Fresnel and Fraunhofer Diffraction

4. 1 Background 4.1.1 The intensity of a wave field 4.1.2 The Huygens-Fresnel principle in rectangular coordinates 4.2 The Fresnel approximation 4. 2. 1 Positive vs. negative phases 4.2.2 Accuracy of the Fresnel approximation 4.2.3 The Fresnel approximation and the angular spectrum 4.2.4 Fresnel diffraction between confocal spherical surfaces 4. 3 The Fraunhofer approximation 4.4 Examples of Fraunhofer diffraction patterns 4.4.1 Rectangular aperture 4.4.2 Circular aperture 4. 4. 3 Thin sinusoidal aperture grating 4.4.4 Thin sinusoidal phase grating 4.5 Examples of Fresnel diffraction calculations 4.5.1 Fresnel diffraction by a square aperture 4. 5 2 Fresnel diffraction by a sinusoidal amplitude grating-Talbot images 4.1 Background

• In chapter 3 we dealt with most general form of the diffraction theory. • In ch apt er 4 we will d eal w ith – Intensity of a wave field – Huygens-Fresnel principle – Certain approximations to reduce the problem to a simpler mathematical form. These approximations are: • Fresnel • Fraunhofer – We consider the wave propagation phenomenon as a system. – The approximations will be valid for certain class of inputs. – Preparation for the calculations related to the approximations

Spring 2010 Eradat Physics Dept. SJSU 2 4.1.1 The intensity of a wave field

Intensity is the physically measurable attribute of an optical wavefield Intensity and power density are not the same but proportional Intensity of a scalar monochromatic wave at point P IP()|= UP ()|2 For a narrow-band (not perfectly monochromatic) intensity is given by IP( )== | uPt ( ,)|2 and IPt ( ,) | uPt ( ,)|2 An infinite Instantaneous time average Intensity In calculating a diffraction pattern, we are looking for the intensity of the pa ttern.

Spring 2010 Eradat Physics Dept. SJSU 3 4.1.2 The Huygens-Fresnel principle in rectangular coordinates

According to the first Rayleigh-Somefeld solution the diffracted field

on the xy plane due to the aperture on the ξη plane is UI ( P0 ) and the Huygens-Fresnel principle can be written as: 1 e jkr01 z UP( )= UP() cosθθdsnrwhere cos== cos()( , ) I 01∫∫Σ 01 jλ rr01 01 zejkr01 Uxy( , )= U()(ξη, ξ) η dd, ∫∫Σ 2 jrλ 01 η 222 y where rzx01 =+−+−()()ξη y ξ x Two approximations are used in this result: P0 1) inherent approximation in the scalar theory Σ z θ 2) r01 >> λ

P Spring 2010 Eradat Physics Dept. SJSU 1 4 4.2 The Fresnel approximation I Goal: to reduce the H-F principle to a simple and usable expression

WhithibWe achieve this by approx itifimations for r01 : nn(1)− Binomial expansion: (xy+=+ )nn x nxy n−−122 + x n y ++= ... yn n , 1,2,3,... 2! 11 (1+=±bbb )±1/2 1 ∓ 2 ±−<≤... for 1b 1: the higher order terms are negligable 28 22 2 2 ⎛⎞⎛⎞xy−−ξη⎡⎤ ξ11 ⎛⎞⎛⎞ x − η y − rz01 =+11⎜⎟⎜⎟ + ≈+ z⎢⎥ ⎜⎟⎜⎟ + +... ⎝⎠⎝⎠zz⎣⎦⎢⎥22 ⎝⎠⎝⎠ z z

Where do we cut the series? We will use r01 in the diffracted field equation zeλ jkr01 Ux( ,(,)y)= Uddξ ηξη λλ∫∫Σ 2 jr01

jkr01 The term er is very sensitive to the values of 01 specially since it is multiplied blby a very large num ber k = 2/πλλ. IthiiblIn the visible ofthf the ord er 10 7. W e k eep t wo 2 terms for the exponent. For rz01 error introduced by dropping all terms but is small. ⎧⎫k jkr01 jkz ∞ ⎡⎤22 zee ⎨⎬jx⎣()()−+−ξη y⎦ UUddUU (x, yU)==()(ξξξ,η)( ddUξ η ξ ξ ()ξ,η)e⎩⎭2z ddξ η ∫∫Σ 2 ∫ ∫ jrjz01 −∞ The integration limit is let to ∞ using the usual boundary conditions. Spring 2010 Eradat Physics Dept. SJSU 5 4.2 The Fresnel approximation II

jkz ∞ ⎧⎫k 22 ⎨⎬jx⎡()()−+−ξη y⎤ e λ 2z ⎣⎦ Uxy(),(,)= ∫∫ U ξξη η e⎩⎭ dd this looks like a convolution jz −∞

∞ jkz ⎧⎫k ⎡⎤22 e ⎨⎬jx⎣⎦()()−+ξη+ y− Uxy( ,(,)(,)(,))=−−= Uξη ξ η hx ξ η y dd where hxy e⎩⎭2z ∫∫ jz −∞ First form of the Fresnel diffraction integral Another form of the Fresnel diffraction integral is expressed as the following

k 22 j (ξη+ ) Fourier transform of the U (ξ ,)η e 2 z which is complex field λ just to the right of aperture multipξηlied by a quadratic phase factor λ jkz kk22∞ 22 2π e jxy()++⎧⎫ j()−+jxy() Uxy(),(,)= e22zz⎨⎬ ξ U ηξη e eλz ξη dd jz ∫∫ −∞ ⎩⎭ Seond form of the Fresnel diffraction integral Observation in the near field of the aperture or Fresnel diffraction region xy--ξη Where r >>λ,1,1, < < 01 zz and scalar theory approximation are assumed

Spring 2010 Eradat Physics Dept. SJSU 6 4.2.1 Positive vs. negative phases

Goal: to understand the meaning of the signs of the y phase exponentials: in the sppqherical wave e jkr01 and its equivalent k jxy()22+ in the quadratic approximation ez2z (0) for > z Sign convention: our phasors rotate in the clockwise direction (the angle becomes more negative as time goes) and their time dependence is e− jt2πν We move in space in such a way that we encounter Wavefront Wavefront emitted emitted portions of the wavefield that were emitted earlier in time. later earlier The phase must become more positive since the y phasor had not have time to rotate as far in clockwise. k We move in space in such a way that we encounter θ z portions of the wavefield that were emitted later in time. The phasor will have advanced in the clockwise direction, therefore the phase must become more negative.

