WAVE NATURE OF LIGHT 1.1 Light Waves in a Homogeneous Medium 1.2 Refractive Index

Chapter 1 Wave Nature of Light

1.3 Group Velocity and Group Index 1.4 Magnetic Field, Irradiance and Poynting Vector 1.5 Snell's Law and Total Internal Reflection (TIR) 1.6 Fresnel's Equations 1.7 Multiple Interference and Optical Resonators 1.8 Goos-Hnchen Shift and Optical Tunneling 1.9 Temporal and Spatial Coherence1

1.10 Diffraction Principles901 37500

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A. Plane Electromagnetic Wave

Ex

1.1 Light Waves in a Homogeneous Mediumy

Direction of Propagation

k

x z By z

An electromagnetic wave is a travelling wave which has time varying electric and magnetic fields which are perpendicular to each other and the direction of propagation, z.3

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Monochromatic plane wave

Ex = Eo cos(t kz + o )propagation constant (wave number): phase:

z

E and B have constant phase in this xy plane; a wavefront E E B k Propagation

k =/cEx

= t kz + 0

The optical field refers the electric field-k Ex = Eo sin(t z )

Wavefront A surface over which the phase of a wave is constant

Ex ( z , t ) = Re[ Eo exp( jo ) exp j (t kz )] or Ex ( z , t ) = Re[ Ec exp j (t kz )]5

z

A plane EM wave travelling along z, has the same E x (or By) at any point in a given xy plane. All electric field vectors in a given xy plane are therefore in phase. The xy planes are of infinite extent in the x and y directions.901 37500

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y

Direction of propagation

E (r , t ) = E0 cos(t k r + 0 )Wave vector

kk r = kx x + k y y + kz zr E(r,t) z

O

= t kz + o = constantPhase velocity

r

v =

A travelling plane EM wave along a directionk k?1999 S.O. Kasap, Optoelectronics (Prentice Hall)7

dz = = v dt k8

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B. Maxwells Wave Equation and Diverging WavesNo divergence Amplitude is independent of x and yWave fronts (constant phase surfaces) Wave fronts k Wave fronts

Maxwells EM wave equation

2E 2E 2E 2E + 2 + 2 = r 2 x 2 y z t

(5)

P

k

P

E r

O

Spherical wave (Wavefronts are spheres.) A E = c o s ( t k r ) (6) r Optical divergence: the angular separation of wavevectors on a given wavefront (spherical wave: 360 deg.)

z A perfect plane wave (a) A perfect spherical wave (b) Examples of possible EM waves ?1999 S O Kasap Optoelectronics (Prentice Hall) A divergent beam (c)9 10

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w0 : waist radius 2w0: spot sizeWave fronts

Gaussian wave beam divergence 4 2 = (7) ( 2 wo )y (b) x

1.2 Refractive Index

2wo O

z Beam axis Intensity

(a)

Gaussian

(c)

e

r2

2

r 2w

(a) Wavefronts of a Gaussian light beam. (b) Light intensity across beam cross section. (c) Light irradiance (intensity) vs. radial distance r from beam axis (z).?1999 S.O. Kasap, Optoelectronics (Prentice Hall)11 12

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The stronger the interaction between the field and the dipoles, the slower the propagation of the wave. (relative permittivity) Phase velocity

Example 1.2.1 Relative permittivity and refractive index

v =

1

r o oc = v

(1)

Refractive index

n =

r

(2)

Isotropic: material structure is the same in all directions, n independent of directione.g. cubic and non-crystalline

Anisotropic: Structure => physical properties depend on direction

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http://cst-www.nrl.navy.mil/lattice/13 14

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Example 1.2.1 Relative permittivity and refractive index

Permittivity due to different polarizations Molecular (polarized) + atomic + electronic Atomic (ionic or covalent) + electronic

1.3 Group Velocity and Group Index

Electronic Vacuum

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vg = + ?

