36
Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same . Let A = the source strength per unit area. Then each infinitesimal area element dS emits a spherical wave that will contribute an amount dE to the field at P (X, Y, Z) on the screen dS e r dE kr t i A ) ( The distance from dS to P is 2 2 2 ) ( ) ( z Z y Y X r which must be very large compared to the size (a) of the aperture and greater than a 2 / in order to satisfy conditions for Fraunhofer diffraction. Therefore, as before, for OP , we can expect A /r A /R as before (i.e., the 2 2 2 2 Z Y X R

Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let A = the source strength per unit

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Page 1: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same.

Let A = the source strength per unit area. Then each infinitesimal area element dS emits a spherical wave that will contribute an amount dE to the field at P (X, Y, Z) on the screen

dSer

dE krtiA )(

The distance from dS to P is

222 )()( zZyYXr

which must be very large compared to the size (a) of the aperture and greater than a2/ in order to satisfy conditions for Fraunhofer diffraction. Therefore, as before, for OP , we can expect A/r A/R as before (i.e., the behavior is approximated as that of a plane wave far from the source).

2222 ZYXR

Page 2: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Fig. 10.19 A rectangular aperture.

2/

2/

/2/

2/

/)(~ a

a

RikZzb

b

RikYykRtiA dzedyeeR

E

At point P (X, Y), the complex field is calculated as follows:

22

2

)(

sinsin)0(

~Re),(

sinsin~

IEZYI

isirradianceaveragedtimetheand

R

eAE

T

kRtiA

R

kaZand

R

kbY

22

Page 3: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

For x, y << R, we can approximate r as follows:

Aperture

RZzYyikkRtiA

Aperture

krtiA

Aperture

dseeR

dser

dEE

Therefore

RZzYyR

RZzYyRzyRZzYyzyR

zzZZyyYYXzZyYXr

/)()()(

2

2/122222/1222

2/1222222/1222

~

,

.../1

]/)(2/)(1[)](2[

22)()(

Thus, for the specific geometry of the rectangular aperture:

sin

2

sin

2

22

~

2/

2/

/2/

2/

/

2/

2/

/2/

2/

/)(

ai

eeadzeandb

i

eebdyeThen

R

kaZand

R

kbYLet

dzedyeeR

E

iia

a

RikZziib

b

RikYy

a

a

RikZzb

b

RikYykRtiA

Page 4: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Therefore, the resulting complex field at point P on the screen for a rectangular aperture having area A = ab is given by

22

2

)(

sinsin)0(

~Re),(

sinsin~

IEZYI

isirradianceaveragedtimetheandR

eAE

T

kRtiA

where I(0) is the intensity at the center of the screen at point P0 (Y = 0, Z = 0).

A typical far-field diffraction pattern is shown in Fig. 10.20. Note that when = 0 or = 0, we get the familiar single slit pattern.

The approximate locations of the secondary maxima along the -axis (which is the Y- axis when = 0 or Z = 0) is given by m 3/2, 5/2, 7/2... Since sin = 1 at these maxima, the relative irradiances along the -axis are approximated as

22

)0(),0(,

)0()0,(

mm

IZIaxisZortheforsimilarlyand

IYI

where m 3/2, 5/2, 7/2... 122

1~

62

1~

22

1~1 I

Page 5: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Square aperture in which a = b.

R

kaZand

R

kbY

22

where m= m 3/2, 5/2, 7/2... are the positions of the secondary maxima and

122

1~

62

1~

22

1~1 I

(Z)

(Y)

Distribution of irradiance, I(Y,Z)

Distribution of electric field, E(Y,Z)

Page 6: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Fig. 10.21 Circular Aperture Geometry

A very important aperture shape is the circular hole, as this involves the natural symmetry for lenses. Such a symmetry suggests the need for cylindrical coordinates:

)cos()/(

0

2

0

)(~

sin,cos,sin,cos

RqkiakRti

A eddR

eE

isfieldcomplextheanddddydzds

qYqZyz

RZzYyike /)(

)cos(

)coscossin(sin

:/)(

q

qzZyY

involvesetermthethatNote RZzYyik

We are calculating the field E on the screen as a function of the screen’s radial coordinate q.

