Chapter 15S Fresnel diffraction - Binghamton...

$: Chapter 15S Fresnel diffraction - Binghamton Universitybingweb.binghamton.edu/~suzuki/Math-Physics/LN-15S_Fresnel_diffraction.pdfChapter 15S Fresnel-Kirchhoff diffraction Masatsugu$
Chapter 15S Fresnel-Kirchhoff diffraction

Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton

(Date: January 22, 2011) Augustin-Jean Fresnel (10 May 1788 – 14 July 1827), was a French physicist who contributed significantly to the establishment of the theory of wave optics. Fresnel studied the behaviour of light both theoretically and experimentally. He is perhaps best known as the inventor of the Fresnel lens, first adopted in lighthouses while he was a French commissioner of lighthouses, and found in many applications today.

http://en.wikipedia.org/wiki/Augustin-Jean_Fresnel ______________________________________________________________________________ Christiaan Huygens, FRS (14 April 1629 – 8 July 1695) was a prominent Dutch mathematician, astronomer, physicist, horologist, and writer of early science fiction. His work included early telescopic studies elucidating the nature of the rings of Saturn and the discovery of its moon Titan, the invention of the pendulum clock and other investigations in timekeeping, and studies of both optics and the centrifugal force. Huygens achieved note for his argument that light consists of waves, now known as the Huygens–Fresnel principle, which became instrumental in the understanding of wave-particle duality. He generally receives credit for his discovery of the centrifugal force, the laws for collision of bodies, for his role in the development of modern calculus and his original observations on sound perception (see repetition pitch). Huygens is seen as the first theoretical physicist as he was the first to use formulae in physics.

http://en.wikipedia.org/wiki/Christiaan_Huygens ____________________________________________________________________________

Green's theorem Fresnel-Kirchhhoff diffraction Huygens-Fresnel principle Fraunhofer diffraction Fresel diffraction Fresnel's zone Fresnel's number Cornu spiral

________________________________________________________________________ 15S.1 Green's theorem

First we will give the proof of the Green's theorem given by

AV

dd a)()( 22 .

This theorem is the prime foundation of scalar diffraction theory. However, only an prudent choice of the Green's function and a closed surface A will allow its direct

application to the diffraction theory: d is volume element and da is the surface element. ((Proof)) In the Gauss's theorem, we put

ξ

Then we have

AVV

dddI aξ )()(1 .

Noting that

2)( ,

we have

AV

ddI a)()( 21 .

By replacing , we also have

AV

ddI a)()( 21 ,

Thus we find the Green's theorem

AV

ddII a)()( 2221 ,

or

AV

dad n)()( 22 ,

15S.2 Fresnel-Kirchhoff diffraction theory

According to the Huygen's construction, every point of a wave-front may be considered as a center of a secondary disturbance which gives rise to spherical wavelets, and the wave-front at any later instant may be regarded as the envelop of these wavelets. Fresnel was able to account for diffraction by supplementing Huygen's construction with the postulate that the secondary wavelets mutually interfere. This combination of the Huygen's construction with the principle of interference is called the Huygens-Fresnel principle. (Born and Wolf, Principles of optics, 7-th edition).

The basic idea of the Huygens-Fresnel theory is that the light disturbance at an observation point P arises from the superposition of secondary waves that proceed from a surface (aperture) situated between the point P and the light source S.

Fig. Illustrating the deviation of the Fresnel-Kirchhoff diffraction formula. r' is the

position vector from the source point S (at the origin). The aperture is denoted by the green line. The volume integral is taken over a region bounded by red lines

(consisting a screen with an aperture and the surface C). rSP . 'rr PQ . The

green line denotes an aperture.

We now start with the Green's theorem given by

AV

dad ')]'(')'()'(')'([')]'(')'()'(')'([ 322 nrrrrrrrrr ,

Here we assume that

)'()'( rr u

S

PQ

u=0

r'

r

u=0

u=0

s=r-r'

Aperture

Screen

Cn'

and

)',(|'|4

)'('

rrrr

rrr

Geik

, )'()',()'( 22 rrrr Gk ,

with

s

esik

sik

s

esG

iksiks

4

ˆ)

1(

4

ˆ)',(' rr ,

where k (= 2/) is the wavenumber, )',( rrG is the Green's function, )()',( srr GG with

'rrs , and s is the unit vector of s. Then we have

CscreenApertureV

dauGGuduGGu ')]'(')',()',(')'([')]'(')',()',(')'([ 322 nrrrrrrrrrrrrr ,

or

CscreenApertureV

dauGGuduGu ')]'(')',()',(')'([')]'(')',()'()'([ 32 nrrrrrrrrrrrrr .

Further we assume that u(r') = 0 everywhere but in the aperture. In the aperture, u(r') is

the field due to a point source at the point S. So we get

Aperture

ikik

Aperture

daee

u

daGGuu

'')]'('|'||'|

')'([4

1

'')]'(')',()',(')'([)(

''

nrurrrr

r

nrurrrrrr

rrrr

where n' (=-n) is an inwardly directed normal to the aperture surface. This is one form of the integral theorem of Helmholtz and Kirchhoff. It is reasonable to suppose that everywhere on the aperture A, )'(ru and )'(' ru will not appreciably differ from the

values obtained in the free space. So we assume that )'(rincu satisfies the Maxwell's

equation in the free space. )'(rincu is the component of the electric field for the spherical

wave (incident wave);

0)'()'( 22 rincuk . The solution of this differential equation is given by

r

eEuu

ikr

inc

'

0)'()'( rr ,

and

)'('ˆ4'4

'ˆ)'

1(

'4

'ˆ)'('

'

0

'

0 rr inc

ikrikr

inc urik

r

erikE

rik

r

erEu

,

on the aperture surface, where E0 is a electric field constant amplitude and 'r is the unit vector of r'. Then we have

Aperture

inc

iks

Aperture

inc

iks

Aperture

ikriksiksikr

daIus

eik

daus

eik

dar

ikr

e

s

e

sik

s

e

r

eEu

')ˆ,'ˆ()'(2

''ˆ)'ˆˆ)('(4

')]}'

1(

''ˆ'ˆ[)]

1('ˆ[

'{

4

1)(

''

0

srr

nrsr

nrnsr

(Fresnel-Kirchhoff diffraction formula) where the inclination factor is defined as

'ˆ)'ˆˆ(2

1)ˆ,'ˆ( nrssr I .

for the convenience. This expression is consistent with the Huyghens' principle. u(r) is the superposition of the spherical waves eiks/s emanating from the wavefront eikr/r produced by a point source.

