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
(2 2 )d ( ) da . V A
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
I ξd ()d ()da . 1 V V A
Noting that
() 2 , we have
I (2 )d () da . 1 V A
By replacing , we also have
I (2 )d ( ) da , 1 V A
Thus we find the Green's theorem
I I (2 2 )d ( ) da , 1 2 V A or
(2 2)d ( ) nda , V A
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. Screen
u=0
Q s=r-r' P r' u=0 r Aperture
S n' C u=0
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). SP r . PQ r r' . The green line denotes an aperture.
We now start with the Green's theorem given by
[(r')'2 (r') (r')'2 (r')]d 3r' [(r')' (r') (r')'(r')]nda' , V A
Here we assume that
(r') u(r')
and
eik rr' (r') G(r,r') , ('2 k 2 )G(r,r') (r r') , 4 | r r'|
with
sˆ eiks 1 sˆ eiks 'G(r,r') (ik ) ik , 4 s s 4 s
where k (= 2/) is the wavenumber, G(r,r') is the Green's function, G(r,r') G(s) with s r r' , and sˆ is the unit vector of s. Then we have
[u(r')'2 G(r,r') G(r,r')'2 u(r')]d 3r' [u(r')'G(r,r') G(r,r')'u(r')]nda' , V AperturescreenC or
[u(r') (r r') G(r,r')'2 u(r')]d 3r' [u(r')'G(r,r') G(r,r')'u(r')]nda' . V AperturescreenC
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
u(r) [u(r')'G(r,r') G(r,r')'u(r')]n'da' Aperture 1 eik rr' eik rr' [u(r')' 'u(r')]n'da' 4 Aperture | r r'| | r r'|
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, u(r') and 'u(r') will not appreciably differ from the values obtained in the free space. So we assume that uinc (r') satisfies the Maxwell's
equation in the free space. uinc (r') is the component of the electric field for the spherical wave (incident wave);
2 2 (' k )uinc (r') 0 .
The solution of this differential equation is given by
eikr' u(r') u (r') E , inc 0 r
and
rˆ' eikr' 1 rˆ' eikr' ik 'u (r') E (ik ) ikE rˆ'u (r'), inc 0 4 r' r' 0 4 r' 4 inc
on the aperture surface, where E0 is a electric field constant amplitude and rˆ' is the unit vector of r'. Then we have
1 eikr' eiks 1 eiks eikr' 1 u(r) E { [sˆ n' (ik )] [rˆ'nˆ' (ik )]}da' 0 4 Aperture r' s s s r' r'
ik eiks u (r')(sˆ rˆ')nˆ'da' inc 4 Aperture s ik eiks u (r')I(rˆ',sˆ)da' inc 2 Aperture s (Fresnel-Kirchhoff diffraction formula)
where the inclination factor is defined as
1 I(rˆ',sˆ) (sˆ rˆ')nˆ'. 2
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.
Screen
n`' A Q c ` s r0 s dA ` r0' C S P r0 Aperture
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
eik0 u(Q) E0 , 0
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
ik eik0 eiks du(P) I()E0 dA 2 0 s
where s =QP. The inclination factor I() is given by
1 1 I() (sˆ ρˆ )nˆ' (1 cos) 2 0 2 ,
where is the angle (often called the angle of diffraction) between the normal at Q and
the direction of propagation, and ρˆ 0 nˆ' in the present case. The total disturbance at the point P is given by
ik eik0 eiks u(P) E I() dA . 0 2 0 Aperture s
15S.4 Fresnel zone
A Q c s
q C S P r0 r0
Fresnel's 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,
2 3 4 5 r , r , r , r , r , r ,..... 0 0 2 0 2 0 2 0 2 0 2
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 ( r - r ), the Fresnel's 1st zone 0 0 2 2 Z2 ( r - r ), the 2-nd zone 0 2 0 2 2 3 Z3 ( r - r ), the 3-rd zone 0 2 0 2 ...... ( j 1) j Zj ( r - r ), the j-th zone 0 2 0 2 ......
Fig. Fresnel's zone (n = 1, 2, 3, ---). The distance PQ is equal to r0 n / 2
r0 (n 1) / 2 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
2 2 2 s (0 r0 ) 0 20 (0 r0 )cos .