Spring 2010 Eradat Physics Dept. SJSU 7 4.2.2 Accuracy of the Fresnel approximation I Fresnel approximation replaced the spherical secondary wavelets with parabolic wavefronts in the Huygens-Fresnel principle

jkr jkz ∞ ⎧⎫k 2 2 01 ⎨⎬jx⎡⎤()()−+−ξη y ze e ⎩⎭2z ⎣⎦ Uxy( , )= U()(ξη ξη, ξ η )(2 dd→ U (), ) e ddξη jrjzλλ∫∫Σ ∫∫ 01 −∞ Parabolic wavelets Spherical wavelets The accuracy of this approximation depends on the size of the higher order terms in binomial expansion. A sufficient condition for accuracy is: ⎡⎤ ⎢⎥ 22 222 ⎢⎥1 ⎛⎞⎛⎞x −ξ 11yxy−−−ηξη⎪⎪⎧⎫ ⎛⎞⎛⎞ rz01 ≈+⎢⎥1 ⎜⎟⎜⎟++⎨⎬ ⎜⎟⎜⎟ + ... ⎢⎥2 ⎝⎠⎝⎠z 28zzz⎩⎭⎪⎪ ⎝⎠⎝⎠ ⎢⎥ maximum phase change due to dropping ⎢⎥2 ⎣⎦b /8 term must be much less than one radian 2 φ φ ⎧⎫2 2 jkr 2 21πξ⎪⎪⎛⎞x − ⎛⎞y −η eO01 →Δ() = ⎨⎬⎜⎟+<<⎜⎟ 1 λ 8 ⎝⎠z ⎝⎠z ⎩⎭⎪⎪Max 2 3 π 22 zxy>>{}()() − + − 4 Max Spring 2010λ ξη Eradat Physics Dept. SJSU 8 4.2.2 Accuracy of the Fresnel approximation II

Example: calculate the safe distance to use the Fresnel approximation for a circular aperture of size 1cm and a circular observation ragion of 10.525cm withπ a light of λμ=>> m . (Answere: z cm ) 2 3 22 zx>>{()() −ξη+−yxy} ξη hint: − and − should have their 4λ Max maximum possible values to evaluate the condition. If the higher order terms do not change the value of the Fresnel integral substantially, we can use the approximation. They do not need to be small in this case.

Spring 2010 Eradat Physics Dept. SJSU 9 4.2.3 The Fresnel approximation and the angular spectrum I

Goal: understand the implications of the Fresnel approximations from the point of view of angular spectrum method of analysis. We compare the transfer function of propagation through free space, predicted by RS scalar diffraction theory, with the transfer function predicted by the Fresnel analysis

General spatial phase dispersion ⎧representing propagation ⎪ ⎪ ⎛⎞2π z 2 2 Hff(,)= ⎜⎟jff1()−−()XY ← RS theory RS X Y ⎨ ef⎝⎠λ λλ 22+<1f / λ ⎪ X Y ⎩⎪0 otherwise A constant Qadratic phase phase delay dispersion due to traveling jkz ⎧⎫k ⎡⎤22 ⎨⎬jx()()−+−ξη y jzf22 f e ⎩⎭2z ⎣⎦FT jkz −+πλ ()XY hx(,yyff )=⎯⎯→=eHFXY (ff , ) e e jzλ All plane waves Different plane-wave suffer equally components suffer Fresnel approximation impulse response different phase delays

Spring 2010 Eradat Physics Dept. SJSU 10 4.2.3 The Fresnel approximation and the angular spectrum II

⎛⎞2π z 2 2 ⎧ ⎜⎟jff1()−−()XY ⎪ ⎝⎠λ λλ 22 eff XY+<1/ λ HffRS(,) X Y =←⎨ RS theory ⎩⎪0 otherwise 22 jkz −+jzfπλ ()XY f HffFXY(,)= ee

We can see that HffFXY(,) is an approximation to the H RSXY (,) ff λλ λ λ Applying the binomial expansion to the H RS(,)ff X Y we get:

π 22 2222 ()λλffXY() 1()1−−≈−−λ()fλλ f if () f 1 and f () 1 XY22 X Y π ⎛⎞⎡⎤22 2 z ()λλffXY() ⎛⎞2 z 2 2 ⎜⎟j λ ⎢⎥1−− ⎜⎟jff1()−−()XY ⎜⎟⎢⎥22 jzf22 f ⎝⎠⎝⎠⎣⎦jkz −+()XY ee≈== eeHffFXY(,) Conclusion: πλ 2 2 2 ⎛⎞kx HffRS(,) X Y≈== Hff F (,) X Y When the conditions: ()λα f X ⎜⎟ 1, ⎝⎠||k 2 2 2 ⎛⎞kY ()λβfY ==⎜⎟1 are satisfied. So Fresnel approxilation is equivalent ⎝⎠||k to the paraxial approximation that is limitted to small propagation angles. Spring 2010 Eradat Physics Dept. SJSU 11 4.2.4 Fresnel diffraction between confocal spherical surfaces I Goal: analysis of diffraction between two confocal spherical surfaces Confocal spheres: center of each lies on the surface of the other. We set the spheres tangant to the plance we used before.