1 d = dk dk dd = c = phase velocity dk

(1) (2)

v g (vacuum) =

Emax

Emax

k

v g = k = ( k ) 1

Wave packet

Two slightly different wavelength waves travelling in the same direction result in a wave packet that has an amplitude variation which travels at the group velocity.?1999 S.O. Kasap, Optoelectronics (Prentice Hall)901 37500 17

= vk = [

c 2 ][ ] n ( )

(3)

Material dispersion: waverform changes18

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Dispersive mediumv g ( medium ) = d c = dn dk n d(approximation) Around 1300nm minimal Ng zero dispersion

1.49 1.48 1.47 1.46 1.45 1.44 500 700 900 1100 1300 1500 1700 1900

Ng n

v g ( medium ) =

c Ng

(4)Wavelength (nm)

Group index

Ng

dn = n d

(5)

Refractive index n and the group index Ng of pure SiO2 (silica) glass as a function of wavelength.19

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Example 1.3.1 Group velocityConsider two sinusoidal waves that are close in frequency: frequencies = - and + Their wave vectors are: k - k and k + k 1. Resultant wave: Ex(z,t) = ? 2. What is the group velocity? (the speed of propagation of the maximum electric field along z)

Example 1.3.2 Group and phase velocityConsider a light wave traveling in a pure SiO2 (silica) glass medium. If the wavelength of light is 1 m and the refractive index at this wavelength is 1.450, what is the phase velocity, group index and group velocity?

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E x = vB y =

c By n

vt z

1.4 Magnetic Field, Irradiance and Poynting VectorB

E

Area A

kPropagation direction

A plane EM wave travelling along k crosses an area A at right angles to the direction of propagation. In time t, the energy in the cylindrical volume Avt (shown dashed) flows through A .?1999 S.O. Kasap,23

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Ex = vBy =

c By n

(1)

Poynting vector

electrical energy density = magnetic energy density (i.e. energy per unit volume) 1 1 2 o r Ex2 = By 2 2 o Total Energy:

S = v 2 0 r E B(2)Average irradiance (intensity)

(4)

0 r E x2

S = Energy flow per unit time per unit areaS= ( Avt )( o r Ex2 ) = v o r Ex2 = v 2 o r Ex By At

I = S average =

1 v o r Eo2 21 c o nEo2 = (1.33 103 )nEo2 2

(5) (6)

(3)

I = S average =

(instantaneous irradiance)901 37500

S = v 2 0 r E B25 26

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Example 1.4.1 Electric and magnetic fields in lightThe intensity (irradiance) of the red laser beam from a HeNe laser at a certain location has been measured to be about 1mW/cm2. What are the magnitudes of the electric and magnetic fields? What are the magnitudes if this beam is in a glass medium with a refractive index n = 1.45?

1.5 Snells Law and Total Internal Reflection (TIR)

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tRefracted Light A

At kt Bt

ReflectionAB ' = v1t vt = 1 sin i sin r

y

tz A B

t

i = rn2 n1

O

i rA B ki

i r

ikr Br

t r

RefractionAB ' = or v 1t v 2t = s in i s in t

(1)Snells Law

s in i v n = 1 = 2 s in t v2 n1

Ai Ar

Critical angle (if n1 > n2)sin c = n2 n1

Incident Light

Bi

Reflected Light

(2)30

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A light wave travelling in a medium with a greater refractive index (n1 > n2) suffers 29901 37500

TIR and evanescent waveTransmitted (refracted) light kt n2 kr n 1 > n2

tki Incident light (a)

Evanescent wave

i

i

c c

i >c

1.6 Fresnels EquationsA. Amplitude reflection and Transmission coefficients (r and t)

Reflected light (b)

TIR(c)