So that
sds 0 (0 r0 )sind ,
and the surface element dA is given by
2 2 sds dA 20 sind 20 . 0 (0 r0 )
((Note-1)) The area of the end cap is given by
A dA 2 2 sind 2 2 (1 cos ) 0 0 0 0 2 2 2 2 2 (0 r0 ) 0 s 20 20 20 (0 r0 )
The area of the j-th zone is obtained as
2 2 j 2 (0 r0 ) 0 (r0 ) 2 2 2 Aj [20 20 ] 20 (0 r0 )
2 2 ( j 1) 2 (0 r0 ) 0 (r0 ) [2 2 2 2 2 ] 0 0 20 (0 r0 ) 2 20 j 2 ( j 1) 2 [(r0 ) (r0 ) ] 20 (0 r0 ) 2 2
0 (2 j 1) [r0 ] 0 r0 4
The mean distance from the j-th zone to the point P is denoted as s j
j ( j 1) (r ) (r ) 0 0 (2 j 1) s 2 2 r j 2 0 4
Then we have
A j 0 , s j 0 r0 which is independent of j.
((Note-2)) Inclination factor We note that
2 r r 2 s 2 sˆ nˆ' cos 0 0 0 . 20 s
Then the inclination factor is obtained as
1 2 r r 2 s 2 (s r )(s r 2 ) I() (1 0 0 0 ) 0 0 0 , 2 20s 40s since
SP (0 r0 ,0) , SQ (0 cos, 0 sin ) , nˆ' (cos,sin )
s QP SP SQ (0 r0 0 cos,0 sin ) ,
s nˆ' ( r )cos 2 r r 2 s2 cos 0 0 0 0 0 0 . s s 20s
______Then the contribution of the j-th Fresnel's zone to u(P) is
ik eik0 eiks sds u (P) 2 E I() 2 j 2 0 s 0 ( r ) 0 Z j 0 0 0 r j / 2 eik0 0 ikE I eiks ds 0 ( r ) j 0 0 r0 ( j1) / 2 j / 2 eik (0 r0 ) ikE I eiks'ds' 0 j (0 r0 ) ( j1) / 2
where s' s r , and I() is approximated by Ij, the value of I() at s r ( j 1) ; 0 0 2
[2r0 ( j 1) ][20 ( j 1) ] I 2 2 j 4 [r ( j 1) ] 0 0 2
and I1 = -1. Noting that
j / 2 i(1) j eiks'ds' ( j1) / 2 we get
ik (0 r0 ) e j u j (P) 2E0 I j (1) (0 r0 )
The total effect at the point P is obtained by summing all the contributions:
n ik (0 r0 ) n e j u(P) u j (P) 2E0 I j (1) j1 (0 r0 ) j1
For n = 2m+1 (odd)
ik (0 r0 ) e I1 I1 I3 I 2m3 I 2m1 u(P) 2E0 [ ( I 2 ) ... ( I 2m2 ) (0 r0 ) 2 2 2 2 2 I I I ( 2m1 I 2m1 ) 2m1 ] 2 2m 2 2
ik (0 r0 ) e I1 I 2m1 2E0 ( ) (0 r0 ) 2 2 eik (0 r0 ) E0 (I1 I 2m1 ) (0 r0 )
since Ij changes slowly with j, and
I I I j1 j1 . j 2
For n = 2m (even)
ik (0 r0 ) e I 2 I 2 I 4 u(P) 2E0 [I1 ( I3 ) ... (0 r0 ) 2 2 2 I I I ( 2m2 I 2m ) 2m ] 2 2m1 2 2
ik (0 r0 ) e I 2 I 2m 2E0 (I1 ) (0 r0 ) 2 2 eik (0 r0 ) E0 (I1 I 2m ) (0 r0 )
where I 2 I1 . 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
eik (0 r0 ) eik (0 r0 ) u(P) E0 I1 E0 0 r0 0 r0
since I1 = 1. We note that the disturbance at P only due to the first zone is
eik (0 r0 ) eik (0 r0 ) u1 (P) 2E0 I1 2E0 (0 r0 ) (0 r0 )
Therefore we have
1 u(P) u (P) . 2 1
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. O1Q r0 .
O1P r1 . PQ r1 r0 . 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
r' ρ0 r0 ,
where r0 is a position vector located on the aperture. The vector r is expressed by
r ρ0 r1 ,
where ρ0 is the vector connecting the source point S and the new origin O1 in the aperture, and r1 is a position vector from the new origin O1 to the position on the observation plane. The vector s is given by
2 2 2 s r r' r1 r0 , s | r1 r0 | (x x0 ) (y y0 ) z .