Located at zzzr==0. and 01 is the distance between two spherical caps. We write the equationsξη for both surfaces and find the distance between them. Make paraxial approximation by using the binomial expansion. Assuming the extend of the spherical capsξη about the z axis is small compared to the radii of the spheres, i.e. for z>> x − ξ and y −η , we get xy η rzx=+−+−→≈222()() y rz -- y 01 01 zz ξ x zejkr01 Uxy(),(,)= λ Uξη dξηd , ∫∫Σ 2 j r01 r01

jkz ∞ ⎧⎫2π e λ ⎨−+jxy[ ]⎬ Uxy( ,(,)) = Uξ ηξη e⎩⎭λz ξη dd ∫∫ jz −∞ Field on the right hand Fourier transform of the Field on the z spherical cap left hand spherical cap Spring 2010 Eradat Physics Dept. SJSU 12 η 4.2.4 Fresnel diffraction between y ξ confocal spherical surfaces II x r Comparing the two form: 01

jkz ∞ ⎧⎫2π e ⎨⎬−+jxy[] Uxy(),(,)= U ξξη η e⎩⎭λz ξη dd ∫∫ jzλ −∞ Field on the right hand Fourier transform of the Field on the spherical cap left hand spherical cap z Compared with the Fresnel diffractionintegral:

k 22 j (ξη+ ) Fourier transform of the Ue(,)ξ η 2 z which is complex field just λ to the right of aperture multiplied by a quadratic phase factor

jkz kk22∞ 22 2 e jxy()++⎪⎪⎧⎫ j()−+jxy() Uxy(),(,)= e22zz⎨⎬ ξ U ηξη eξη ez dd ∫∫ π jz −∞ ⎩⎭⎪⎪ λ ξη Seond form of the Fresnel diffraction integral We see that by replacing the two plates with spherical caps, the quadratic

kk22 22 jxy()++ j()ξη factor in (xy , ), e22zz , and (ξη , ), e , have been elminated. In fact these two phase factors are paraxial representations of spherical phase surfaces. By having a spherical observation plane, they are gone. On derivation of the Fresnel diffraction integral we approximated the spherical waves with plane waves. Now getting back to spherical surfaces, there is no approximation. Sherical surface will see the spherical wave like a flat surface sees the plane wave. Spring 2010 Eradat Physics Dept. SJSU 13 4.3 The Fraunhofer approximation I Goal: applying another more stringent approximation to the Fresnel diffraction integral to simplify the calculations for valid cases.

k 22 j (ξη+ ) Fourier transform of the quadratic phase function, Ue(,)ξη 2 z , which is the k j 22+ λ 2 z( ) apertditibtiture distribution U ()(ξ ,η ), mu ltilidbltiplieξηd by a quadtihdratic phase fac tor e ξη jkz kk22∞ 22 2π e jxy()++⎧⎫ j()−+jxy() Uxy(),(,)= e22zz⎨⎬ ξ U ηξη e eλz ξη dd jz ∫∫ −∞ ⎩⎭ Seond form of the Fresnel diffraction integral k ()ξη22+ Appl yin g the Fraunhofer a pproximation: z max the quadratic 2

k 22 j ()ξη+ 2z η y phase factorλ e ≈ 1 jkz k 22 ∞ 2λπ x e j ()x + y − j (xξ + yη ) ξ Uxy(),(,)= e2z Uξη ξ η ez dd ∫∫ P0 jz −∞ Σ z Fourier transform of the aperture distribution xy θ evaluated at ff= and = XYλλzz Spring 2010 Eradat Physics Dept. SJSU

P1 14 4.3 The Fraunhofer approximation II 22 k (ξ +η ) z Fraunhofer approximation:π z max or aperture size, λ 2 λ 22 Fresnel approximation: zx ()()( -ξ )(+ y-η) π 2 Since k = is a large number Fraunhofer approximation is much

stringent than Fresnel approximation At optical frequencies: λμ==0.6m ; aperture width 2.5 cm ; z 1600 m A less stringent condition is called the "antenna designer formula": for an aperture with linear dimension ofD , the Fraunhofer approximation 2D2 will be valid if zz>> now 2000 meters (>> is replaced with >) λ The Frauhofer diffraction pattern will form at very far distances but we can bring the pattern by using a proper lens or proper ilumination. Will see in the problems.

Spring 2010 Eradat Physics Dept. SJSU 15 Bessel functions I The Bessel functions or cylinder functions or cylinderical harmonics of the first kind , Jn ()x , are defined as the solu tions to the dy2 dy Bessel differential equation: xxxny222++−=()0 dx2 dx These functions are nonsingular at the origin. ⎧ ∞ (-1)l 1 xm2||lm+ || ⎪∑ 2||lm+ ≠ ⎪ l=0 2!(||)!lml+ 2 ⎪ 21 Jxm ()⎨ sin x m= ⎪ π x 2 ⎪ 21 ⎪ cosx m = - ⎩⎪ π x 2 m Jx−mm()=−(1) Jxm () = 0,1,2,3,... d A derivative identity: ⎡⎤xJm () x = xJm () x dx ⎣⎦m m−1 u An integral identity: uJ'(')' u du= uJu () ∫0 01 ∞ Bessel function addition theorem: Jyz()+= JyJ ()() z nmnm∑ m=−∞ − ∞∞ Jx()==+ 1; eizcosθ Jz () 2 jJz n ()cos() nθ Spring 2010 ∑∑k =∞- knEradat0 Physics Dept. −∞ SJSU

There are more of these identities. Check you favorite math handbook. 16 Bessel functions of the first kind (MATLAB)

u = (0:0.1:15) BJ0=besselj(0,u); BJ1=besselj(1,u); BJ2=besselj(2, u); BJ3=besselj(3,u) plot(u,BJ0,u,BJ1,u,BJ2,u,BJ3); Bessel functions of first kind legend('J0','J1','J2','J3') 1 title('Bessel functions of first kind' ); J0 xlabel('u'),ylabel('J') J1 J2 grid on J3

0.5 J

0

-050.5 0 5 10 15 u

Spring 2010 Eradat Physics Dept. SJSU 17 Bessel functions II

Various integrals expressed in terms of the Bessel functions: 1π π Jz()=− cos( zsinnd θθθ ) Bessel's first integral n ∫0

−πn i π Jz()= eizcosθ cos( ndθθ ) n ∫0

2 π 1 izcos in Jzn ()=→= e edθθ with n 0 2πin ∫0 θ k 2 1 2π ∞ ()Z /4 Ja()= eiacos dJzθ or ()=− (1)k 0 ∫0 θ 0 ∑ 2 2π k =0 (k!) n 2 z π /2 Jz( )== sin2n u cos( z cos udu ) for n 1,2,... n π ()21!!n − ∫0 11x z−1 Jx()=> e2 z z−−n 1 dzforn - n 22πi ∫γ ∞ The Bessel functions are normaized: Jxdx() =10 for n= ,,,,1,2,... ∫0 n 22 ∞∞⎡⎤Jx()41 ⎡⎤ Jx () Integrals involving Jx(): 11 dx= and xdx= 1 ∫∫00⎢⎥xx32π ⎢⎥ Spring 2010 Eradat Ph⎣⎦ysics Dept. SJSU ⎣⎦ 18 Fourier transform of a circularly symmetric function I Most apertures and lenses have circular symmetry for example

