k is in incident onto planar mirror located inA plane wave of wavevector k1 is in incident onto planar mirror located in free space in the z = 0. Certain boundary conditions must be satisfied at the surface of the mirror (all points (x,y)).It is necessary that the wavefronts of two waves match

r ( 0)

z = 0

r = (x, y, 0), k1 = (k0 sin 1, 0, k0 cos ), k (k i 0 k )k2 = (k0 sin , 0, - k0 cos 2),

k i k i k0 x sin k0 x sin

Reflection and Refraction at a Planar Dielectric BoundaryBoundary

A plane wave of wavevector k, incident on a planar boundary between two homogeneous media of refractive indices n1 and n2.between two homogeneous media of refractive indices n1 and n2.

The boundary lies in the z=0 plane The refracted plane waves of wavevectors k2 emerge The refracted plane waves of wavevectors k2 emerge

1 2k r k r since (at z=0 in the x-z plane)

r = (x y 0)r (x, y, 0), k1 = (n1k0 sin 1, 0, n1k0 cos 1), k = (n k sin 0 n k cos )k2 = (n2k0 sin , 0, n2k0 cos 2),

sin sinn n1 1 2 2

B. Transmission Through Optical Components

The surfaces of the plate are the planes z = 0 and z = d and the incident wave travels in the z direction.

Complex amplitude transmittance t(x, y) = U(x, y, d)/U(x, y, 0) :( , y) ( , y, ) ( , y, )

The plane introduce a phase shift

nk0d =2(d/)z=0 z=d

Inclined Incident The incident plane wave makes an angle with respect to the z axis and

has wavevector k. The refracted and transmitted plane wave with wavevectors k1 and k and The refracted and transmitted plane wave with wavevectors k1 and k and

angle and The complex amplitude inside the plane is

Complex amplitude transmittance (U(x, y, d)/U(x, y, 0))

If the angle of incident is small If the angle of incident is small

=>2

1 11cos 12

>=>

2

1 / n 2

0 0( , ) exp( )exp( / 2 ) t x y jnk d jk d n

Thin Transparent Plate of Varying Thickness The amplitude transmittance of a thin transparent plate

whose thickness d(x, y) varies smoothly as a function of x and y.A hi l f h i f h thi k d d ( ) A thin layer of the air of the thickness: d0-d (x, y)

The plate lies between the planes z = 0 and z = d, the incident wave is an arbitrary paraxial waveis an arbitrary paraxial wave.

In paraxial approximation and d0 is sufficient small Complex amplitude transmittance Complex amplitude transmittance

0 0 0( , ) exp[ ( , )] exp[ ( ( , ))] t x y jnk d x y jk d d x y

h0 = exp (-jk0d0) :constant phase factor0 p ( j 0 0) p

Thin Lens The thickness at point (x, y) of planoconvex thin lens

or

By considering the points x and y are sufficient small in comparison with R (x2+y2

From eqs. 2.4-5=>2 22 2

0 0 0 0

2 2

x + y(x, y) exp[-jk d ]exp[-j(n -1)k (d - )]2R

t 2 2

0 0 0 0

2 2

x + y exp[-jk d ]exp[jk (-(n -1)d + )]2(R/(n -1))

2 2

0 0 0x + y exp[-jnk d ]exp[jk ]

2(R/(n -1))

Where

constant phase factorp

EXERCISE 2.4-2

D bl C L Sh h h l li dDouble-Convex Lens. Show that the complex amplitude transmittance of the double convex lens shown in Fig. 2.4-8 is given bygiven by

You can prove this by cascaded of two planconvex lense.

EXERCISE 2.4-4

Imaging Property of a Lens. Show that a paraboloidal wave centered at the point P1 (Fig. 2.4-10) is converted by a lens of focal length f into a paraboloidal wave centered about P2 , where 1/z1+1/z2=1/f . (imaging equation)

Diffraction Gratings A diffraction grating is used to periodically modulate the A diffraction grating is used to periodically modulate the

phase or the amplitude of the incident wave. It can be classified into It can be classified into

Transparent diffraction grating: Transparent plate:Transparent plate:

periodically varying thickness periodically graded refractive index

Repetitive arrays of diffracting elements such as apertures, obstacles, or absorbing element are used.