For simplicity we assume that
s z , and I(rˆ',s) 1, for sˆ //rˆ'//nˆ' . Then we get the formula of the Fresnel-Kirchhoff diffraction,
ik u(r ) eik|r1 r0 |u (r ρ )dA , 1 inc 0 0 0 2z Aperture
where dA0 is the surface area element in the aperture; dA0 = dx0dy0. We assume that
uinc (r0 ρ0 ) uinc (ρ0 ) ,
Then we get
ik i u(r ) u (ρ ) eik|r1 r0 |dA u (ρ ) eik|r1 r0 |dA . 1 inc 0 0 inc 0 0 2z Aperture z Aperture
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). OP r (x, y, z)
The electric field E(x, y, z) at the point P is expressed by
ik E(x, y, z) E (ρ ) eik|rr0 |dx dy , inc 0 0 0 2z Aperture
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.
2 2 2 r r0 (x x0 ) (y y0 ) z
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 2 2 2 z (x x0 ) (y y0 )
Then we can expand r r0 around = 0, as
2 4 6 r r z .... 0 2z 8z 3 16z 5
Then we have
k 2 k 4 k 6 k r r kz .... 0 2z 8z 3 16z 5
We assume that
k 4 2 , 8z3 or
4 8. z3
When is on the same order as the size of aperture a, this condition can be rewritten as
a4 8. z3
((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 a4 k 4 above condition; 8 . The Fresnel diffraction may o ccur when the factor exp(i ) z3 8z 3 k 2 slowly changes compared to the other factors exp[i(kz )]. 2z
Under such a condition, the distance r r0 is approximated by
2 r r z . 0 2z
This is called the Fresnel approximation.
eikz (x x )2 (y y )2 E(x, y, z) i E (ρ ) dx dy exp[ik 0 0 ] z inc 0 0 0 2z Aperture eikz (x x )2 (y y )2 i E (ρ ) (x , y ,0)dx dy exp[ik 0 0 ] z inc 0 0 0 0 0 2z
where (x0 , y0 ,0) 1 for the inside of the aperture, and 0 for the outside of the aperture.
Mathematically, this integral corresponds to the convolution of (x0 , y0 ,0) and h(x0 , y0 ) , and can be expressed by
E(x, y, z) (x, y,0) exp h(x, y) where
eikz x 2 y 2 h(x, y) i exp(ik ) . z 2z
15S.7 Fraunhofer diffraction The above equation can be rewritten as
eikz (x2 y 2 ) x 2 y 2 k E(x, y, z) i exp[ik ] dx dy (x , y ,0)exp[ik 0 0 ]exp[i (x x y y)] z 2z 0 0 0 0 2z z 0 0
Suppose that
x 2 y 2 0 0 k 1. 2z
x 2 y 2 Then the quadratic phase factor exp(ik 0 0 ) is approximately unity over the entire aperture. 2z So we get
eikz (x2 y2 ) k E(x, y, z) i exp[ik ] dx dy (x , y ,0)exp[i (x x y y)]. z 2z 0 0 0 0 z 0 0
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.
1 dx dy (x , y ,0)exp[i(k x k y )] F[(x , y ) 2 0 0 0 0 x 0 y 0 0 0 2 with
k 2 k 2 k x x , k y y x z z y z z 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
ik E(x, y, z) E (ρ ) (r )eik|rr0|d 2r 2z inc 0 0 0
As shown in Fig., the distance s is approximated by
s r r0 r (rˆ r0 ) ,
for r>>r0.
P
H
Q O
Fig. OQ r OP r . QP s r r . OH (rˆr )rˆ 0 0 0
QP s | r r0 | QH r (rˆr0 ) .
Then we get the expression,
ik ˆ E(x, y, z) E (ρ ) (r )eikrik(rr0 )d 2r 2z inc 0 0 0
ik ˆ E (ρ )eikr (r )eik(rr0 )d 2r . 2z inc 0 0 0
Here we note that
k k x y krˆr (r r ) (xx yy ) k( )x k( )x . 0 r 0 z 0 0 z 0 z 0
Then E(x, y, z) is proportional to the Fourier transform of the aperture,
1 ˆ x y (r )eik (rr0 )d 2r F[(r ),{k k ,k k }] 2 0 0 0 x y 2 z z with the wave vector given by
x y k k ,k k . x z y z
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
a2 F , z 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
a 2 1 F = 0.318. z
On the other hand, the condition for the appearance of the Fresnel diffraction can be rewritten as
a 2 z 2 F 8 . z a 2
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 ( a x a , b x b ) is normally illuminated by a monochromatic plane wave of unit amplitude and the wavelength .