15S.3 The Huygens-Fresnel principle

In order to explain the essence of the Huygens-Fresnel principle, we use the simple model as shown below.

Fig. Illustrating the diffraction formula. We assume that the shape of the aperture in

the screen is a sphere.

Let A be the instantaneous position of a spherical monochromatic wave-front of

radius 0 which proceeds from a point source S, and let P be a point at which the light disturbance at a point Q on the wave-front may be represented by

r0'

n'

Screen

Aperture

A

dAr0

r0

ss

S

Qc

PC

00

0

)(

ikeEu Q ,

where E0 is the amplitude at unit distance from the source S. According to the Huygens-Fresnel principle, each element of the wave-front is the center of a secondary disturbance, which is propagated in the form of spherical wavelets. The contribution du(P) due to the element dA at the point Q is expressed by

dAs

eeEI

ikPdu

iksik

00

0

)(2

)(

where s =QP. The inclination factor )(I is given by

)cos1(2

1'ˆ)ˆˆ(

2

1)( 0 nρsI ,

where is the angle (often called the angle of diffraction) between the normal at Q and

the direction of propagation, and 'ˆˆ 0 nρ in the present case. The total disturbance at the

point P is given by

Aperture

iksik

dAs

eI

eikEPu )(

2)(

00

0

.

15S.4 Fresnel zone

Fig. Fresnel's zone construction (Born and Wolf, Principles of optics, p.413) In order to evaluate u(P), we use the so-called Fresnel's zone. With the center at P, we construct the spheres of radii,

0r , 20

r ,

2

20

r ,

2

30

r ,

2

40

r ,

2

50

r ,.....

where r0 = CP. C is the point of the intersection of SP with the wave-front S. The spheres divide A into a number of zones

Z1 ( 0r - 20

r ), the Fresnel's 1st zone

Z2 (20

r -

2

20

r ), the 2-nd zone

Z3 (2

20

r -

2

30

r ), the 3-rd zone

................................................................................................

Zj (2

)1(0

jr -

20

jr ), the j-th zone

................................................................................................

A

r0 r0

s

S

Qc

qP

C

Fresnel's zone

Fig. Fresnel's zone (n = 1, 2, 3, ---). The distance PQ is equal to 2/0 nr

2/)1(0 nr and for the n-th zone. This figure is drawn by using Graphics3D of

the Mathematica.

Since 0 and r0 are much larger than the wavelength , the inclination factor may be assume to have the same value (Ij), for points for the j-th zone. From the figure, we have

cos)(2)( 0002

02

002 rrs .

So that

drsds sin)( 000 ,

and the surface element dA is given by

)(2sin2

000

20

20 r

sdsddA

.

((Note-1))

The area of the end cap is given by

)(2

)(22

)cos1(2sin2

000

220

2002

02

0

20

0

20

0

r

sr

ddAA

The area of the j-th zone is obtained as

]4

)12([

])2

)1(()

2[(

)(2

2

])(2

)2

)1(()(

22[

])(2

)2

()(22[

000

0

20

20

000

20

000

20

20

200

20

20

000

20

20

200

20

20

jr

r

jr

jr

r

r

jrr

r

jrr

Aj

The mean distance from the j-th zone to the point P is denoted as js

4

)12(

2

)2

)1(()

2(

0

00

j

r

jr

jr

s j

Then we have

00

0

rs

A

j

j

,

which is independent of j. ((Note-2)) Inclination factor

We note that

s

srr

0

22000

2

2cos'ˆˆ

ns .

Then the inclination factor is obtained as

s

rsrs

s

srrI

0

000

0

22000

4

)2)(()

2

21(

2

1)(

,

since

)0,( 00 rSP , )sin,cos( 00 SQ , )sin,(cos'ˆ n

)sin,cos( 0000 rSQSPQPs ,

s

srr

s

r

s 0

22000000

2

2cos)('ˆcos

ns

.

______________________________________________________________________ Then the contribution of the j-th Fresnel's zone to u(P) is

2/

2/)1(

'

00

)(

0

2/

2/)1(000

000

20

00

')(

)(

)()(

22)(

00

0

0

0

0

j

j

iksj

rik

jr

jr

iksj

ik

Z

iksik

j

dseIr

eikE

dseIr

eikE

r

sds

s

eI

eE

ikPu

j

where 0' rss , and )(I is approximated by Ij, the value of )(I at 2

)1(0

jrs ;

]2

)1([4

]2

)1(2][2

)1(2[

00

00

jr

jjrI j

and I1 = -1. Noting that

jj

j

iks idse

)1('

2/

2/)1(

'

we get

jj

rik

j Ir

eEPu )1(

)(2)(

00

)(

0

00

The total effect at the point P is obtained by summing all the contributions:

n

j

jj

rikn

jj I

r

eEPuPu

100

)(

01

)1()(

2)()(00

For n = 2m+1 (odd)

)()(

)22

()(

2

]2

)22

(

)22

(...)22

(2

[)(

2)(

12100

)(

0

121

00

)(

0

12122

12

1222

3232

11

00

)(

0

00

00

00

m

rik

mrik

mmm

m

mm

mrik

IIr

eE

II

r

eE

III

I

II

III

II

r

eEPu

since Ij changes slowly with j, and

211

jjj

III .