Page 7: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

The circular shape of the aperture results in complete axial symmetry. Therefore, the solution must be independent of . Therefore, we are permitted to set = 0. We are therefore led to evaluate the integral:

,2

1)(

2

0

cos0

2

0

cos)/( dveuJde viuRqki

where J0(u) is the Bessel function (BF) of order zero. More generally the BF of order m, Jm(u), is represented by the following integral:

,2

)(2

0

)cos()(

dvei

uJ vumvim

m

Bessel functions are slowly decreasing oscillatory functions very common in mathematical physics.

Therefore the field is expressed as

akRti

A dRqkJR

eE

0

0

)(

)/(2~

Rqkuand

vwith

/

Fig. 10.22

Page 8: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Another important property of the Bessel Function is the recurrence relation that connects BFs of consecutive orders:

wdwwJkqRdRqkJ

anddwkq

RdthenRqkwLet

dummyuJuJ

dJd

ddJ

uJuuJudu

dmwhen

uJuuJudu

d

Rkaq

w

a

u

u u

mm

mm

)()/(/

,/

.var)()(

)()(

)()(,1

)()(

/

0

02

0

0

101

1

0 0

0

01

1

Page 9: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

akRti

A dRqkJR

e

0

0

)(

)/(2

RkaqJkaq

Ra

R

e

eddR

eE

kRtiA

RqkiakRti

A

/2

~

12

)(

)cos()/(

0

2

0

)(

We can express the field, using the recurrence relation, as

Therefore, the irradiance at point P on the screen is

2

2

1

2* ;

/

/2

~~

2

1)( aA

Rkaq

RkaqJ

R

AEEqI A

It is useful to examine the series representation of the Bessel Functions:

)!1()(

...22

1

2)(

)1(!

)2/()1()(

3

10

2

nnwhere

xxxJ

knk

xxJ

k

knk

n

Page 10: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Consider the limit near x = 0: 2

1)(lim 1

0

x

xJx

Therefore, the irradiance at P0 when q = 0 is

222

2

1

2

12)0(

R

A

R

AI AA

2

1

2

1

sin

sin2)0()(

sin;/

/2)0()(

ka

kaJII

R

qwithand

Rkaq

RkaqJIqI

and the Irradiance becomes

(Fig. 10.21)

We usually express the irradiance as a function of the angular deviation from the central maximum at point P0. The large central maximum is called the Airy Disk, which is surrounded by the first dark ring corresponding to the first zero of J1(u). J1(u) = 0 when u = 3.83 or kaq1/R = kasin1= 3.83.

Page 11: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

DRqa

R

a

Rq

R

aqRkaq

/22.1/sin2

22.12

83.383.3

2/

11

11

1

D = 2a is the diameter of the circular hole

Note that ~84% of the light is contained in the Airy Disk (i.e. 0 kasin 3.83)

Airy Disk

Airy Disk

Page 12: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

D = 0.5 mm

D = 1.0 mm

Airy Rings with different hole diameters

Airy Disk

Suppose that the aperture is a lens which focuses light on a screen:

Entrance Pupil (Aperture)

screen

f

Converging lens

D

DfqRfSince /22.11

which gives the radius of the Airy disk on the screen.

Page 13: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Fig. 10.24 Overlapping Images

Suppose that we image two equal irradiance point sources (e.g., stars) through the objective lens of a telescope. The angular half-width of each image point is q1/f = sin . If the angular separation of the stars is and >> the

images of the stars will be distinct and well resolved .

Rays arriving from two stars and striking a lens

Analysis of overlapping images using Airy rings

Page 14: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

)l(min

D

22.11

Half-angle of an airy disk:

Rays from two stars

If the stars are sufficiently close in angle so that the center of the Airy disk of star 1 falls on the first minimum (dark ring) of the Airy pattern of star 2, we can say that the two stars are just barely resolved. In this case, we have 1 = ( )min = q1/f 1.22 / D (l)min 1.22 f/D . This is Rayleigh’s criterion for angular or spatial resolution.