eikz k U (x, y) i exp{i [(x x)2 (y y)2 ]}dx dy z 2z 1 1 1 1
eikz a k b k i exp[i (x x)2 ]dx exp[i (y x)2 ]dy 1 1 1 1 z a 2z b 2z 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,
a k F(x) exp[i (x x)2 ]dx 2z 1 1 a b k F(y) exp[i (y y)2 ]dy 1 1 b 2z
These integrals are simplified by the change of variables,
k k (x x) , (y y) z 1 z 1 yielding
z 2 F(x) exp(i 2 )d k 2 1 z 2 F(y) exp(i 2 )d k 2 1 where the limits of integration are
k k (a x) , (a x) 1 z 2 z
k k (b y) , (b y) 1 z 2 z
The integrals F(x) and F(y) can be evaluated in terms of the Fresnel integrals, which are defined by
C() cos( t 2 )dt 2 0 S() sin( t 2 )dt 0 2
We note that
exp[i( 2 )]d [cos( 2 ) isin( 2 )]d 0 2 0 2 2 C() iS() and
2 2 1 exp[i( 2 )]d exp[i( 2 )]d exp[i( 2 )]d 2 2 2 1 0 0
C(2 ) iS(2 ) C(1) iS(1)
C(2 ) C(1) i[S(2 ) S(1)]
Finally we have the complex field distribution
eikz U (x, y) {C( ) C( ) i[S( ) S( )]}{C( ) C( ) i[S( ) S( )]} 2i 2 1 2 1 2 1 2 1 and the corresponding intensity distribution
1 I(x, y) {[C( ) C( )]2 [S( ) S( )]2}{[C( ) C( )]2 [S( ) S( )]2} 4 2 1 2 1 2 1 2 1
Note that
1 1 C() , C() , C( ) C( ) 2 2
1 1 S() , S() S( ) S( ) 2 2
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. S x
0.6
FresnelS x 0.4
0.2 x -10 -5 5 10
-0.2
-0.4
-0.6
Fig. Plot of S(x) as a function of x. C x
0.5 FresnelC x
x -10 -5 5 10
-0.5
Fig. Plot of C(x) as a function of x.
S a
0.6
0.4
0.2
C a -0.5 0.5
-0.2
-0.4
-0.6
Fig. ParametricPlot of {C(), S()} when a is changed as parameter. S a
0.7 a=1.4 a=1.5 a=1.3 a=1.6 a=2.5 a=2.4 a=1.2 0.6 a=3.2 a=3.1 a=3.8a=4.7a=4.23.7 1.7 2.6a=4.3 a=2.3 a= a= a=1.1 a=3.3 a=4.6 0.5 a=4.8 a=a=5.a=3.63. a=3.9 a=4.1 a=1.8a=2.7a=4.4 a=2.2 a=3.4a=a=4.94.5 a=1 a=a=4.a=3.52.9 0.4 a=2.8 a=1.9 a=2.1 a=2. a=0.9 0.3
a=0.8
0.2 a=0.7
0.1 a=0.6 a=0.5 a=0.4 a=0.3 a=0. a=0.1 a=0.2 C a 0.2 0.4 0.6
Fig. ParametricPlot of {C(), S()} around the point Z (1/2, 1/2) when a is changed as parameter.
Si a Ci a a=-0.3a=-0.2 a=-0.1 a=0 -0.6 a=--0.40.4 -0.2 a=-0.5 a=-0.6 -0.1
a=-0.7 -0.2 a=-0.8 -0.3 a=-0.9 a=-2. a=-2.1 a=-1.9 a=-2.8 -0.4 a=-a=-2.9a=-3.54. =-1. a=-a=-4.54.9a=-3.4 a=-2.2 a=-2.7a=-1.8 a=-4.1 a=-a=-4.43.9 a=-a=-a=-3.3.65. a=-4.8 -0.5 a=-4.6 a=-3.3 a=-1.1 a=-2.6a=-1.7 a=-2.3a=-4.2a=-4.33.8 a=-a=-3.1a=-3.7 4.7 a=-3.2 -0.6 a=-2.4a=-2.5 a=-1.2 a=-1.6
a=-1.3 a=-1.5 -0.7 a=-1.4
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) 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).