For n = 2m (even)

)()(

)22

()(

2

]2

)22

(

...)22

(2

[)(

2)(

2100

)(

0

221

00

)(

0

2212

22

43

221

00

)(

0

00

00

00

m

rik

mrik

mmm

m

rik

IIr

eE

III

r

eE

III

I

II

III

r

eEPu

where 12 II . Suppose that we allow m to become large enough so the entire spherical

wave is divided into zones: I2m = 0 and I2m+1 = 0. Then we get

00

)(

0100

)(

0

0000

)(r

eEI

r

eEPu

rikrik

since I1 = 1. We note that the disturbance at P only due to the first zone is

)(2

)(2)(

00

)(

0100

)(

01

0000

r

eEI

r

eEPu

rikrik

Therefore we have

)(2

1)( 1 PuPu .

In other words, the total disturbance at P is equal to half of the disturbance due to the first zone. ________________________________________________________________________ 15S.5 Reformulation of the Fresnel-Kirchhoff diffraction:

We consider the new coordinate system for the above Fresnel-Kirchhoff diffraction formula, where the origin of the system is moved from the source point S to a specific point in the screening with an aperture (in the above figure, we put the new origin O1 at the center of the aperture). Note that the shape of the aperture is two-dimensional (such as a square aperture or a circle aperture).

Fig. New coordinate system. The origin is moved to the center of aperture. 01 rQO .

11 rPO . 01 rr PQ . The point Q is on the aperture and the point P is on the

observation plane. We define the new origin O1 in a specific point in the screen with an aperture. The vector vector r' is expressed by

00' rρr ,

where r0 is a position vector located on the aperture. The vector r is expressed by

10 rρr ,

where 0ρ is the vector connecting the source point S and the new origin O1 in the

aperture, and 1r is a position vector from the new origin O1 to the position on the

observation plane. The vector s is given by

01' rrrrs , 220

2001 )()(|| zyyxxs rr .

For simplicity we assume that

zs , and

1),'ˆ( srI ,

for 'ˆ//'ˆ//ˆ nrs . Then we get the formula of the Fresnel-Kirchhoff diffraction,

Aperture

incik dAue

z

iku 000

||1 )(

2)( 01 ρrr rr

,

where dA0 is the surface area element in the aperture; dA0 = dx0dy0. We assume that

)()( 000 ρρr incinc uu ,

Then we get

Aperture

ikinc

Aperture

ikinc dAeu

z

idAeu

z

iku 0

||00

||01

0101 )()(2

)( rrrr ρρr

.

15S.6 Fresnel diffraction

Imagine that we have an opaque shield with a single small aperture which is illuminated by plane waves from a very distance point source. First we also have a screen parallel with, and very close to the aperture. In this case, an image of the aperture is projected on the screen. As the screen moves further away from the aperture, the image of the aperture becomes increasing more structured as the fringes become more prominent. This is called the Fresnel diffraction. The moving of the screen at a very great distance from the aperture results in the drastic chage of the projected pattern which is the two-dimensional (2D) Fourier transform image of the aperture pattern. This is called the Fraunhofer diffraction.

Fig. Rectangular aperture for the Fresnel diffraction and Fraunhofer diffraction. r0 = (x0, y0, 0).

),,( zyxOP r

The electric field E(x, y, z) at the point P is expressed by

Aperture

ikinc dydxeE

z

ikzyxE 00

||0

0)(2

),,( rrρ

,

where r =(x, y, z) at the point P of the observation plane, r0 = (x0, y0,0) at the point Q in the aperture, as shown in Fig.

220

200 )()( zyyxx rr

This equation describes how the light travels from the aperture to the observation plane, a distance z apart. All points from the aperture contribute to the intensity at the point P. Suppose that

2

02

022 )()( yyxxz

Then we can expand 0rr around = 0, as

....1682 5

6

3

42

0 zzz

z

rr

Then we have

....1682 5

6

3

42

0 z

k

z

k

z

kkzk

rr

We assume that

2

8 3

4

z

k,

or

83

4

z

.

When is on the same order as the size of aperture a, this condition can be rewritten as

83

4

z

a

.

((Note))

Instead of this condition, we use the Fresnel's number F. When F>>1, the Fresnel diffraction can occur. On the other hand, when F<<1, the Fraunhofer diffraction can occur. The Definition of F will be given below. Note that the condition (F>>1) is not so restrictive compared to the

above condition; 83

4

z

a

. The Fresnel diffraction may o ccur when the factor )

8exp(

3

4

z

ki

slowly changes compared to the other factors )]2

(exp[2

z

kkzi

.

Under such a condition, the distance 0rr is approximated by

zz

2

2

0

rr .

This is called the Fresnel approximation.

]2

)()(exp[)0,,()(

]2

)()(exp[)(),,(

20

20

0000

20

20

000

0 z

yyxxikdydxyxE

z

ei

z

yyxxikdydxE

z

eizyxE

inc

ikz

Aperture

inc

ikz

ρ

ρ

where 1)0,,( 00 yx for the inside of the aperture, and 0 for the outside of the aperture.

Mathematically, this integral corresponds to the convolution of )0,,( 00 yx and ),( 00 yxh , and

can be expressed by

),(exp)0,,(),,( yxhyxzyxE

where

)2

exp(),(22

z

yxik

z

eiyxh

ikz

.