22 gxy(, )= {1 xya+≤ expresses a circular aperture with radius of a . 0 xya22+> The circular symmetry justifies usage of cylindical coordinates.

xr===+=cosθθ ; yr sin ; r θ x22 y ; tan− 1 ( yx / )

22 -1 ffxy===+=ρφcos ; ρφρ sin ; φ ffff XYYX ; tan ( / )

dxdy== rdrd;; dfXY df d d θρρφ∞ F {}g(, x y )== G ( f , f ) g( x , y) e−+jfxfy2π ( XY) dxdy XY ∫∫−∞ Now apply change of varaibles: 2π ∞ F {}gr(,θρφθθ )== G ( , ) d gr (, ) e−+jr2coscossinsinπρ() φ θ ρ φ θ r rdr 0 ∫∫00 For circularly symmetric functions gr is only function of . So we write:

gr(,θ )= gR () r 2π ∞∞2π G(,)ρφ θ == d gre ()−−jr2cos() rdr grrdre () − j2cos()r − dθ 0 ∫∫00RR ∫ 0 ∫0 πρ θφ Spring 2010 Eradat Physics Dept. SJSU 19 πρ θφ Fourier transform of a circularly symmetric function II ∞ 2π φθGgrrdred(,)ρφ = () −−jr2cos() 0 ∫∫00R πρ θφ this relation is correct for any value of including = 0 , π 2π 1 − jacos( ) Value of the integral edJaθ = 0 () is own known as the 2 ∫0 θ zeroth order Bessel function of the first kind. Withρρπ substituting ar==20πρ φ πρand we φφget:

∞ Fourier-Bessel transform, B, or B()==GrgrJrdr () 2 ()(2 ) ← 00∫0 R Hankel transform of zero order The inverse Fourier-Bessel transform is then: ∞ B −1gr(,θπρρπρρ )== g () r 2 G ( ) J (2 r ) d R ∫0 00 Conclusions: 1) Fourier transform of a circularly symmetric function is a circularly summetric function itself. 2) There is no difference between the direct and inverse transform operations.

Spring 2010 Eradat Physics Dept. SJSU 20 Fourier transform of a circularly symmetric function III Following the Fourier integral theorem. and simmilarity theorem, we get: −−11 BB{}grRRRRR()===← B B gr () {} BB gr () {} gr ()when gr () is continuous. 1 ⎛⎞ρ B{}garR ()= 2 G0 ⎜⎟ aa⎝⎠ B for Fourier-Bessel transform. All other Fouri er transform theorems apply since this isjustaspecials just a special case of the general two-dimensional Fourier transforms.

Spring 2010 Eradat Physics Dept. SJSU 21 Fourier transform of a circular aperture with radius a 22 ⎪⎧1 xya+≤ ⎧1 ra≤ gxy(, )=→=⎨⎨ gR () r 22 0 ra> ⎩⎪0 xya+> ⎩

Substituting grR () in ∞ GG(,)ρφ π ρ ρπ == () 2 rgrJrdr () (2 ) 00∫0 R 0 a GrJrdr()ρπ= 2 πρ (2 ) 00∫0 u Using the the integral identity: uJ' (() u') du' = uJu () ∫0 0'1 rr'2= πρ r=→0'0 r = and r = a r '2 =πρ a 11aa2πρ GrJrdrrJrdr()ρπρπρπρ== 2 (2 )(2 ) ' ()( ') ' 0022πρπρ22∫∫00 0

1 Ja11(2 )2 Ja (2 ) GaJaaa()ραα== 2πρ πρ πρ π (2 )= 2 with k = 2 ρρπ ρ01 ρ 22πρ 2 a α

2 ⎡⎤Jka1()α Gk01()== Fk() 2π a⎢⎥ whereπρ πρ J is a first order Bessl function. ka Spring 2010⎣⎦ Eradatα Physics Dept. SJSU 22 Circular aperture with Bessel functions in MATLAB 22 ⎪⎧1 xya+≤ ⎧1 ra≤ x=(-15.1:0.5:14.9); gxy(, )=→= g () r αα⎨⎨R y=(-15.1:0.5:14.9); ⎪0 xya22+> ⎩0 ra> A=y.'*x; ⎩ i_ index= 0; 2 ⎡⎤Jka1 ()α for i=-15.1:0.5:14.9 Gk0 ()== Fk( ) 2π a⎢⎥ j_index=0; ⎣⎦kaα i_index=i_index+1; for j=-15.1:0.5:14.9 j_index=j_index+1; r=sqrt(i^2+j^2); if r <=5 A(i_index,j_index)=1; else A(i_index,j_index)=0; end end end subplot(2,1,1); mesh(x,y,A); title('Circular Aperture') axis([-15.1 14.9 -15.1 14.9 0 1]); a=1; kx=(-15.1:0.5:14.9); ky=(-15.1:0.5:14.9); [kax,kay]=meshgrid(kx,ky);

ka=sqrt(kax.^2+kay.^2); Gka=2*pi*a^2.*besselj(1,ka)./(ka*a); subplot(2,1,2); mesh(kx,ky,Gka); xlabel('kx'); ylabel('ky'); Spring 2010 axis([-15.1 14.9 -15.1 14.9 -1 4]);Eradat Physics Dept. SJSU title('Fourier Bessel of Circular Aperture') 23 Circular aperture with Bessel functions in MATLAB

Spring 2010 Eradat Physics Dept. SJSU

24 Circular aperture with FFT in MATLAB %PHYS 258 sppgring 07, Na yer Eradat %A program to plot a circular aperture function %and its Fourier transform using fft and shift fft function x=(-2:0.05:2); y=(-20052)2:0.05:2); A=y.'*x; i_index=0; for i=-2:0.05:2 j_index=0; i_ index= i_ index+1; for j=-2:0.05:2 j_index=j_index+1; r=sqrt(i^2+j^2); if r <=0.2 A(i_index,j_index)=1; else A(i_index,j_index)=0; end end end subplot(2,1,1); mesh(x,y, A); %3D p lo t xlabel('x'); ylabel('y'); zlabel('E'); title('Circular aperture'); fft_v=abs(fft2(A)); fft_val=fftshift(fft_v); %shift zero -frequency component to center of spectrum subplot(2,1,2); Spring 2010 mesh(x,y,fft_val); Eradat Physics Dept. SJSU xlabel('fx'); ylabel('fy'); zlabel('E'); title('fft of Circular aperture'); 25 4.4 Examples of Fraunhofer diffraction patterns • We can apply the results of Fraunhofer approximation to calculate the complex field distribution pattern across any given aperture. • The physically observable quantity is the intensity of the radiation rather than the field strength. • In the following examples we will calculate the intensity distributions across the apertures.