Light wave travelling in a more dense medium strikes a less dense medium. Depending on the incidence angle with respect to c, which is determined by the ratio of the refractive indices, the wave may be transmitted (refracted) or reflected. (a) i < c (b) i = c (c) i > c and total internal reflection (TIR).?1999 S.O. Kasap, Optoelectronics (Prentice Hall) TIR: i > c sint > 1901 37500

t : imaginary angle31 32

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y

E t,//tz

Transmitted wave kt

Transverse electrical field (TE) Ei, , Er, , Et, : electrical field components perpendicular to z-directiont=90n2

E t, E r,kr Reflected wave Incident wave

E t, Evanescent wave E r,Reflected wave n1 > n 2

x into paper

E i,//

ki

i r

E i,// E i,

i r

Transverse magnetic field (TM) Ei,// , Er,// , Et,// : magnetic field components perpendicular to z-direction

E i,Incident wave

E r,//

E r,//

Ei = Eio exp j ( t ki r ) Er = Ero exp j ( t k r r ) Et = Eto exp j ( t kt r )Boundary conditions

(a) i < c then some of the wave is transmitted into the less dense medium. Some of the wave is reflected.

(b) i > c then the incident wave suffers total internal reflection. However, there is an evanescent wave at the surface of the medium.

Light wave travelling in a more dense medium strikes a less dense medium. The plane of incidence is the plane of the paper and is perpendicular to the flat interface between the two media. The electric field is normal to the direction of propagation . It can be resolved into perpendicular () and parallel (//) components 33901?1999 S.O. Kasap, Optoelectronics (Prentice Hall) 37500

E tangential (1) = Etangential (2) Btangential (1) = Btangential (2)901 37500

r = i n1sin1 = n2sin234

Define: n = n2 / n1r = Er 0, Ei 0, cos i [ n 2 sin 2 i ]1/ 2 = cos i + [ n 2 sin 2 i ]1/ 2

3/19/2009

r =

Er 0, Ei 0, Er 0,// Ei 0,//

=

(1a) (1b)r// =

cos i [ n 2 sin 2 i ]1/ 2 cos i + [ n 2 sin 2 i ]1/ 2

(1a)

t =

Et 0, Ei 0, Er 0,// Ei 0,//

=

2 cos i cos i + [ n 2 sin 2 i ]1/ 2

[ n 2 sin 2 i ]1/ 2 n 2 cos i = 2 [ n sin 2 i ]1/ 2 + n 2 cos i

(2a)

r// =

[ n 2 sin 2 i ]1/ 2 n 2 cos i = 2 [ n sin 2 i ]1/ 2 + n 2 cos i

(2a)

Normal incidence

r// = r =

n1 n2 n1 + n2

(4)

t // =

Et 0,// Ei 0,//

=

2 n cos i n cos i + [ n 2 sin 2 i ]1/ 22

(2b) (3)35

Brewsters angle or polarization angle

tan p =

r// + nt // = 1901 37500

and

r + 1 = t

n2 n1

(5)36

=> linearly polarized wave901 37500

General formula

Magnitude of reflection coefficients

Phase changes in degrees 180 TIR (b) 120 60 0

c1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 (a)

p c

| r |p

60

//

| r // |

120 180 0 10 20 30 40 50 60 70 80 90

0 10 20 30 40 50 60 70 80 90

Incidence angle, i

Incidence angle, i

Internal reflection: (a) Magnitude of the reflection coefficients r// and r vs. angle of incidence i for n1 = 1.44 and n2 = 1.00. The critical angle is 44? (b) The corresponding phase changes // and vs. incidence angle.37

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Applications Camera polarization filters (PL) Sunglasses Laser tubes (Brewster angle)

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r =

Er 0, Ei 0,

=

cos i [ n 2 sin 2 i ]1/ 2 cos i + [ n 2 sin 2 i ]1/ 2

(1a)

r// =

Er 0,// Ei 0,//

[n 2 sin 2 i ]1/ 2 n 2 cos i = 2 [n sin 2 i ]1/ 2 + n 2 cos i

(2a)

Total internal reflection: i > c

sin i >

n2 >n n1

r = 1.0 exp( j ) r// = 1.0 exp( j// )http://health.howstuffworks.com/sunglass.htm/printable901 37500