F=22.7848 Intensity
1.8
1.6
1.4
1.2
1.0
0.8
x -1.0 -0.5 0.5 1.0
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 = 632 nm, z = 400 mm, L = a = 2.1 mm x(mm), y(mm)
F=17.4446 Intensity
2.0
1.5
1.0
x -1.0 -0.5 0.5 1.0
(c) F = 12.8165 = 632 nm, z = 400 mm, L = a = 1.8 mm x(mm), y(mm)
F=12.8165 Intensity
2.0
1.5
1.0
0.5
x -1.0 -0.5 0.5 1.0
(d) F = 8.9032 = 632 nm, z = 400 mm, L = a = 1.5 mm x(mm), y(mm)
F=8.90032 Intensity
1.5
1.0
0.5
x -1.0 -0.5 0.5 1.0
(e) F = 5.6962 = 632 nm, z = 400 mm, L = a = 1.2 mm x(mm), y(mm)
F=5.6962 Intensity
2.0
1.5
1.0
0.5
x -1.0 -0.5 0.5 1.0
(f) F = 3.20411 = 632 nm, z = 400 mm, L = a = 0.9 mm x(mm), y(mm)
F=3.20411 Intensity
3.0
2.5
2.0
1.5
1.0
0.5
x -1.0 -0.5 0.5 1.0
(g) F = 1.42405 = 632 nm, z = 400 mm, L = a = 0.6 mm x(mm), y(mm)
F=1.42405 Intensity
1.5
1.0
0.5
x -1.0 -0.5 0.5 1.0
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)
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
r z2 x 2 (y y )2 0 0
where P at (x = 0, y, z) and Q at (x0, y0, z = 0). The distance r is approximated as
x 2 (y y)2 r z 0 0 . 2z
Then the wave arriving at the point P (x = 0, y, z) on the screen is expressed in the form
k k exp(ikz) exp(i x 2 )dx exp[i (y y)2 ]dy 0 0 0 0 2z 0 2z
We put
k t (y y) z 0 and
k dt dy z 0
Then we have
k z t 2 exp[i (y y)2 ]dy exp(i )dt 0 0 0 2z k w 2
where
k w y . z
The resultant electric field at the point P is given by
z t 2 t 2 exp(ikz) exp[i ]dt exp[i ]dt k 2 w 2 where
t 2 exp[i ]dt C() iS() 0 2
t 2 t 2 w t 2 exp[i ]dt exp[i ]dt exp[i ]dt w 2 0 2 0 2 1 1 i [C(w) iS(w)] 2 2 1 1 i [C(w) iS(w)] 2 2 1 1 [C(w) ] i[S(w) ] 2 2
The intensity is given by
1 1 1 I I ([C(w) ]2 [S(w) ]2 ) . 2 0 2 2
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.
Intensity 1.4
1.2
1.0
0.8
0.6
0.4
0.2
y -2 2 4
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.
Intensity
1.3 P1 1.2
1.1 P2 P 3 P 1.0 4 Q P5 Q5 Q3 4 P6 Q2
0.9 Q1
y 0.4 0.6 0.8 1.0 1.2 1.4 1.6
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).
Fig. DensityPlot of the intensity vs the distance y from y = 0.
Using the value of y in the units of mm, we get
25 w y . 79
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. S a
y=0.5 P1
0.6 y=0.9 P2 y=1.1 y=0.6 y=1.5 y=1.3 y=0.4 Z y=0.8 y=y1.2=1.4 0.4 Q2 y=1. Q1 y=0.7
y=0.3
0.2
y=0.2
y=0. y=0.1 C a -0.5 0.5
-0.2
-0.4
Z'
-0.6
Fig. Cornu plot. Z' at (-1/2, -1/2). Z at (1/2, 1/2). y=0.5
P1
y=0.9 P2 y=1.1 y=1.5 P3 y=0.6 P4 y=0.4 y=1.3
Z y=0.8
y=1.2y=1.4 Q3 Q2 y=1. Q1
y=0.7
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)
Fig. Single slit aperture. The width of the single slit is a.