Fig. 10.25

f

Rayleigh’s criterion for the minimum resolvable angular separation or angular limit of resolution

Page 15: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Another criterion for resolving two objects has been proposed by C. Sparrow.

At the Rayleigh limit there is a central minimum between adjacent peaks.

Further decrease in the distance between two point sources will cause this minimum (dip) to disappear such that

0)(

2

2

Sprdr

rIdSp

The resultant maximum will therefore have a broad flat top when the distance between the peaks is r = Sp, and serves as the Sparrow criterion for resolving two point objects.

Page 16: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Diffraction Gratings

Transmission Grating is made by scratching rulings or notches onto a clear flat plate of glass. Each notch serves as a source of scattering that affects radiating secondary sources, in much the same way as for a multiple-slit diffraction array. When the phase conditions are met through OPD = = m, constructive

interference is observed .

Oblique incidence (i > 0)

maCDABTherefore

ABa

CDa

im

m

i

)sin(sin

sin

sin

m = 0 (zeroth order), m = 1 (first order),

m = 2 (second order), m = 3 (third order)

from the geometry.

Page 17: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

maCDAB im )sin(sin

The diffraction grating can also be constructed as a reflection grating. The principals and conditions for constructive interference are the same as that for a transmission grating.

Most commercial gratings for spectroscopy are constructed with a Blaze angle to control the efficiency of diffraction for a particular and order m.

Page 18: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Most of the incident light undergoes specular reflection, similar to a plane mirror, and this occurs when i = m and m = 0 for the zeroth order beam. The problem is that most of the irradiance is wasted for the purpose of spectroscopy. It is possible to shift the reflected energy distribution into a higher order (m = 1) in which m depends on . It is possible to change the distribution of the specular reflection by changing the blaze angle so that the first order diffraction is optimized for a particular range of wavelengths.

Consider the situation such that: i = 0 so m = 0, 0 = 0. For specular reflection i - r = 2 and so most of the diffracted irradiance is concentrated near r = -2. This will correspond to a particular non-zero order in which m = -2 and asin(-

2) = m .

Controlling the irradiance distribution of diffracted orders using a Blazed grating.

Page 19: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Schematic from the Fluorescence Group, University of California, Santa Barbara, USA

Page 20: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

•Exit slit determines spectral resolution of the instrument •Resolution is determined by the product of the monochromator linear

dispersion (nm/mm) and the slit width •Monochromator resolution depends on the grating pitch and the

(focal) length of the monochromator •For PTI monochromator with 1200 groove/mm grating the reciprocal

linear dispersion is 4 nm/mm. •1 turn of the slit micrometer = 0.5 mm slit opening = 2 nm

spectral resolution. Note that since E = h = hc/ E = (-hc/2)

Monochromators with micrometer adjustable entrance and exit slit widths

320 360 400 440 480

P=200 W/cm2

Sample

x 2.4

x 1.04

x 1

S2

S8

S4

S1

GaN

x 1.08CL

Int

ensi

ty (

a.u.

)

(nm)

T =50 K

CRSQW

(b)Example of luminescence spectra measured with a grating monochromator on the left for GaN/InGaN multiple quantum well (QW) samples. The maximum spectral resolution is obtained for the narrowest slit widths. Sit widths narrow (top)

and wide (bottom)

Page 21: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Schematic diagram of optical collection system and data acquisitionsetup in our CL system.

S E M Im a g in g & C o n tro l

E x te rn a lXY

C O M Im a gi n g a n dS p e c tr os co py

D A C YD A C X

C o ntr o lle r

L iq u id H e S ta g e

E ll ip so id a l M i rr or

C o h e re n t O p ti ca l F ib e r B u n dl e

S ca n C o i ls

T h re e -a x i sM a n ip u la to r

C o n d e ns er L e n s

S E M Va cu u m S y s tem

Sa m p le

E le c tr on B e am

ADC/C ounterL o c k- i n A m p / P h o to n C o u n t in g Un it

M on o ch ro m a to ro r S p e c tro g ra p h

D e te c tor

P o la r ize r

Computer

Cathodoluminescence (CL) - Light emitted by the injection of high-energy electrons.