Spring 2010 Eradat Physics Dept. SJSU 26 Screen amplitude transmittance function

complex field amplitude imediately behind the screen Screen amplitude transmittance function= incident complex field amplitude Screen amplitude transmittance for an infinite opaque screen: ⎧1 in the aperture tA ()(ξη, ) = ⎨ ⎩0 outside the aperture It is possible to introduce for example a) Phase mask: spatial patterns of phase shift by means of transparent plates of various thickness b) Amplitude mask: spatial attenuation by placing an absorbing photographic

transparency with real values between 01≤≤tA

These two techniques extend all realizable values of tA over the complex planes within the unit circle.

Spring 2010 Eradat Physics Dept. SJSU 27 4.4.1 Rectangular aperture I Goal: calculate the intensity of the Fraunhofer diffraction pattern at a distance z from a rectangular aperture located on an infinite opaque screen. Appperture amplitude transmittance: ξη ⎛⎞⎛⎞ξ η tA (,)= rect ⎜⎟⎜⎟rect ⎝⎠⎝⎠2wX 2wY

where wwXY and are the half widths of the aperture in ξ and η directions. Illumination: a unit-amplitude, normally incident, monochromatic plane wave: For such an illumination the field distribution jjpust across the aperture is the

transmittance function tA , and the Fraunhofer diffraction pattern is:

jkz k ∞ 2π e jxy()22+ −+jxy() UUy(x,(,)y)= e 2z Uξη ξ ηeddλz ξη jλz ∫∫ −∞ Fourier transform of the aperture distribution xy evaluated at ff== and λ XYλλzz jkz k e jxy()22+ Uxy(),,= e2z F {} U()ξη jz fxzfyzXY==/,λλ / Spring 2010 Eradat Physics Dept. SJSU 28 4.4.1 Rectangular aperture II

jkz k e jxy()22+ Uxy(),(,)= e2z F {} tξη A fxzfyz==/,λλ / ξη jλz XY ⎛⎞⎛⎞ξη tA (,)= rect⎜⎟⎜⎟ rect ⎝⎠⎝⎠22wwXY

F {}twcwfwcwfAwwA (,)ξη == 2XXXYYYXY sin2 () 2 sin2 with () 4 jkz k jjy()x22+ y e 2z Uxy( ,s)= e Aiiin2 c( wXXfff ) sin2cw( YYf ) jzλ fxzfyzXY==/,λλ / jkz k jxy()22+ e ⎛⎞⎛⎞22wxXY wy U (xy, )= e 2z Asiiinscc⎜⎟⎜⎟in jzλλλ⎝⎠⎝⎠ z z 2 22A ⎛⎞⎛⎞22wxXY 2 wy Ixy()|()|( , )|== Uxy (, )|22 sin c⎜⎟⎜⎟ sin c λλzz⎝⎠⎝⎠ λ z

Spring 2010 Eradat Physics Dept. SJSU 29 4.4.1 Rectangular aperture III

2 22A ⎛⎞⎛⎞22wxXY 2 wy Ixy(, )== | Uxy (, )|22 sin c⎜⎟⎜⎟ sin c λλzz⎝⎠⎝⎠ λ z

Exercise: prove that width of the maine lobe or distance between the λz first t wo zeros i s Δx = . wX Solution: λλwe need to find roots of the I, when y=0, we have λ 2 ⎛⎞2wyY sinc ⎜⎟= 1 so weπ need to require ⎝⎠λz ππ

⎛⎞2wxX sin ⎜⎟π ⎛⎞22wxz wx mzλ sincmx2 XX=→ 0⎝⎠ =→ 0 = →= ⎜⎟ 2wx ⎝⎠zzwX 2 X λz λλλzzz with mx=±1 we get +− = and x =− →Δ x = 22wwwXXX Spring 2010 Eradat Physics Dept. SJSU 30 Spring 2010 Eradat Physics Dept. SJSU 31 4.4.2 Circular aperture I

Goal: calculate the intensity of the Fraunhofer diffraction pattern at a distance zq from a circular aperture of radius located on an infinite opaque screen. Aperture amplitude transmittance: ⎛⎞q tqA ()= circ⎜⎟ ⎝⎠w Circular symmetry suggests using the cylinderical coordinates and the Fourier-Besselλ transformation. The Fraunhofer diffraction pattern is:

jkz k 22 ∞ 2λπ e j ()xy+ − j (xyξ + η ) Uxy(),(,)= e2z Uξη ξ η ez dd jz ∫ ∫ −∞ Fourier-Bessel transform of the aperture distribution xyy evaluated at ff= and = XYλλzz jkz k jr2 e 2z 22 Uxy(), ==+ eB{} Uq() ρλ where r x y is the radius in the jzλ =rz/

22 aperture plane and ρ =+ffXY is the radius in the spatial frequency plane.