15S.7 Fraunhofer diffraction

The above equation can be rewritten as

)](exp[]2

exp[)0,,(]2

)(exp[),,( 00

20

20

0000

22

yyxxz

ki

z

yxikyxdydx

z

yxik

z

eizyxE

ikz

Suppose that

12

20

20

z

yxk .

Then the quadratic phase factor )2

exp(2

02

0

z

yxik

is approximately unity over the entire aperture.

So we get

)](exp[)0,,(]2

)(exp[),,( 000000

22

yyxxz

kiyxdydx

z

yxik

z

eizyxE

ikz

.

In other words, the field distribution can be found directly from a Fourier transform of the aperture distribution itself. Aside from the multiplicative factors, this expression is simply the Fourier transform of the aperture.

),([)](exp[)0,,(2

1000000002

yxykxkiyxdydx yx F

with

xz

xz

kkx

2 , y

zy

z

kky

2

where F is the operation of Fourier transform. This is called the Fraunhofer diffraction. 15S.8 Direct derivation for the Fraunhofer diffraction

We start with the formula given by

02||

000)()(

2),,( rrρ rr deE

z

ikzyxE ik

inc

As shown in Fig., the distance s is approximated by

)ˆ( 00 rrr rrs ,

for r>>r0.

Fig. 0rOQ rOP . 0rrs QP . rrOH ˆ)ˆ( 0r

)ˆ(|| 00 rrr rrQHsQP .

Then we get the expression,

02)ˆ(

00

02)ˆ(

00

0

0

)()(2

)()(2

),,(

rrρ

rrρ

r

r

deeEz

ik

deEz

ikzyxE

rikikrinc

rikikrinc

.

Here we note that

000000 )()()()(ˆ xz

ykx

z

xkyyxx

z

k

r

krk rrr .

Then E(x, y, z) is proportional to the Fourier transform of the aperture,

}],{),([)(2

100

2)ˆ(02

0

z

ykk

z

xkkde yx

rik rFrr r

with the wave vector given by

OQ

P

H

z

ykk

z

xkk yx , .

The x and y coordinates of the observation point P are proportional to the wave numbers kx, and ky, respectively. 15S.9 Fresnel's number; F

Fresnel's number F is a dimensionless number occurring in optics, in particular in diffraction theory. For an electromagnetic wave passing through an aperture and hitting a screen, the Fresnel number F is defined as

za

F2

,

where a is the characteristic size (the side of the square aperture) of the aperture, z is the distance of the observation plane from the aperture and is the incident wavelength. Depending on the value of F the diffraction theory can be simplified into two special cases: Fraunhofer diffraction for F<<1 and Fresnel diffraction for F>>1. We note that the condition for the appearance of the Fraunhofer diffraction is evaluated as

12

z

aF = 0.318.

On the other hand, the condition for the appearance of the Fresnel diffraction can be rewritten as

2

22

8a

z

z

aF

.

Suppose that z = 100 mm and a = 5 mm. Then we get the inequality as

F<<3200. 15S.10 Fresnel diffraction with a rectangular aperture

The Fresnel diffraction occurs when the condition z is satisfied.

Suppose that a rectangular aperture of ( axa , bxb ) is normally illuminated by a

monochromatic plane wave of unit amplitude and the wavelength .

b

b

a

a

ikz

ikz

dyxyz

kidxxx

z

ki

z

ei

dydxyyxxz

ki

z

eiyxU

12

112

1

112

12

1

])(2

exp[])(2

exp[

]})()[(2

exp{),(

where k is the wavenumber and is given by k = 2/. It follows that the expression can be separated into the product of two integrals over x1 and y1,

b

b

a

a

dyyyz

kiyF

dxxxz

kixF

12

1

12

1

])(2

exp[)(

])(2

exp[)(

These integrals are simplified by the change of variables,

)( 1 xxz

k

, )( 1 yy

z

k

yielding

2

1

2

1

)2

exp()(

)2

exp()(

2

2

dik

zyF

dik

zxF

where the limits of integration are

)(1 xaz

k

, )(2 xa

z

k

)(1 ybz

k

, )(2 yb

z

k

The integrals F(x) and F(y) can be evaluated in terms of the Fresnel integrals, which are defined by

0

2

0

2

)2

sin()(

)2

cos()(

dttS

dttC

We note that

)()(

)]2

sin()2

[cos()]2

(exp[0

22

0

2

iSC

didi

and

)]()([)()(

)()()()(

)]2

(exp[)]2

(exp[)]2

(exp[

1212

1122

0

2

0

22122

1

SSiCC

iSCiSC

dididi

Finally we have the complex field distribution

)]}()([)()()]}{()([)()({2

),( 12121212 SSiCCSSiCCi

eyxU

ikz

and the corresponding intensity distribution

})]()([)]()(}{[)]()([)]()({[4

1),( 2

122

122

122

12 SSCCSSCCyxI

Note that

2

1)( C ,

2

1)( C , )()( CC

2

1)( S ,

2

1)( S )()( SS

15S.11 Cornu spiral Marie Alfred Cornu (March 6, 1841 – April 12, 1902) was a French physicist. The French generally refer to him as Alfred Cornu. Cornu was born at Orléans and was educated at the École polytechnique and the École des mines. Upon the death of Émile Verdet in 1866, Cornu became, in 1867, Verdet's successor as professor of experimental physics at the École polytechnique, where he remained throughout his life. Although he made various excursions into other branches of physical science, undertaking, for example, with Jean-Baptistin Baille about 1870 a repetition of Cavendish's experiment for determining the gravitational constant G, his original work was mainly concerned with optics and spectroscopy. In particular he carried out a classical redetermination of the speed of light by A. H. L. Fizeau's method (see Fizeau-Foucault Apparatus), introducing various improvements in the apparatus, which added greatly to the accuracy of the results. This achievement won for him, in 1878, the prix Lacaze and membership of the French Academy of Sciences (l'Académie des sciences), and the Rumford Medal of the Royal Society in England. In 1892, he was elected a member of the Royal Swedish Academy of Sciences. In 1896, he became president of the French Academy of Sciences. In 1899, at the jubilee commemoration of Sir George Stokes, he was Rede lecturer at Cambridge, his subject being the wave theory of light and its influence on modern physics; and on that occasion the honorary degree of D.Sc. was conferred on him by the university. He died at Romorantin on April 12, 1902. The Cornu spiral, a graphical device for the computation of light intensities in Fresnel's model of near-field diffraction, is named after him. The spiral is also used in geometrical road design. The Cornu depolarizer is also named after him.

http://en.wikipedia.org/wiki/Marie_Alfred_Cornu

Here we make a plot of Cornu's spiral.