R fl ti diff ti ti Reflection diffraction gratings: Fabricated by use of periodically ruled thin films of aluminum that

have been evaporated onto glass substratehave been evaporated onto glass substrate.

Incident m = 1 First-orderZero-order

First-order

light wavem = 0

m = -1

mZero-order

First-orde

Incidentlight wave First-order

m -1 First-orde

(a) Transmission grating (b) Reflection grating

(a) Ruled periodic parallel scratches on a glass serve as a transmission grating. (b) Areflection grating. An incident light beam results in various "diffracted" beams. Thezero-order diffracted beam is the normal reflected beam with an angle of reflection equalto the angle of incidenceto the angle of incidence.

?1999 S.O. Kasap, Optoelectronics (Prentice Hall)

Diffraction grating made of a thin transparent plateplate A diffraction grating made of a thin transparent plate placed

in the z = 0 plane The thickness varies periodically in the x direction with

period In the paraxial approximation (>>)

i: angle of incident waveq: angle of diffraction wave

q = 0, 1, 2, 3, :diffraction order

Thin diffraction gratings without paraxial approximationapproximation

Without use of the paraxial approximation,

Diffraction gratings are used as filters and spectrum analyzers, and numerous applications in spectroscopy.

The diffraction angle are dependent on the wavelength polychromatic wave is separated by the grating into its spectral components

Principle of superposition

Principle of superposition => Principle of superposition => u1(r, t) and u2(r, t) represent optical wave

> ( t) ( t) + ( t) t ti l=> u(r, t) = u1(r, t) + u2(r, t) represents a optical wave. The resultant electric field at given place and time due to simultaneous action

of two or more sinusoidal waves is algebraic sum of the electric fields of the gindividual waves

(b)

(a)

1(b)

(a)

1

2(c) 2(c)

(b) 1

Two waves with the same frequency Two waves with the same frequency, Two waves with the same frequency, amplitude and phase

q y,amplitude and but or 180 phase difference

Phasors diagrams

Shows the electric field components and can be used to find the resultant magnitude of the electric fieldelectric field

The sinusoidal wave can be represented graphically by a phasor of magnitude Eo rotating about the origin counterclockwise with an angular frequency q y E1 = Eo sin t It makes an angle of t with the horizontal

iaxis E1 is the projection on the vertical axis

The second sinusoidal wave is The second sinusoidal wave is- E2 = Eo sin (t + )

It has the same amplitude and frequency as E1but having phase with respect to E1

Phasor Addition of Waves

The resultant is the sum of E1and E2

ER rotates with the same angular frequency angular frequency

The projection of ER along the e p oject o o R a o g t evertical axis equals the sum of the projections of the other two vectorsvectors

1 2E = E + Ep 1 2E E E

= sin(t) + sin(t + )= sin(t + )

E EE R = sin(t + )E

If the two waves have the same wavelength and different initial phase 1 and 2

sin ( ) sin ( ) 1 1 2 2x xy A ft A ft

( )2 xif ft( ) ( )Im[ ]1 21 2i i y A e A e

( )2if ft

( )

Im [ ( )]

Im [ ( ) / ]

1 2

1 2

1 2i ii

i ii i

e A e A ee A e A e e

( )

( )

Im [ ( ) / ]

Im[( ) / ]Im[ ]

1 2

1 2

1 2

1 2

i i i i

e A e A e e

A e A e e e sin ( ) ]2 xy B ft

sin ( )B Im [( ) / ]1 2i i iB A e A e e

Im [( ) / ]sin sin

sin

1 2

1 1 2 2

B A e A e eA A

sin

Conditions for Interference

To observe interference in light waves, the following two conditions

t b tmust be met:1) The sources must be coherent

()() They must maintain a constant

phase with respect to each other

Two light bulbs in a lamp do not demonstrate anot demonstrate a superposition

2) The sources should be2) The sources should be monochromatic ()