Spring 2010 Eradat Physics Dept. SJSU 32 4.4.2 Circular aperture II Illumination: a unit-amplitude, normally incident, monochromatic plane wave: For such an illumination the field distribution just across the aperture is the

transmittance function tA ,

jkz k jkz k jr2 jr2 e 2z eq2z ⎧⎫⎛⎞ Ur()= e B{}tqA () = eB ⎨⎬ circ⎜⎟ jzλ ρλ=rz/ jzλ w ⎩⎭⎝⎠ρ =rz/ λ

⎧⎫⎛⎞qrkwrJw1(2πρ ) 2 B ⎨⎬circ⎜⎟==== A where A w.;2 With w ⎩⎭⎝⎠ww λπρ z z k jr2 A jkz 2z ⎡⎤Jk1(/)kwr z πρπρ Ur()= e e ⎢⎥2 jzλ ⎣⎦ kwrz/ 2 2 ⎛⎞A ⎡⎤J1(()kwr/ z) I (r)=←⎜⎟⎢⎥2 The Airy pa ttern. ⎝⎠λzkwrz⎣⎦/ λz Width of the central lobe measured along the xy and axis: d=1.22 w

Spring 2010 Eradat Physics Dept. SJSU 33 4.4.2 Circular aperture III EiPExercise: Prove thtidthfthtthat width of the centralll lo be measure d a long the xy an d ax is λz on the Airy pattern is: d = 1.22 w 2 2 ⎛⎞A ⎡⎤Jkwrz1 (/) we start from the Airy pattern: Ir()==⎜⎟20 for the roots*. λzkwrz⎢⎥/ λ ⎝⎠⎣⎦ J (/)kwr z kwr2πλ wr 3.8317 z 1 =≠0 f0for r 0 so ==→= 3.8317 r kwr/3.14 z zλ z w z r = 1.2203 w Using the table we an calculate the other zeros

*First few roots of the Bessel functions for the first kind using BesselZeros[n, k} in Mathematica

zero

1 2.4048 3.8317 5.1356 6.3802 7.5883 8.7715

2 5.5201 7.0156 8.4172 9.7610 11.0647 12.3386

3 8.6537 10.1735 11.6198 13.0152 14.3725 15.7002

4 11.7915 13.3237 14.7960 16.2235 17.6160 18.9801 Spring5 201014.9309 16.4706 17.9598 19.4094 Eradat20.8269 Ph22.2178ysics Dept. SJSU 34 The grating equation CditifthCondition for the cons truc tiitftive interference flihtifor a light passing through a transmission grating:

nnm2211Λ−Λ=sinθθλ sin The gratingθθλ equation λ nnmsin=+ sin 2211 Λ A "positive" diffraction order (m>0) ←→θθ > 2 1 Grating

A "nagative" diffraction order (m<0) ←→θθ2 < 1 θθ and are signed angles poitive counterclocckwise 2 1 θ1

θθ2 > 1 corresponds to the zeroth order

For a reflection grating bothe incident and reflected θ2 rays are on the same

side so nnn12==

n1 n2

Spring 2010 Eradat Physics Dept. SJSU 35 4.4.3 Thin sinusoidal amplitude grating I

Goal: calculate the intensity of the Fraunhofer diffraction pattern at a ηπξη ξ distance z from a thin sinusoidal amplitude grating. The amplitude trnsmittance function is: ⎡⎤1 m ⎛⎞ξ ⎛⎞η tA (,)=+⎢⎥ cos(2 f0 ) rect ⎜⎟rect ⎜⎟ ⎣⎦22 ⎝⎠22ww ⎝⎠ We have assumed that the grating structure is bounded by a square aperture of width 2.w tA m is the pppeak-to-peak chan ge of am plitude 1 transmittance across the screen.

f0 is the sptial freuency of the grating. thin means the structure can be modeled by a simple 0.5 m amplitude transmittance (no effect on the phase). Illumination: a unit-amplitude plane wave

tA: the fi eld di s tr ibu tion across ththe aperture. x Figure: cross section of the grating amplitude transmittance function.

Spring 2010 Eradat Physics Dept. SJSU 36 4.4.3 Thin sinusoidal amplitude grating II

The Fraunhofer diffraction pattern is the Fourier transfor of tA :

jkzkk22 jkz 22 jxy()++ jxy() ee22zz Uxy(),(,)(,)== eFF ξη{} Uξη e{} tA jzλλffXY,, jz ff XY but first: ⎧⎫11mmm F ⎨⎬+cos(2πξfffffffff000 δ ) δ δ =()XY , +++−() X , Y() X , Y ⎩⎭22 2 4 4

⎧⎫⎛⎞ξη ⎛⎞ 2 F ⎨⎬rect⎜⎟ rect ⎜⎟== Asnic()()22 wfXY snic wf where A 4 w is the ⎝⎠22ww ⎝⎠ ⎩⎭λλλλ λ area of aperture bounding the grating. Amm⎧⎫ F {}tAYXXX(,)ξη =+++− snic() 2 wf⎨⎬ snic 2 wf ( snic ) (2( w f f00 ) snic (2( w f f ) 222⎩⎭

With fxzXY==// and fyz

λλk 22 jxy()+ Awywxmwmwjkz 2z ⎛⎞22⎧ ⎛⎞ 2 2 ⎫ Ux(),y=+++− e e snic⎜⎟⎨ snic ⎜⎟ sin c ( ( fXX fλλ z ) sin c ( ( f f0 z )⎬ jz⎝⎠ z⎩⎭ ⎝⎠ z22 z z

Spring 2010 Eradat Physics Dept. SJSU 37 4.4.3 Thin sinusoidal amplitude grating III AdfillAnd finally 2 k 22 ⎡⎤Awy⎡⎤jxy()+ ⎛⎞2 Ixy(), = ejkz e2z snic2 ⎢⎥⎢⎥λλ⎜⎟ λ ⎣⎦jzλλ⎣⎦⎝⎠ z 2 ⎧⎫⎛⎞22wxmw mw 2 ⎨⎬snic⎜⎟+++−sin cffzcffz ( (XX00λλ ) sin ( ( ) ⎩⎭⎝⎠zz22 z 2 2 ⎡⎤Awywxmw2 ⎛⎞22⎧⎫ ⎛⎞ 2 mw2 I() x,sin( y=+⎢⎥ snic⎜⎟⎨⎬ snic ⎜⎟ c ()sin(()ffzXX++00λλ c ffz − ⎣⎦jzλλλ⎝⎠ z⎩⎭ ⎝⎠ z 2 λzz2 λ

For fw0 1/ or for a very fine grating rulling the overlap between the sinc functions is negligible and I is approximately eual to the sum of squared amplitudes.