Fig. Plot of S(x) as a function of x.

FresnelSx

-10 -5 5 10x

-0.6

-0.4

-0.2

0.2

0.4

0.6

Sx

Fig. Plot of C(x) as a function of x.

Fig. ParametricPlot of {C(), S()} when a is changed as parameter.

FresnelCx

-10 -5 5 10x

-0.5

0.5

Cx

-0.5 0.5Ca

-0.6

-0.4

-0.2

0.2

0.4

0.6

Sa

Fig. ParametricPlot of {C(), S()} around the point Z (1/2, 1/2) when a is changed as parameter.

a=0. a=0.1 a=0.2 a=0.3a=0.4

a=0.5

a=0.6

a=0.7

a=0.8

a=0.9

a=1

a=1.1

a=1.2

a=1.3a=1.4

a=1.5

a=1.6

a=1.7

a=1.8

a=1.9a=2.

a=2.1

a=2.2

a=2.3

a=2.4a=2.5

a=2.6

a=2.7

a=2.8a=2.9

a=3.

a=3.1a=3.2

a=3.3

a=3.4 a=3.5

a=3.6

a=3.7a=3.8

a=3.9

a=4.

a=4.1

a=4.2a=4.3

a=4.4a=4.5

a=4.6

a=4.7

a=4.8

a=4.9

a=5.

0.2 0.4 0.6Ca

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Sa

Fig. ParametricPlot of {C(), S()} around the point Z' (-1/2, -1/2) when a is changed as parameter.

15S.12 Fresnel diffraction with a square aperture

We calculate the intensity of the Fresnel diffraction with the square aperture by using the

Mathematica (Plot3D, ContourPlot). and z are fixed. The size of the square (L = a)

a=-5.

a=-4.9

a=-4.8

a=-4.7

a=-4.6

a=-4.5a=-4.4

a=-4.3a=-4.2

a=-4.1

a=-4.

a=-3.9

a=-3.8a=-3.7

a=-3.6

a=-3.5a=-3.4

a=-3.3

a=-3.2a=-3.1

a=-3.

a=-2.9a=-2.8

a=-2.7

a=-2.6

a=-2.5a=-2.4

a=-2.3

a=-2.2

a=-2.1a=-2.

a=-1.9

a=-1.8

a=-1.7

a=-1.6

a=-1.5a=-1.4a=-1.3

a=-1.2

a=-1.1

=-1.

a=-0.9

a=-0.8

a=-0.7

a=-0.6

a=-0.5a=-0.4

a=-0.3a=-0.2 a=-0.1 a=0-0.6 -0.4 -0.2

Cia

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

Sia

(a) F = 22.7848

= 632 nm, z = 400 mm, L = a = 2.4 mm x(mm), y(mm)

Fig. Plot3D of the intensity I(x, y).

Fig. Plot of intensity vs x (mm). y (mm) is changed as a parameter. y = -1, -0.8, -0.6, -0.4, -0.2,

0, 0.2, 0.4, 0.6, 0.8, and 1.0 mm

Fig. DensityPlot. The distribution of the intensity in the x (mm) and y (mm) plane. (b) F = 17.4446


-1.0 -0.5 0.5 1.0x

0.8

1.0

1.2

1.4

1.6

1.8

IntensityF=22.7848

-1.0 -0.5 0.5 1.0x

1.0

1.5

2.0

IntensityF=17.4446

(c) F = 12.8165


(d) F = 8.9032


-1.0 -0.5 0.5 1.0x

0.5

1.0

1.5

2.0

IntensityF=12.8165

-1.0 -0.5 0.5 1.0x

0.5

1.0

1.5

IntensityF=8.90032

(e) F = 5.6962


(f) F = 3.20411


-1.0 -0.5 0.5 1.0x

0.5

1.0

1.5

2.0

IntensityF=5.6962

-1.0 -0.5 0.5 1.0x

0.5

1.0

1.5

2.0

2.5

3.0

IntensityF=3.20411

(g) F = 1.42405


15S.13 A semi-infinite planar opaque screen with infinte 0 We calculate the intensity of the Fresnel diffraction with the semi-infinite planar opaque

screen by using the Mathematica (Plot3D, ContourPlot). and z are fixed. The size of the square (L = a)

-1.0 -0.5 0.5 1.0x

0.5

1.0

1.5

IntensityF=1.42405

Fig. A semi-infinite planar opaque screen with finte 0. r0 = z. We assume that the plane wave arrives at a semi infinite opaque screen. As shown in the above figure, the distance r (=QP) is given

20

20

2 )( yyxzr

where P at (x = 0, y, z) and Q at (x0, y0, z = 0). The distance r is approximated as

z

yyxzr

2

)( 20

20

.

Then the wave arriving at the point P (x = 0, y, z) on the screen is expressed in the form

0

02

002

0 ])(2

exp[)2

exp()exp( dyyyz

kidxx

z

kiikz

We put

)( 0 yyz

kt

and

0dyz

kdt

Then we have

w

dtt

ik

zdyyy

z

ki )

2exp(])(

2exp[

2

0

02

0

where

z

kyw

.