(amplitude of the ref. is equal to that of inc. with phase change)43 44

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Magnitude of reflection coefficients

Phase changes in degrees 180 TIR (b) 120 60 0

c1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 (a)

Phase change in TIR[sin 2 i n 2 ]1/ 2 1 tan( ) = 2 cos i[sin 2 i n 2 ]1/ 2 1 1 tan( // + ) = n 2 cos i 2 2

(6) (7)

p c

| r |p

60

//

| r // |

120 180 0 10 20 30 40 50 60 70 80 90

i > cEvanescent wave Attenuation coef. Penetration depth901 37500

0 10 20 30 40 50 60 70 80 90

Et , ( y, z , t ) = e 2 y exp j (t kiz z ) (8)2 =2 n2

Incidence angle, i

Incidence angle, i

Internal reflection: (a) Magnitude of the reflection coefficients r// and r vs. angle of incidence i for n1 = 1.44 and n2 = 1.00. The critical angle is 44? (b) The corresponding phase changes // and vs. incidence angle.?1999 S.O. Kasap, Optoelectronics (Prentice Hall)45

[(

n1 2 2 ) sin i 1]1/ 2 n2

(9)46

= 1/ 2

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Internal reflection n1 > n2 External reflection n1 < n21 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 0

no phase change at normal incidence phase shift of 180 at normal incidence

Example 1.6.1 Evanescent waveTotal internal reflection (TIR) of light from a boundary between a more dense medium n1 and a less dense medium n2 is accompanied by an evanescent wave propagating in medium 2 near the boundary. Find the functional form of this wave and discuss how its magnitude varies with the distance into medium 2.

External reflection

p

r//

No phase change for transmitted lightr10 20 30 40 50 60 70 80 90

Incidence angle, i

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The reflection coefficients r// and r vs. angle of incidence i for n1 = 1.00 and n2 = 1.44.

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B. Intensity, Reflectance and TransmittanceLight intensity or irradiance Reflectance R

Normal incidence

I =

1 v r o E o2 2| Ero,// |2

(10)

Reflectance

R = R = R// = (=| r |2 and R// = | Eio,// |2 =| r// |2

R =

| Ero, | | Eio, |

2

n1 n2 2 ) n1 + n2

(13)

2

(11)

e.g. air-glass: R = (1.5-1)2/(1.5+1)2 = 4% TransmittanceT = T = T // = 4 n1 n 2 ( n1 + n 2 ) 2

r: complex R: real Transmittance T 2 cos t n2 Eto ,T = cos i n1 Eio , cos t n2 Eto , // cos i n1 Eio , //2 2 2

| Ero ,// |2 = ( Ero ,// )( Ero ,// )*

(14)

cos t n2 = cos n t i 1 cos t n2 = cos n t // i 1

2

(12)2

Power conservation (valid for arbitrary incident angle)

T// =

R + T =149

(15)50

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Example 1.6.2 Reflection of light from a less dense medium (internal reflection)A ray of light is traveling in a glass medium of n1 = 1.45 becomes incident on a less dense glass medium of n2 = 1.43. Suppose the free space wave length of the light ray is 1m. (a) What should the minimum incidence angle for TIR be? (b) What is the phase change in the reflected wave when i = 85 and i = 90 (c) What is the penetration depth of the evanescent wave into medium 2 when i = 85 and i = 90

Example 1.6.3 Reflection at normal incidence. Internal and external reflectionConsider the reflection of light at normal incidence on a boundary between a glass medium of refractive index 1.5 and air of refractive index 1. (a) If light is traveling from air to glass, what is the reflection coefficient and the intensity of the reflected light w.r.t. that of the incident light? (b) If light is traveling from glass to air, what is the reflection coefficient and the intensity of the reflected light w.r.t. that of the incident light? (c) What is the polarization angle in the external reflection in (a) above? How would you make a polariod device that polarizes light based on the polarization angle?