2 22 ⎡⎤Awywxmwmw22⎛⎞22⎧⎫ ⎛⎞ 2 2 2 2 I (x, y) ≈+++−⎢⎥snic⎜⎟⎨⎬ snic ⎜⎟ si((in c ((ffXX00λλzc ) si i((n ((ff z ) ⎣⎦jzλλ⎝⎠ z⎩⎭ ⎝⎠ λ z44 λ λ z z η ==Diffraction efficiency fraction of the power in a single order of the Fraunhofer diff. patteren. ⎧⎫11mmm It can be found from: F ⎨⎬+cos(2πξfffffffff000 δ ) δ δ =()XY , +++−() X , Y() X , Y ⎩⎭22 2 4 4 Since the delta functions determine power of the pattern and sinc functions only spread them. 22 2 ⎛⎞1 ⎛m ⎞ mmm22⎛⎞ 1 ηη01==⎜⎟0.25, = ⎜ ⎟===, ηη−1 ⎜⎟ so 1 max == 6.25% and total power ⎝⎠24 ⎝ ⎠ 16⎝⎠ 4 16 16 in 3 orders is 3/8. The rest of the power is lost by absorption of the grating. Spring 2010 Eradat Physics Dept. SJSU 38 4.4.4 Thin sinusoidal phase grating I Goal: calculate the intensity of the Fraunhofer diffraction pattern at a distance z from a thin sinusoidal phase grating. The amplitude trnsmittance function is:

⎡⎤m ⎢⎥jfsin(2πξ0 ) ⎣⎦2 ⎛⎞ξη ⎛⎞ teA (,)ξη = rect⎜⎟ rect ⎜⎟ Sinusoidal phase difference ⎝⎠22ww ⎝⎠ introduced by the grating average phase delay caused by grating is elliminated by proper choice of teference. We have assumed that the grating structure is bounded by a square aperture of width 2.w m is the peak-to-peak excursion of the phase delay.

f0 is the sptial freuency of the grating. thin means the structure can be modeled by a simple phase transmittance (no effect on amplitude). Illum ina tion: a un it-amp litu de p lane wave

tA: the field distribution across the aperture.

Spring 2010 Eradat Physics Dept. SJSU 39 ξη 4.4.4 Thin sinusoidal phase grating II ⎡⎤m ⎢⎥j sin(2π f0ξ ) ⎣⎦2 ⎛⎞ξ ⎛⎞η tA (,)= e rect⎜⎟ rect ⎜⎟ λλ⎝⎠22ww ⎝⎠

jkzkk22 jkz 22 jxy( ++) jxy( ) ee22z z Uxy( ,(,)(,))== eFF{ Uξ ηξη} e{ tA } jzf XY,,fff jz XY

⎡⎤m ∞ ⎢⎥jfsin(2πξ0 ) ⎣⎦2 ⎛⎞m jqf2πξ0ξ Using the Bessel function identity: eJe= ∑ q ⎜⎟ q=−∞ ⎝⎠2 ⎧⎫∞ ⎛⎞m jqf2πξ0 ⎧⎫ ⎛⎞ξη ⎛⎞ FF{tJerectrectA (,)ξη}= ⎨⎬⎨⎬∑ q ⎜⎟⊗ F ⎜⎟ ⎜⎟ ⎩⎭q=−∞ ⎝⎠222⎩⎭ ⎝⎠ww ⎝⎠ ⎡⎤∞ ⎛⎞m F {tJfqffAcwfcwfAqXYXY δ(ξη , )}=−⊗⎢⎥∑ ⎜⎟ (0 , )[ sin (2 )sin (2 )] ⎣⎦q=−∞ ⎝⎠2 ∞ ⎛⎞m F {}tAJcwfqfcwfAq(ξη , )=−∑ ⎜⎟ sin⎣⎦⎡⎤ 2()() X0 sin 2 Y q=−∞ ⎝⎠2

k 22 ∞ jxy()+ A jkz 2z ⎛⎞mw ⎡22λλ ⎤ ⎛ wy ⎞ Uxy(), = e eJcxqfzc∑ q ⎜⎟sin⎢⎥()− 0λ sin ⎜ ⎟ Spring 2010jzλ Eradatq=−∞ Physics⎝⎠2 Dept. SJSU ⎣zz ⎦ ⎝ ⎠ 40 4.4.4 Thin sinusoidal phase grating III ⎡⎤m ⎢⎥jfsin(2πξ0 ) ⎣⎦2 ⎛⎞ξη ⎛⎞ tA (,)ξη = e rect⎜⎟ rect ⎜⎟ 22ww λλλ⎝⎠ ⎝⎠ k 22 ∞ jxy()+ Amwwyjkz 2z ⎛⎞ ⎡22 ⎤ ⎛ ⎞ Uxy(),sinsin=− e e∑ Jq ⎜⎟ c⎢⎥() x qfz0 c ⎜ ⎟ jzλλλq=−∞ ⎝⎠2 ⎣ z ⎦ ⎝ z ⎠ 2 k 22 ∞ ⎛⎞jxy()+ Amwwyjkz 2z ⎛⎞ ⎡22 ⎤ ⎛ ⎞ IeeJ=−⎜⎟∑ q ⎜⎟sin cxqfzc⎢⎥()0λ sin ⎜ ⎟ ⎝⎠jzq=−∞ ⎝⎠2 ⎣ z ⎦ ⎝ z ⎠

For fw0 1/ or for a very fine grating rulling the overlap between the sinc functions is negligible and I is approximately eual to theλ sum of squared amplitudes. ⎡⎤⎛⎞ 2 ∞ ⎢⎥⎜⎟ ⎛⎞Amw22 ⎛⎞ ⎢⎥2 ⎜⎟2 ⎛⎞2wy IJcx≈ ⎜⎟q ⎜⎟sin − qf0λ zsin c ⎜⎟ λλzz∑ 2 ⎢⎥⎜⎟ z ⎝⎠q=−∞ ⎝⎠ Displacement ⎝⎠ ⎢⎥⎜⎟of the order ⎢⎥fromthe center ⎣⎦⎝⎠λ We see that introduction of the sinusoidal phase grating has deflected power from the zeroth order to the higher ordes. 2 ⎡⎤Am⎛⎞ Peak intensity of the qJ th order= ⎢⎥q ⎜⎟ ⎣⎦λz ⎝⎠2

It happens when y=−=→=00 and x qf00λλ z x qf z Spring 2010 Eradat Physics Dept. SJSU 41 4.4.4 Thin sinusoidal phase grating IIII

Displacement of qqfz th order from the center of the pattern = 0λ For qzeroy===000 or th order and x

For qyxfz=±10 or first order = and =± 0λ function of frequency of the grating lining, wavelength, and distance of observation.