The resultant electric field at the point P is given by

w

dtt

idtt

iikzk

z]

2exp[]

2exp[)exp(

22

where

)()(]2

exp[0

2

iSCdtt

i

]2

1)([]

2

1)([

)]()([2

1

2

1

)]()([2

1

2

1

]2

exp[]2

exp[]2

exp[0

2

0

22

wSiwC

wiSwCi

wiSwCi

dtt

idtt

idtt

iw

w

The intensity is given by

)]2

1)([]

2

1)(([

2

1 220 wSwCII .

15S.14 Example: a semi-infinite planar opaque screen

= 632 nm (He-Ne laser) and z = 400 mm (the distance between the screen and the aperture). The intemsity oscillates with the distance y. The intensity has a local maximum at the points P1, P2, P3, P4, P5, P6, and so on, and a local minimum at the points Q1, Q2, Q3, Q4, and so on.

Fig. Intensity vs the distance y. y<0 (shadow region). I0 = 1. = 632 nm (He-Ne laser) and z = 400 mm. y is in the units of mm. The geometrical shadow edge is at y = 0 (the intensity = 1/4); At w = 0, C(w) = S(w) = 0.

-2 2 4y

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Intensity

Fig. Intensity I vs y (mm). I0 = 1.

P1( y = 0.432748, I = 1.37044), P2( y = 0.83353, I = 1.19927), P3( y = 1.09572, I = 1.15088), P4( y = 1.30625, I = 1.12606), P5( y = 1.48725, I = 1.11039), P6( y = 1.6485, I = 1.09937).

Q1( y = 0.665733, I = 0.778251), Q2( y = 0.973793, I = 0.843162), Q3( y = 1.20571, I = 0.871942), Q4( y = 1.39975, I = 0.889064), Q5( y = 1.56999, I = 0.900735).

P1

P2P3 P4 P5 P6

Q1

Q2Q3

Q4Q5

0.4 0.6 0.8 1.0 1.2 1.4 1.6y

0.9

1.0

1.1

1.2

1.3

Intensity

Fig. DensityPlot of the intensity vs the distance y from y = 0. Using the value of y in the units of mm, we get

yw79

25 .

The intensity corresponds to the square of the distance between (-1/2, -1/2) point and (C(),S()). The intensity has a maximum at y = 0.432748, 0.83353, 1.09572, 1.30625 (mm). In the Cornu spiral.

Fig. Cornu plot. Z' at (-1/2, -1/2). Z at (1/2, 1/2).

y=0. y=0.1

y=0.2

y=0.3

y=0.4

y=0.5

y=0.6

y=0.7

y=0.8

y=0.9

y=1.

y=1.1

y=1.2

y=1.3

y=1.4

y=1.5

Z

Z'

P1

P2

Q1

Q2

-0.5 0.5Ca

-0.6

-0.4

-0.2

0.2

0.4

0.6

Sa

Fig. Cornu plot (enlarged part of the above figure). P1, P2, P3, and P4 are the local

maximum points of the intensity vs y, and Q1, Q2, and Q3 are the local minimum points of the intensity I vs y.

15S.15 Fresnel diffraction with a single slit aperture

We calculate the intensity of the Fresnel diffraction with a single slit aperture by using the

Mathematica (Plot3D, ContourPlot). and z are fixed. The size of the square (L = a)

y=0.4

y=0.5

y=0.6

y=0.7

y=0.8

y=0.9

y=1.

y=1.1

y=1.2

y=1.3

y=1.4

y=1.5

Z

P1

P2

P3P4

Q1

Q2

Q3

Fig. Single slit aperture. The width of the single slit is a.

The electric filed distribution is given by

)]}()([)()()]}{()([)()({2

),( 12121212 SSiCCSSiCCi

eyxU

ikz

The intensity is proportional to |U(x, y)|2, and is given by the form

})]()([)]()(}{[)]()([)]()({[4

),( 212

212

212

212

0 SSCCSSCCI

yxI

where

1 , 2 , )2

(1 ya

z

k

,

y

a

z

k

22

Using these values of x1, x2, h11, and h2, we get

)]}()([)()({22

),( 11224

iSCiSCei

eyxU

iikz

(a) a = 2.4 mm

(b) a = 2.1 mm

-1.0 -0.5 0.5 1.0y

0.8

0.9

1.0

1.1

1.2

1.3

1.4

Intensitya= 2.4

y=0.

y=0.2

y=0.4

y=0.6 y=0.8

y=1.

y=1.2

y=1.4

y=1.6 y=1.8

y=2.

y=2.2

y=2.4y=2.6y=2.8

y=3.

-0.5 0.5Ca

-0.6

-0.4

-0.2

0.2

0.4

0.6

Saa = 2.4

(c) a = 1.8 mm

-1.0 -0.5 0.5 1.0y

0.6

0.8

1.0

1.2

1.4

Intensitya= 2.1

y=0.

y=0.2

y=0.4

y=0.6

y=0.8

y=1.y=1.2

y=1.4

y=1.6

y=1.8

y=2.

y=2.2y=2.4

y=2.6

y=2.8

y=3.

-0.5 0.5Ca

-0.6

-0.4

-0.2

0.2

0.4

0.6

Saa = 2.1

(d) a = 1.5 mm

-1.0 -0.5 0.5 1.0y

0.4

0.6

0.8

1.0

1.2

1.4

Intensitya= 1.8

y=0.

y=0.2

y=0.4

y=0.6

y=0.8y=1.

y=1.2

y=1.4

y=1.6

y=1.8y=2.

y=2.2y=2.4

y=2.6

y=2.8

y=3.