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Example 1.6.4 Antireflection (AR) coatings on solar cellsConsider wavelength around 700-800 nm No AR coating: n1 (air) = 1 and n2 (Si) = 3.5 With AR coating: n1 (air) = 1, n2 (Si3N4) = 1.9, n3 (Si) = 3.5

Example 1.6.5 Dielectric mirrorsDefinition: a stack of dielectric layers of alternating refractive indices 1. n1 < n2 2. t1 = 1 / 4 and t2 = 2 / 4 (quarter wavelength of the light in the layer) where 1 = 0/n1 and 2 = 0/n2 At 0: reflected waves from the interfaces interfere constructively mirror

n1 A B

d n2

n3A B C

1/4

2/4

Reflectance 1

Surface Antireflection Semiconductor of photovoltaic device coating

1 n1

2 n2

1 n1

2 n2

0 330

(nm)

o

550

770

reflected 901 37500

Illustration of how an antireflection coating reduces the light intensity

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Schematic illustration of the principle of the dielectric mirror with many low and high 54 refractive index layers and its reflectance.

1.7 Multiple Interference and Optical Resonators

LC optical resonators (narrow band around f0) Store energy Filtering light at certain frequencies (wavelengths) Used in laser, interference filter, and spectroscopic applications

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Stationary or standing EM waves

Fabry-Perot optical resonator

m( ) = L 2Cavity mode:M1 M2

m = 1, 2,3...

(1)

vm = m (

c ) = mv f ; 2L

v f = c / 2L

(2)

Lowest frequency = ?m=1 m=2 Relative intensity 1

Free spectral range = ?R ~ 0.8 R ~ 0.4

A

f

mB L(a)

A + B = A + Ar 2 exp( j 2 kL )

m=8(b)

m - 1

m(c)

m + 1

Ecavity = A + B + ... = A + Ar 2 exp( j2kL) + Ar 4 exp( j 4kL) + Ar6 exp( j6kL) + ...

Schematic illustration of the Fabry-Perot optical cavity and its properties. (a) Reflected waves interfere. (b) Only standing EM waves, modes, of certain wavelengths are allowed in the cavity. (c) Intensity vs. frequency for various modes. R is mirror reflectance57 and lower R means higher 901 37500 loss from the cavity.

Ecavity =

A 1 r exp( j 2kL)258

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Ecavity =I cavity = I max =

A 1 r exp( j 2kL)22

Io (1 R ) + 4 R sin 2 (kL) km L = m

(3) (4)

I transmitted = I incident

(1 R ) 2 (1 R) 2 + 4 R sin 2 (kL)

(6)

Io ; (1 R) 2

Partially reflecting plates

Transmitted light

Spectral width (FWHM)

Finesse

Input light

Output light

vm =

vf F

;

F=

R1/ 21 R

(5)to spectral width m59

L

Finesse: the ratio of mode separation f901 37500

Fabry-Perot etalon

m - 1

m

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Transmitted light through a Fabry-Perot optical cavity.

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Example 1.7.1 Resonator modes and spectral widthConsider a Fabry-Perot optical cavity of air of length 100 micros with mirrors that have a reflectance of 0.9. (a) Calculate the cavity mode nearest to 900 nm. (b) Calculate the separation of the modes and the spectral width of each mode.

1.8 Goos-Hanchen Shift and Optical Tunneling

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Lateral shift of the reflected beam due to TIR

Optical tunneling Frustrated total internal reflection (FTIR)n2

z = 2 tan iByPenetration depth, Virtual reflecting plane

yz

C B A

AIncident light

iz

r

n 1 > n2

n1 n2d

Goos-Haenchen shift Reflectedlight

i r

n1 > n2

z

The reflected light beam in total internal reflection appears to have been laterally shifted by an amount z at the interface.?1999 S.O. Kasap, Optoelectronics (Prentice Hall) In Ex. 1.6.2 n1=1.45, n2=1.43, inc. at 85 deg.