The electric filed distribution is given by
eikz U (x, y) {C(2 ) C(1) i[S(2 ) S(1)]}{C(2 ) C(1) i[S(2 ) S(1)]} 2i
The intensity is proportional to |U(x, y)|2, and is given by the form
I0 2 2 2 2 I(x, y) {[C(2 ) C(1)] [S(2 ) S(1)] }{[C(2 ) C(1)] [S(2 ) S(1)] } 4 where
k a k a 1 , 2 , 1 ( y) , 2 y z 2 z 2
Using these values of x1, x2, h11, and h2, we get
ikz i e 4 U (x, y) 2e {C(2 ) iS(2 ) [C(1) iS(1)]} 2i (a) a = 2.4 mm
a= 2.4 Intensity
1.4
1.3
1.2
1.1
1.0
0.9
0.8
y -1.0 -0.5 0.5 1.0 a = 2.4 S a
0.6 y=0.6 y=0.8 y=0.4 y=0. 0.4 y=0.2
0.2
y=1.
y=1.2 C a -0.5 0.5
y=1.4
-0.2
y=2.2 -0.4 y=2.8 y=y2.6=2.4 y=2. y=1.6 y=3. y=1.8 -0.6
(b) a = 2.1 mm a= 2.1 Intensity
1.4
1.2
1.0
0.8
0.6
y -1.0 -0.5 0.5 1.0 a = 2.1 S a
y=0.6
0.6 y=0.2
y=0. y=0.4 0.4
0.2 y=0.8
y=1. C a -0.5 0.5 y=1.2
-0.2
y=1.8 -0.4 y=1.4 y=2.8 y=2. y=2.6 y=3. yy==2.42.2 -0.6
y=1.6
(c) a = 1.8 mm
a= 1.8 Intensity
1.4
1.2
1.0
0.8
0.6
0.4
y -1.0 -0.5 0.5 1.0
a = 1.8 S a
y=0.4
0.6 y=0.
0.4 y=0.2 y=0.6
0.2
y=0.8 C a -0.5 y=1. 0.5
-0.2
y=1.2 y=1.6 -0.4
y=2.8 yy=3.2.6 y=2.2 y=y2.=2.4 y=1.8 -0.6
y=1.4
(d) a = 1.5 mm a= 1.5 Intensity
1.4
1.2
1.0
0.8
0.6
0.4
y -1.0 -0.5 0.5 1.0 0.0
a = 1.5 S a
y=0.2
0.6
y=0.4 0.4 y=0.
0.2
y=0.6 y=0.8 C a -0.5 0.5
y=1. -0.2
-0.4 y=2. y=1.4 y=y1.8=2.2y=3. y=2.6 y=2.8 y=2.4 -0.6 y=1.6 y=1.2
(e) a = 1.2 mm a= 1.2 Intensity
1.5
1.0
0.5
y -1.0 -0.5 0.5 1.0 0.0
a = 1.2 S a
0.6 y=0. y=0.2
0.4
0.2
y=0.4
y=0.6 C a -0.5 0.5
y=0.8
-0.2
y=1.6 -0.4 y=2.2 y=y2.6y==2.1.8 y=1.4 y=3. y=1. y=2.4y=2.8 y=1.2 -0.6
(f) a = 0.9 mm a= 0.9 Intensity
1.5
1.0
0.5
y -1.0 -0.5 0.5 1.0
a = 0.9 S a
y=0.
0.6
0.4
0.2 y=0.2
y=0.4 C a -0.5 0.5 y=0.6
-0.2
y=1.2 -0.4 y=0.8 y=2.2 y=2.8 y=2.6 y=3. y=1.4 y=2. y=2.4 yy==1.81.6 -0.6
y=1.