-0.5 0.5Ca

-0.6

-0.4

-0.2

0.2

0.4

0.6

Saa = 1.8

(e) a = 1.2 mm

-1.0 -0.5 0.5 1.0y

0.0

0.4

0.6

0.8

1.0

1.2

1.4

Intensitya= 1.5

y=0.

y=0.2

y=0.4

y=0.6y=0.8

y=1.

y=1.2

y=1.4

y=1.6

y=1.8y=2.

y=2.2

y=2.4

y=2.6y=2.8

y=3.

-0.5 0.5Ca

-0.6

-0.4

-0.2

0.2

0.4

0.6

Saa = 1.5

(f) a = 0.9 mm

-1.0 -0.5 0.5 1.0y

0.0

0.5

1.0

1.5

Intensitya= 1.2

y=0. y=0.2

y=0.4

y=0.6

y=0.8

y=1. y=1.2

y=1.4

y=1.6

y=1.8y=2.y=2.2

y=2.4

y=2.6

y=2.8y=3.

-0.5 0.5Ca

-0.6

-0.4

-0.2

0.2

0.4

0.6

Saa = 1.2

(g) a = 0.6 mm

-1.0 -0.5 0.5 1.0y

0.5

1.0

1.5

Intensitya= 0.9

y=0.

y=0.2

y=0.4y=0.6

y=0.8

y=1.

y=1.2

y=1.4

y=1.6y=1.8

y=2.

y=2.2

y=2.4

y=2.6y=2.8y=3.

-0.5 0.5Ca

-0.6

-0.4

-0.2

0.2

0.4

0.6

Saa = 0.9

_________________________________________________________________________ REFERENCES M. Born and E. Wolf, Principles of optics, 7-th edition (Cambridge University Press. 1999). J.W. Goodman, Introduction to Fourier Optics (McGraw-Hill Book Company, San Francisco,

1968). E. Hecht and A. Zajac, Optics (Addison Wesley, Reading, Massachusetts, 1979). E. Hecht, Theory and Problems of Optics, Schaum's Outline Series (McGraw-Hill, New York,

1978). F.A. Jenkins and H. E. White, Fundamentals of Optics 3rd edition (McGraw-Hill Book Company,

New York, 1957). R.G. Wilson, Fourier Series and Optical Transform Techniques in Contemporary Optics;

Introduction (John Wiley & Sons, New York, 1995). K.M. Abedin, M.R. Islam, and A.F.M.Y.Haider, Optics & Laser Technology 39, 237 (2007). L.D. Landau and E.M. Lifshitz, The Classical Theory of Fields (Pergamon Press, 1975). F.L. Pedrotti, S.J. and L.S. Pedrotti, Introduction to Optics, second edition (Printice Hall,

Upper Saddle River, New Jersey, 1993). M.V. Klein, Optics (John Wiley & Sons, Inc., New York, 1970). __________________________________________________________________________ APPENDIX Appendix. Square aperture in the case of finite distance 0 (Landau and Lifshitz)

-1.0 -0.5 0.5 1.0y

0.2

0.4

0.6

0.8

1.0

1.2

Intensitya= 0.6

2

02

02

0 yx

20

20

20 )()( yyxxrr

21 xxx , 21 yyy (square aperture)

The total path is given by

)(

)(2])()[(

2

2

)()(

2

00

00

222

00

00

2

00

00

00

00

0

20

20

0

20

20

00

r

r

yx

r

yry

r

xrx

r

r

r

yyxxyxrr

2

1

2

1

02

00

00

00

00

02

00

00

00

00

00

22

00

]))(2

(exp[

])(2

(exp[

])(2

exp[)](exp[

y

y

x

x

dyr

yry

r

rik

dxr

xrx

r

rik

r

yxikrik

We note that

2

00

00

00

00

00

2

00

0

00

000

00

00

00

00

2

00

00

00

00

2

00

00

2

00

00

00

00

)(2

]22

[

2

1

)(

2

2)(

2

xx

r

r

r

r

r

xr

r

rx

r

r

r

r

r

xrx

r

rr

r

r

xrx

r

r

20

20

20 yx

20

20

20 )()( yyxxrr

21 xxx , 21 yyy (square aperture)

The total path is given by

)(

)(2])()[(

2

2

)()(

2

00

00

222

00

00

2

00

00

00

00

0

20

20

0

20

20

00

r

r

yx

r

yry

r

xrx

r

r

r

yyxxyxrr

2

1

2

1

02

00

00

00

00

02

00

00

00

00

00

22

00

]))(2

(exp[

])(2

(exp[

])(2

exp[)](exp[

y

y

x

x

dyr

yry

r

rik

dxr

xrx

r

rik

r

yxikrik

We note that

2

00

00

00

00

00

2

00

0

00

000

00

00

00

00

2

00

00

00

00

2

00

00

2

00

00

00

00

)(2

]22

[

2

1

)(

2

2)(

2

xx

r

r

r

r

r

xr

r

rx

r

r

r

r

r

xrx

r

rr

r

r

xrx

r

r

Then we have

2

2

2

1

02

00

00

00

00

00

02

00

00

00

001

})(2

1exp{

]))(2

(exp[

x

x

x

x

dxx

xr

r

r

rik

dxr

xrx

r

rikI

We put

])1[()(

])1[()(

)(

00

0

000

0

00

02

000

00

00

00

00

00

00

xxrr

rk

xxrr

rk

xx

r

r

r

rkt

and

000

000

000

2

0

00 )(

)(

)1(

dxr

rkdx

r

rkr

dt

2

1

)2

exp(2

00

001

dtt

ir

r

kI

where

])1[()(

])1[()(

20

0

000

02

10

0

000

01

xxrr

rk

xxrr

rk

.