Incident light

Reflected light

=0.78m, => z ~ 18 m63

When medium B is thin (thickness d is small), the field penetrates to the BC interface and gives rise to an attenuated wave in medium C. The effect is the tunnelling of the incident beam in A through B to C.901 37500 ?1999 S.O. Kasap, Optoelectronics (Prentice Hall)64

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Frustrated total internal reflection Beam splitter Gap thickness and Refractive indexReflected

Reflected light n2 n1

B = Low refractive index transparent film ( n 2) n1

1.9 Temporal and Spatial CoherenceFTIR

TIR

i > cA Incident light

n1 Transmitted

i > cC A

Incident light

(a) Glass prism

(b)

(a) A light incident at the long face of a glass prism suffers TIR; the prism deflects the light. (b) Two prisms separated by a thin low refractive index film forming a beam-splitter cube. 65 The incident beam is split into two beams by FTIR.901 37500 Kasap, Optoelectronics (Prentice Hall) ?1999 S.O. 901 37500

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Perfect coherence:

any two points such as P and Q separated by any time interval are always correlated.Amplitude

(a)

P Time

Q

Perfect coherence: all points on the wave are predictable Temporal coherence Measures the extent to which two points separated in time at a given location in space can be correlated

Amplitude

e.g.

t

l = ct

(b)Time Field Space spectral width = 1/t

589 nm of sodium lamp, v ~ 5 x 1011 Hz, t ~ 2 x 10-12 s = 2 ps He-Ne laser, v=1.5 x 109 Hz, l ~ 200 mm Laser devices have substantial coherence length, and therefore widely used in wave-interference studies and applications

(c)

P

Q Time

Amplitude

(a) A sine wave is perfectly coherent and contains a well-defined frequency o. (b) A finite wave train lasts for a duration t and has a length l. Its frequency spectrum extends over = 1/ t. It has a coherence time t and a coherence length l. (c) White light exhibits 67 practically no coherence.901 37500 901 37500

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Mutual temporal coherence over the time intervalNo interference Interference t No interference

(a)

A B Source Time

1.10 Diffraction Principles

(b)

P c Q

Spatially coherent source

(c)Space

c

An incoherent beam

(a) Two waves can only interfere over the time interval t. (b) Spatial coherence involves comparing the coherence of waves emitted from different locations on the source. 69 An (c) incoherent beam. 901 37500

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A. Fraunhofer DiffractionFraunhofer diffraction Fresnel diffractionLight intensity pattern

Huygens-Fresnel principleIncident plane waveA secondary wave source Another new wavefront (diffracted)

Incident light beam: a plane wave Observation (detection): far away from the aperture

New wavefront

z

Diffracted beam Incident light wave Circular aperture

(a)

(b)

A light beam incident on a small circular aperture becomes diffracted and its light intensity pattern after passing through the aperture is a diffraction pattern with circular bright rings (called Airy rings). If the screen is far away from the aperture, this would be a Fraunhofer diffraction pattern.?1999 S.O. Kasap, Optoelectronics (Prentice Hall) 901 37500 71

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(a) Huygens-Fresnel principles states that each point in the aperture becomes a source of secondary waves (spherical waves). The spherical wavefronts are separated by . The new wavefront is the envelope of the all these spherical wavefronts. (b) Another possible 72 wavefront occurs at an angle to the z-direction which is a diffracted wave.

E ( y ) exp( jky sin )E ( ) = C Incident light wave Yy=a y =0

(1) (2)E ( ) = C E ( ) =b cy=a y=0

y exp( jky sin )y

y exp( jky sin )1 a sin( ka sin ) 2

(2)

Incident light wave R = Large

Screen

yy

Ce

1 j ka sin 2

a A

1 ka sin 2

y z

y

ysin

z =0

1 C ' a sin( ka sin ) 2 I ( ) = [ ]2 = I (0) sinc 2 ( ); 1 ka sin 2

=

1 ka sin 2

(3)

= ky sin (a)

Light intensity

(b)

Zero intensity at sin =

m ; a

m = 1, 2, ...