Monochromatic means they h i l l hhave a single wavelength

Youngs Double-Slit Experiment: Schematic

Thomas Young (1801) first demonstrated interference in light

f t iwaves from two sources in

The source pass through a The source pass through a pinhole=> limit the number of emitters in the source (establish spatial coherence)

Th li S d S

s

The narrow slits S1 and S2 act as sources of waves

The waves emerging from the slits originate from the same wave gfront and therefore are always in phase

Diffraction and inteference

From Huygenss principle we know the wavelet spread out from the slits From Huygens s principle we know the wavelet spread out from the slits This divergence of light from its initial line of travel is called diffraction

Constructive interference => the amplitude of the resultant wave isConstructive interference > the amplitude of the resultant wave is the sum of individual wave

Destructive interference =>the amplitude of the resultant wave is less pthan that of either individual wave

Measurement of the wavelength of light

Youngs experiment => measurement the wavelength of light.

The path difference, , is found from the tan triangle

y d y

d L L

This assumes the paths are parallelN t tl t b t Not exactly true, but a very good approximation if L is much greater than dmuch greater than d

2 1 = - = sin r r d2 1 sin r r d

Interference from three Slits

Maximum Secondary maximaI1=|E1+E2+E3|2= I2=|E1+E2+E3|2

Maximum Secondary maxima

=|E+E+E|=9E2 =|E-E+E|=E2

Phasor DiagramsAssume three equally spaced slits, the fields are:

E1 = Eo sin tE2 = Eo sin (t + )E2 = Eo sin (t + )E3 = Eo sin (t + 2)

Primary maxima

ky

the maximum value of field 3E0 at = 0, 2, 4 ...

Secondary maxima the value of filed E0, occur when the wave from one slit exactly cancels the wave from

I1=|E1+E2+E3|2=from one slit exactly cancels the wave from another slit,. at = 0, , 3 ...

=|E+E+E|=9E2

Total destructive interference occurs when the wave from all the slits form a closed triangle, the field at P has a value of 0.

I2=|E1+E2+E3|2

=|E-E+E|=E2closed triangle, the field at P has a value of 0. at = 0, 2/3, 4/3 ...

| |

Three Slits, Phasor Diagrams

Primary maxima Secondary maxima y y

M i k 2destructive interference

Maxima => k=m 2=> =m = d sinm=0,1,2,3

Minima => k=m 2/N

Multiple-Wave interferenceit ik i2k i(M-1)k

0

M-1it ink

() = I e (1+ e + e + L + e )

= I e eE

( ) ikEqual amplitude and phase

0n=0

I e eM

n 2 MM = r = (1 + r + r + L + r )

( ) ikr e

n = 0M

M -1 = + r1 +

M -1 = 1 + r

( ) 11 M M nr r( )

1 01 nM n rriM k iN k /2 -iM k /2 iM k /2

i t i t1 - e e (e - e )E () I ( ) Ii t i t0 0ik ik /2 -ik /2 ik /2

i[ t + (M -1)k /2 ]

( )E () = I e ( ) = I e1 - e e (e - e )

sin(M k /2)= e0 = e sin(k /2)

sin( / )( ) ( )sin( / )

20

22

MkI Ik

Maxima => k=m 2 =m = d sinsin( / )2k =m = d sinMinima => k=m 2/M

As the number of slits increases the primaryAs the number of slits increases, the primary maxima increase in intensity and become narrower

As the number of slits increases, the secondary maxima decrease in intensity with respect to the primary maximaprimary maxima

There are M-1 minima between adjacent principle maxima. There are M-2 minor maxima

0.75

1

0.25

0.5

3 2 0 2 3 4

Interference of M wave with Equal amplitude and Equal phase Differencesand Equal phase Differences

At maxima (=2)Imax=M2I0

sin( / )( ) ( )sin( / )

20

22

MI I

Imax M I0 Mean intensity

sin( / )2

2

1 2

I Id MI The peak intensity is

( / ) 00

1 2I Id MI The peak intensity is T