Similarly for the y direction. we have

2

1

)2

exp(2

00

002

dtt

ir

r

kI

])1[()(

])1[()(

20

0

000

02

10

0

000

01

yyrr

rk

yyrr

rk

The resultant electric field at the point P at the coordinate (x, y,z) is given by

2

1

2

1

]2

exp[]2

exp[

])(2

(exp[)(

22

00

22

0000

00

dtt

idtt

i

r

yxrik

rk

r

where

)()(]2

exp[0

2

iSCdtt

i

)]()([)]()([

]2

exp[]2

exp[]2

exp[

1122

0

2

0

22 122

1

iSCiSC

dtt

idtt

idtt

i

The intensity is determined by the square of the field. Thus, we have

2112211220 |)]()([)]()()])([()([)]()(([|

4

1 iSCiSCiSCiSCII

Then we have

2

2

2

1

02

00

00

00

00

00

02

00

00

00

001

})(2

1exp{

]))(2

(exp[

x

x

x

x

dxx

xr

r

r

rik

dxr

xrx

r

rikI

We put

])1[()(

])1[()(

)(

00

0

000

0

00

02

000

00

00

00

00

00

00

xxrr

rk

xxrr

rk

xx

r

r

r

rkt

and

000

000

000

2

0

00 )(

)(

)1(

dxr

rkdx

r

rkr

dt

2

1

)2

exp(2

00

001

dtt

ir

r

kI

where

])1[()(

])1[()(

20

0

000

02

10

0

000

01

xxrr

rk

xxrr

rk

.

Similarly for the y direction. we have

2

1

)2

exp(2

00

002

dtt

ir

r

kI

])1[()(

])1[()(

20

0

000

02

10

0

000

01

yyrr

rk

yyrr

rk

The resultant electric field at the point P at the coordinate (x, y,z) is given by

2

1

2

1

]2

exp[]2

exp[

])(2

(exp[)(

22

00

22

0000

00

dtt

idtt

i

r

yxrik

rk

r

where

)()(]2

exp[0

2

iSCdtt

i

)]()([)]()([

]2

exp[]2

exp[]2

exp[

1122

0

2

0

22 122

1

iSCiSC

dtt

idtt

idtt

i

The intensity is determined by the square of the field. Thus, we have

2112211220 |)]()([)]()()])([()([)]()(([|

4

1 iSCiSCiSCiSCII

A.2 Fresnel diffraction intensity for the square aperture: Mathematica

FresnelS[];

FresnelC[] Plot3D ParametricPlot DensityPlot RegionPlot3D Or[a, b]

Fresnel diffraction with a square aperturel = 632 nm, z = 400 mm, L = 2.4 mm The unit of x and y are mm.

Clear"Gobal`"; F L2

z ;

rule1 k 2

632 109, 632 109, z 400 103, L 2.4 103,

x1 103 x, y1 103 y;

1 k

z

L

2 x1 . rule1; 2

k

z

L

2 x1 . rule1;

1 k

z

L

2 y1 . rule1;

2 k

z

L

2 y1 . rule1;

I11_, 2_, 1_, 2_ :1

4FresnelC2 FresnelC12 FresnelS2 FresnelS12

FresnelC2 FresnelC12 FresnelS2 FresnelS12;

Fresnel number

F1 F . rule1

22.7848

Plot3DI11, 2, 1, 2, x, 1, 1, y, 1, 1, PlotRange All,

AxesLabel "x mm", "ymm", "Intensity",

AxesEdge 1, 1, 1, 1, 1, 1,

AxesStyle Blue, Thick, Blue, Thick, Red, Thick,

PlotLabel "F" ToStringF1

PlotEvaluateTableI11, 2, 1, 2, y, 1, 1, 0.2, x, 1, 1,

PlotRange All, AxesLabel "x", "Intensity", Background LightGray,

PlotStyle TableHue0.2 i, Thick, i, 0, 5,

PlotLabel "F" ToStringF1

-1.0 -0.5 0.5 1.0x

0.8

1.0

1.2

1.4

1.6

1.8

IntensityF=22.7848

DensityPlotI11, 2, 1, 2, x, 1, 1, y, 1, 1, PlotRange All,

AxesLabel "x mm", "ymm", PlotPoints 40,

ColorFunction Hue0.8 &, PlotLabel "F" ToStringF1

A3. Cornu spiral

________________________________________________________________________

Clear"Global`";

f1 ParametricPlotFresnelCx, FresnelSx, x, 0, 10, PlotStyle Blue, Thin,

PlotRange All, AxesLabel "C", "S";

f2 GraphicsBlack, TablePointFresnelC, FresnelS, , 0, 10, 0.1, Red,

TablePointFresnelC, FresnelS, , 0, 10, 0.5,

TableTextStyle"" ToString, Black, 12, FresnelC, FresnelS,

, 0, 5, 0.1;

Showf1, f2, PlotRange All

a=0. a=0.1 a=0.2 a=0.3a=0.4

a=0.5

a=0.6

a=0.7

a=0.8

a=0.9

a=1

a=1.1

a=1.2

a=1.3a=1.4

a=1.5

a=1.6

a=1.7

a=1.8

a=1.9a=2.

a=2.1

a=2.2

a=2.3

a=2.4a=2.5

a=2.6

a=2.7

a=2.8a=2.9

a=3.

a=3.1a=3.2

a=3.3

a=3.4 a=3.5

a=3.6

a=3.7a=3.8

a=3.9

a=4.

a=4.1

a=4.2a=4.3

a=4.4a=4.5

a=4.6

a=4.7

a=4.8

a=4.9

a=5.

0.2 0.4 0.6Ca

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Sa

Chapter 15S Fresnel diffraction - Binghamton...

Documents

Transcript of Chapter 15S Fresnel diffraction - Binghamton...