Photomultiplier Tube (PMT) or Ge p-i-n detector

From Prof. Rich’s Laboratory for Optical Studies of Quantum Nanostructures

Page 22: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Fresnel (Near-Field) Diffraction

The basic idea is to start again with the Huygen’s-Fresnel principle for secondary spherical wave propagation. At any instant, every point on the primary wavefront is envisioned as a continuous emitter of spherical secondary wavelets. However, no reverse wave traveling back toward the source is detected experimentally.

Therefore, in order to introduce a realistic radiation pattern of secondary emitters, we introduce the inclination factor, K() = (1+cos)/2 which describes the directionality of secondary emissions. K(0) = 1 and K() = 0.

cos12

1)( K

Page 23: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

A rectangular aperture in the near-field (Fresnel Diffraction)

The monochromatic point source S and the point P on a screen are placed sufficiently close to the aperture where far-field conditions are no longer applicable. Consider a point A in the aperture whose coordinates are (y,z). The location of the origin O is determined by a perpendicular line from

the source S to the aperture . The field contributions at P from the secondary sources on dS (area element at point A) is given by

,)(cos)(

)(cos)(

0 dStrkr

K

dStrkr

KdE A

P

where 0 is the source strength at S, A is the secondary wavelet source strength per area, and A = 0 is obtained from the Huygen’s-Fresnel formalism.

Page 24: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

In the case where the dimensions of the aperture are small compared to and r, we can assume primarily forward propagation in the secondary spherical waves so that K() 1 and 1/r 1/0r0. Also, from the figure the geometry yields:

2/12220

2/12220 zyrrandzy

Expand both terms in a binomial series for small y and z:

00

002200 2 r

rzyrr

Note that this approximation contains quadratic terms that appear in the phase whereas the Fraunhofer approximation contains only linear terms. Thus, we can expect a greater sensitivity in the phase of the cosine for this near-field treatment. The complex field at point P on the screen is therefore:

2

1

2

1

)(

00

0~y

y

z

z

rikti

P dydzer

eE

Page 25: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

trki

u

v

v

viu

u

uiu

v

v

viu

u

uitrkiP

er

E

dvedueE

dvedueer

E

GivesVariablesofChange

r

rzv

r

ryuLet

)(

00

0

2/2/2/2/)(

00

0

2/1

00

00

2/1

00

00

00

2

1

22

1

22

1

22

1

200

~

2

~

2

~

2;

2

This is the unobstructed disturbance at P.

,)()()()(2

~~

,

)()(

2/sin)(2/cos)(

2

1

2

1

2

0

2/

0

2

0

2

v

v

u

uu

P

wwi

ww

viCvAuiCuAE

EPatTherefore

wiCwAwdeSince

wdwwCandwdwwALet

where A(w) and C(w) are called Fresnel Integrals; note that both are odd functions of w.

Page 26: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Very often, we work in the limit of incoming plane-waves striking the aperture. For example, a laser beam could strike the aperture. In this limit we let the radius from the source to the aperture 0 . This results in an immediate simplification for the change of variables:

2

1

2

1

)(~

)()()(~

)(~

)(~

4

)()()()(

)()()()(42

~~

22;

22

12

2

12

2

120

2

12

2

12

2

12

2

120

*

2/1

0

2/1

00

00

2/1

0

2/1

00

00

u

u

u

u

PPP

uBuiCuAuBwherevBuBI

vCvCvAvA

uCuCuAuAIEE

I

rz

r

rzv

ry

r

ryuThen

Cornu Spiral

Elegant geometrical description of the Fresnel Integrals (Fig. 10.50).

Page 27: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Fig. 10.50 The Cornu Spiral for a graphical representation of the Fresnel integrals.

A(w)

C(w)

Page 28: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

A(w) C(w)

A(w) C(w)

w A(w) C(w) w A(w) C(w)

Page 29: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

dwdldwww

dlIntegralstheFrom

dCdAdlcurvethealonglengthArc

wwiCwAwBLet

22

22

22

222

2sin

2cos:

:

)()()(~

Therefore, values of w in B(w) correspond to arc length on the Cornu spiral.