(g) a = 0.6 mm
a= 0.6 Intensity
1.2
1.0
0.8
0.6
0.4
0.2
y -1.0 -0.5 0.5 1.0
______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)
2 2 2 0 x0 y0
r r 2 (x x )2 (y y )2 0 0 0
x1 x x2 , y1 y y2 (square aperture)
The total path is given by
2 2 2 2 x0 y0 (x x0 ) (y y0 ) r 0 r0 20 2r0 2 2 0 r0 r0 x 2 r0 y 2 x y [(x0 ) (y0 ) ] 20r0 0 r0 0 r0 2(0 r0 )
(0 r0 )
x2 y2 exp[ik(0 r0 )]exp[ik ] 2(0 r0 ) x2 r xr exp[ik( 0 0 (x 0 )2 ]dx 2 r 0 r 0 x1 0 0 0 0 y2 r yr exp[ik( 0 0 )(y 0 )2 ]dy 2 r 0 r 0 y1 0 0 0 0
We note that 2 r 0 0 r r x 2 r xr 0 0 (x 0 )2 0 0 (x 0 )2 0 r 0 20r0 0 r0 0 0 0 r0
20r0 1 r r r x [ 0 0 x 0 0 0 ]2 r 0 0 0 20r0 20r0 0 r0
20r0
0 r0 0 r0 x 2 ( x0 ) 20r0 0r0 0
2 2 2 0 x0 y0
r r 2 (x x )2 (y y )2 0 0 0
x1 x x2 , y1 y y2 (square aperture)
The total path is given by
2 2 2 2 x0 y0 (x x0 ) (y y0 ) r 0 r0 20 2r0 2 2 0 r0 r0 x 2 r0 y 2 x y [(x0 ) (y0 ) ] 20r0 0 r0 0 r0 2(0 r0 )
(0 r0 )
x2 y2 exp[ik(0 r0 )]exp[ik ] 2(0 r0 ) x2 r xr exp[ik( 0 0 (x 0 )2 ]dx 2 r 0 r 0 x1 0 0 0 0 y2 r yr exp[ik( 0 0 )(y 0 )2 ]dy 2 r 0 r 0 y1 0 0 0 0
We note that 2 r 0 0 r r x 2 r xr 0 0 (x 0 )2 0 0 (x 0 )2 0 r 0 20r0 0 r0 0 0 0 r0
20r0 1 r r r x [ 0 0 x 0 0 0 ]2 r 0 0 0 20r0 20r0 0 r0
20r0
0 r0 0 r0 x 2 ( x0 ) 20r0 0r0 0
Then we have
x2 r r x I exp[ik( 0 0 )(x 0 )2 ]dx 1 0 0 x 20r0 0 r0 1 x2 1 r r x exp{ik 0 0 ( 0 0 x )2}dx 2 r r 0 0 x2 0 0 0 0 0
We put
k 0r0 0 r0 x t ( x0 ) 0 r0 0r0 0
k 0r0 0 2 [(1 )x0 x] (0 r0 )0 r0
k r0 0 [(1 )x0 x] (0 r0 )0 r0 and
0 2 kr0 (1 ) r0 k(0 r0 ) dt dx0 dx0 0 (0 r0 ) 0r0
r 2 t 2 I 0 0 exp(i )dt 1 k r 2 0 0 1 where
k r0 0 1 [(1 )x1 x] (0 r0 )0 r0 .
k r0 0 2 [(1 )x2 x] (0 r0 )0 r0
Similarly for the y direction. we have
r 2 t 2 I 0 0 exp(i )dt 2 k r 2 0 0 1
k r0 0 1 [(1 )y1 y] (0 r0 )0 r0
k r0 0 2 [(1 )y2 y] (0 r0 )0 r0
The resultant electric field at the point P at the coordinate (x, y,z) is given by
2 2 0r0 x y exp[ik(0 r0 ] k(0 r0 ) 2(0 r0 )
2 t 2 2 t 2 exp[i ]dt exp[i ]dt 2 2 1 1 where
t 2 exp[i ]dt C() iS() 0 2
2 t 2 2 t 2 1 t 2 exp[i ]dt exp[i ]dt exp[i ]dt 2 2 2 1 0 0
[C(2 ) iS(2 )] [C(1) iS(1)]
The intensity is determined by the square of the field. Thus, we have
1 I I | ([C( ) iS( )] [C( ) iS( )])([C( ) iS( )] [C( ) iS( )] |2 4 0 2 2 1 1 2 2 1 1 Then we have x2 r r x I exp[ik( 0 0 )(x 0 )2 ]dx 1 0 0 x 20r0 0 r0 1 x2 1 r r x exp{ik 0 0 ( 0 0 x )2}dx 2 r r 0 0 x2 0 0 0 0 0
We put
k 0r0 0 r0 x t ( x0 ) 0 r0 0r0 0
k 0r0 0 2 [(1 )x0 x] (0 r0 )0 r0
k r0 0 [(1 )x0 x] (0 r0 )0 r0 and
0 2 kr0 (1 ) r0 k(0 r0 ) dt dx0 dx0 0 (0 r0 ) 0r0
r 2 t 2 I 0 0 exp(i )dt 1 k r 2 0 0 1 where
k r0 0 1 [(1 )x1 x] (0 r0 )0 r0 .