(4)74

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(a) The aperture is divided into N number of point sources each occupying y with amplitude y. (b) The intensity distribution in the received light at the screen far away 73 from the aperture: the diffraction pattern

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Rectangular aperture1 C ' a sin( ka sin ) 2 ]2 = I (0) sinc 2 ( ); I ( ) = [ 1 ka sin 2b aS1 A2 S1 s S2 A1 I Screen L

1 = ka sin 2

Rayleigh criterion: two spots are resolvable when the principal max. of one diffraction pattern coincides with the min. of the other.

sin = 1.22y

D

The rectangular aperture of dimensions a b on the left gives the diffraction pattern on the right.?1999 S.O. Kasap, Optoelectronics (Prentice Hall) Airy rings (circular aperture)

S2

Divergence angle of Airy disk901 37500

sin = 1.22

D

(5)75

Resolution of imaging systems is limited by diffraction effects. As points S1 and S2 get closer, eventually the Airy disks overlap so much that the resolution is lost.?1999 S.O. Kasap, Optoelectronics (Prentice Hall)901 37500 76

Example 1.10.1 Resolving power of imaging systemstwo neighboring coherent sources are examined through an imaging system with an aperture of diameter D - angular separation of the two sources at the aperture: - diffraction pattern of the sources: S1 and S2 Q: The human eye has a pupil diameter of about 2 mm. What would be the minimum angular separation of two points under a green light of 550 nm and their minimum separation if the two objects are 30 cm from the eye?y A2 S2 I

B. Diffraction gratingGrating equation Bragg diffraction condition

Periodic series of slits in an opaque screen

-

d sin = m ;One possible diffracted beam Single slit diffraction envelope

m = 0, 1, 2,...y

(7)

y Incident light wave

a

(Normal to the grating)

m=2 m=1 m=0 m = -1dsin

Second-order First-order Zero-order First-order Second-order

S1

S1 s S2

d

z

m = -2

A1 Screen

L

Diffraction grating

Intensity (b )78

Resolution of imaging systems is limited by diffraction effects. As points S1 and S2 get closer, eventually the Airy disks overlap so much that the resolution is lost.?1999 S.O. Kasap, Optoelectronics (Prentice Hall)

(a)77

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pattern. There are 901 37500 distinct beams in certain directions (schematic)

(a) A diffraction grating with N slits in an opaque scree. (b) The diffracted light

d (sin m sin i ) = m ;(not normal to the grating)

m = 0, 1, 2,...

(8)Normal to face First order Normal to grating plane

First-order Zero-order

i : angle of incidence w.r.t. grating normal

Incident light wave

m = 1 First-order m = 0 Zero-order m = -1 First-orde

Incident light wave

First-order

(a) Transmission grating (b) Reflection grating

d

(a) Ruled periodic parallel scratches on a glass serve as a transmission grating. (b) A reflection grating. An incident light beam results in various "diffracted" beams. The zero-order diffracted beam is the normal reflected beam with an angle of reflection equal to the angle of incidence.?1999 S.O. Kasap, Optoelectronics (Prentice Hall)901 37500 79

Blazed (echelette) grating.?1999 S.O. Kasap, Optoelectronics (Prentice Hall)80

901 37500

Blazed diffraction gratingN grating normal N groove face normal

i b = () m + b m = 2 b i a (sin i + sin m ) = m a[sin i + sin( 2 b i )] = mif

b = i m 2a

Monochromator Fancy stickers Nature structures (birds, fish, bugs)

2a sin b = m + _m

b = arcsin

Number of grooves/mm and blazed angle determine the wavelength range of the grating.

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901 37500

Other application Enhancement of light extraction in LED, OLED structures Reduce total internal reflection Inhibit undesired direction

Nature, 2009901 37500

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