Consider a 2-mm square aperture hole:

y

z )1 mm, 1 mm(-)1 mm, 1 mm(

-)1 mm, -1 mm( )1 mm, -1 mm(

We are given that = 500 nm, r0 = 4 m, plane wave approx. is valid. Find the irradiance at a point P on the screen along the axis x, directly behind the center of the aperture.

0.1,0.1,0.1

0.14105

2101

2

212

2/1

73

2/1

011

vvu

mmm

ryuThen

P

r0

O

)y1, z1(

)y2, z2()y1, z2(

)y2, z1(

Page 30: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

000

22220

2

12

2

120

1212

12

21

56.2)64.0(44383.07799.0)4(4

)(~

)(~

4

4383.07799.02)(~

)(~

4383.07799.0)(~

4383.07799.0)(~

1,1

IIII

vBuBI

ITherefore

ivBuB

iuBandiuB

uuFor

P

Notice that there is an increase of the irradiance at the center point P on the screen by 256% relative to the unobstructed intensity due to a redistribution of the energy.

In order to find the irradiance 0.1 mm to the left of center, move the aperture to the right relative to the OP line. While y1 and y2 are shifted, z1 and z2 remain unchanged. Then we have u2 = 1.1, u1 = -0.9, v2 = 1.0, v1 = -1.0.

y

z )1.1 mm, 1 mm(-)0.9 mm, 1 mm(

-)0.9 mm, -1 mm( )1.1 mm, -1 mm(r0

)y2, z1()y1, z1(

)y1, z2( )y2, z2(

0.1,0.1,1.1

9.04105

2109.0

2

212

2/1

73

2/1

011

vvu

mmm

ryu

O

P

Page 31: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

4383.07799.02)(~

:)(~

8763.05286.1)(~

5365.07638.0)1.1(~

)(~

3398.07648.0)9.0(~

)(~

1212

12

2

1

ivBunchangedremainsvBAlso

iuB

iBuB

iBuBThen

022220

2

12

2

120 485.28763.05286.14383.07799.0)4(

4)(

~)(

~

4I

IvBuB

IIThus P

The decrease in the irradiance (2.485I0 < 2.56I0) for a small 0.1 mm shift to the left (or right) of center on the screen shows that the center position is a relative maximum (see Cornu spiral on the next slide).

Note that if the aperture is completely opened:

0

22

0

2211

2

22

2

22

4

,

II

I

vuvu

P

which must equal to the unobstructed intensity as a check.

Page 32: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

)(~

12 uB

)(~

12 uB

789.1)(~

12 uB

762.1)(~

12 uBC(w)

A(w)

The decrease in the complex vector length from the position of the central peak shows that the central position is a maximum.

1.1

-0.9

Page 33: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

We can apply this formalism for Fresnel diffraction by a long narrow slit in which

2)(~

,2)(~

,,, 124/

122211 uBeuBuyuy i

2

120

2

12

2

120 )(

~

2)(

~)(

~

4vB

IvBuB

IITherefore P

b = z1 – z2 = slit width and let v = v2 – v1 which is a string of length v

lying along the Cornu spiral (next slide).

Page 34: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

Suppose that v = 2. At point P, opposite point O in the aperture, the aperture and the screen are centered symmetrically and the string is centered at point Os. If the aperture is moved up or down, the arc length of the string remains constant, but the length of the vector B12(v) changes, as

before .

)(~

12 vB

)(~

12 vB

It should be apparent that the length of B12 (and the intensity at point P on the screen) will oscillate as the string slides around one of the spirals, which is equivalent to the slit moving up or down with respect to a reference point on the screen, as shown in the previous slide.

Page 35: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit

It is also possible to visualize a clear minimum at the center of the near field diffraction pattern on the screen by considering the an arc-length of w = 3.5. Any change in the slit position will give and increase in B12 and therefore an increase in irradiance.)5.2(

~12B

)5.3(~

12B

It is apparent that the slit width has a marked effect on whether the central position is a maximum or local minimum. Also note the oscillation in irradiance for positions beyond the width of the slit in both cases.

Page 36: Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture whose width and height are about the same. Let  A = the source strength per unit