k r0 0 2 [(1 )x2 x] (0 r0 )0 r0
Similarly for the y direction. we have
r 2 t 2 I 0 0 exp(i )dt 2 k r 2 0 0 1
k r0 0 1 [(1 )y1 y] (0 r0 )0 r0
k r0 0 2 [(1 )y2 y] (0 r0 )0 r0
The resultant electric field at the point P at the coordinate (x, y,z) is given by
2 2 0r0 x y exp[ik(0 r0 ] k(0 r0 ) 2(0 r0 )
2 t 2 2 t 2 exp[i ]dt exp[i ]dt 2 2 1 1 where
t 2 exp[i ]dt C() iS() 0 2
2 t 2 2 t 2 1 t 2 exp[i ]dt exp[i ]dt exp[i ]dt 2 2 2 1 0 0
[C(2 ) iS(2 )] [C(1) iS(1)]
The intensity is determined by the square of the field. Thus, we have
1 I I | ([C( ) iS( )] [C( ) iS( )])([C( ) iS( )] [C( ) iS( )] |2 4 0 2 2 1 1 2 2 1 1
A.2 Fresnel diffraction intensity for the square aperture: Mathematica
FresnelS[]; FresnelC[] Plot3D ParametricPlot DensityPlot RegionPlot3D Or[a, b] Fresnel diffraction with a square aperture l = 632 nm, z = 400 mm, L = 2.4 mm The unit of x and y are mm.
L2 Clear "Gobal`" ;F ; z
2 rule1 k , 632 109, z 400 103, L 2.4 103, 632 109 x1 103 x, y1 103 y ; k L k L 1 x1 . rule1; 2 x1 . rule1; z 2 z 2 k L 1 y1 . rule1; z 2 k L 2 y1 . rule1; z 2
I1 1_, 2_, 1_, 2_ : 1 FresnelC 2 FresnelC 1 2 FresnelS 2 FresnelS 1 2 4 FresnelC 2 FresnelC 1 2 FresnelS 2 FresnelS 1 2 ;
Fresnel number
F1 F . rule1 22.7848 Plot3D I1 1, 2, 1, 2 , x, 1, 1 , y, 1, 1 , PlotRange All, AxesLabel "x mm ", "y mm ", "Intensity" , AxesEdge 1, 1 , 1, 1 , 1, 1 , AxesStyle Blue, Thick , Blue, Thick , Red, Thick , PlotLabel "F" ToString F1
Plot Evaluate Table I1 1, 2, 1, 2 , y, 1, 1, 0.2 , x, 1, 1 , PlotRange All, AxesLabel "x", "Intensity" , Background LightGray, PlotStyle Table Hue 0.2 i , Thick , i,0,5 , PlotLabel "F" ToString F1 F=22.7848 Intensity
1.8
1.6
1.4
1.2
1.0
0.8
x -1.0 -0.5 0.5 1.0
DensityPlot I1 1, 2, 1, 2 , x, 1, 1 , y, 1, 1 , PlotRange All, AxesLabel "x mm ", "y mm " , PlotPoints 40, ColorFunction Hue 0.8 & , PlotLabel "F" ToString F1
A3. Cornu spiral
Clear "Global`" ;
f1 ParametricPlot FresnelC x , FresnelS x , x,0,10 , PlotStyle Blue, Thin , PlotRange All, AxesLabel "C ", "S " ; f2 Graphics Black, Table Point FresnelC , FresnelS , , 0, 10, 0.1 ,Red, Table Point FresnelC , FresnelS , ,0,10,0.5 , Table Text Style "" ToString , Black, 12 , FresnelC , FresnelS , ,0,5,0.1 ; Show f1, f2, PlotRange All S a a=1.4 0.7 a=1.5 a=1.3 a=1.6 a=2.5 a=2.4 a=1.2 0.6 a=3.2 a=3.1 a=3.8a=4.7a=4.23.7 1.7 2.6a=4.3 a=2.3 a= a= a=1.1 a=3.3 a=4.6 0.5 a=4.8 a=a=5.a=3.63. a=3.9 a=4.1 a=1.8a=2.7a=4.4 a=2.2 a=3.4a=a=4.94.5 a=1 a=a=4. a=3.52.9 0.4 a=2.8 a=1.9 a=2.1 a=2. a=0.9 0.3
a=0.8
0.2 a=0.7
0.1 a=0.6 a=0.5 a=0.4 a=0.3 a=0. a=0.1 a=0.2 C a 0.2 0.4 0.6
______