So for spectroscopy in the blue region we need hight f0 grating or larger spectrometer. η ==Diffraction efficiency fraction of the power in a single order of the Fraunhofer diff. patteren. ⎡⎤∞ ⎛⎞m It can be found from: F {}tJfqffAcwfcwfAqXYXY δ(ξη , )=−⊗⎢⎥∑ ⎜⎟ (0 , )[ sin (2 )sin (2 )] ⎣⎦q=−∞ ⎝⎠2 Since the delta functions determine power of the pattern and sinc functions only spread them.

2 ⎛ m ⎞ η0 = Jq ⎜⎟ ⎝⎠2

2 ⎛⎞m Plot η0 = Jq ⎜⎟ we see that when m/2 is root of J01 then the central lobe wanishes. max= 33.8% ⎝⎠2 1 which is much greater than that of the sinusoidal amplitude grating which is = 6.25. 16 There is no power absorption and sum of the power in all orders is equal to the total incident power. The distribution of power between the lobes varies as m changes.

Spring 2010 Eradat Physics Dept. SJSU η 42 4.5 Examples of Fresnel diffraction calculations • Based on the example we will choose a different approach to the Fresnel diffraction examples. – convolution representation. – Frequency domain approach

Spring 2010 Eradat Physics Dept. SJSU 43 4.5.1 Fresnel diffraction by a square aperture I

Goal: calculate the intensity of the Fresnel diffraction pattern at a distance zw from a sqqpuare aperture of width 2 located on an infinite oppqaque screen. The amplitude transmittance function: ⎛⎞ξη ⎛⎞ trA (,)ξη = ect⎜⎟ rect ⎜⎟ ⎝⎠22ww ⎝⎠ The ξηcomplex field imediately behind the aperture: ⎛⎞ξη ⎛⎞ U(,) = rect⎜⎟ rect ⎜⎟ z=0 ⎝⎠22ww ⎝⎠ Illumination: a unit-amplitude, normally incident, monochromatic plane wave Using the coλnvolution form of the Fresnel diffraction formula:

jkz w π ⎡⎤22 e jx⎢⎥( −−−) ( y) UddU (x, y) = ∫∫e λz ⎣⎦ξηddξ η jz −w e jkz Uxy(),()()= II x y where j

λλwwππ⎡⎤2 ⎡⎤2 11jx⎢⎥()−− jy ⎢⎥() II()x==∫∫ eλλzz ξη⎣⎦ξη ddξη and () y e ⎣⎦ dd zz−−ww Spring 2010 Eradat Physics Dept. SJSU 44 4.5.1 Fresnel diffraction by a square aperture II Change of variables: 22 αξ=−()xy βη and =−() λzzλ αβ 22jj22 11ππ22 I()xed==∫∫αβαβ and I () yed 22−− 11αβ 22 αα=+()wx and =−() wx 12zz λλ 22 ββ=+(wy) and =−( wy) 12λλzz 2 With the Fresnel number: NwzF = / λ and normalized distance variables in the observation region we have: xy XY== and the limits of the integrals become: λλzz

αα12= 22( NXFF+ ) and =−( NX)

β1 =+2 ()NYF and β2 =−2 ()NYF Spring 2010 Eradat Physics Dept. SJSU 45 4.5.1 Fresnel diffraction by a square aperture III

ααπππ αα αα jjj222 221222 Using αedααα=− ed ed and the Fresnel integrals: ∫∫∫00 1 22 zz⎛⎞ππtt ⎛⎞ C() z==cos dt and S() z sin dt we write ∫∫00⎜⎟ ⎜⎟ ⎝⎠22 ⎝⎠ 1 I()xCCjSS=−+−{⎡⎤⎡⎤(α21) (ααα) ( 21) ( ) } and 2 ⎣⎦⎣⎦ 1 I ()y =−{⎡⎤CC(ββ) ( ) + j ⎡SS(ββ)− ( )⎤} 2 ⎣ 21⎦⎣21 ⎦ e jkz Uxy(), =−+−{}⎡⎤⎡⎤ C()αα C () jS αα S () () 2 j ⎣⎦⎣⎦21 21

×−+−{}⎣⎦⎣⎦⎡⎤⎡⎤CC()ββ21 () jSS ββ 21 () () 1 22 Ixy( , )==+{⎡⎤⎡⎤ C(αα21)− C( ) + αα S( 21)− S( ) } 4 ⎣⎦⎣⎦ 22 ×−+−{}⎣⎦⎣⎦⎡⎤⎡⎤CC()ββ21 () ββ SS 21 () () Spring 2010 Eradat Physics Dept. SJSU 46 Fresnel integrals Fresnel integrals are defined as: π zjt2 Cz( )+= iSz( ) e2 dt ∫0 22 zz⎛⎞ππtt ⎛⎞ C( z)==cos dt and S( z) sin dt we write ∫∫00⎜⎟ ⎜⎟ ⎝⎠22 ⎝⎠

Spring 2010 Eradat Physics Dept. SJSU

47 The Fresnel integrals S(u) and C(u) are entire functions or integral functions i.e. they are analytical at all finite points of the complex plane.

The Fresnel integrals are tabulated and are available in many mathematical computer programs

Spring 2010 Eradat Physics Dept. SJSU

48 4.5.1 Fresnel diffraction by a square aperture III Fresnel integrals:

22 zz⎛⎞ππtt ⎛⎞ CCdtSdt(z)==cosdt ; S (z) sin dt ∫∫00⎜⎟ ⎜⎟ ⎝⎠22 ⎝⎠ 1 22 22 Ixy(), =−+−×−+−{}⎡⎤⎡⎤⎡⎤⎡⎤ C()αα21 C () αα S 21 S() ββ{} () C() 21 Cββ () S 21 S () () 4 ⎣⎦⎣⎦⎣⎦⎣⎦ 2 NwzF = /,λ for a fixed w and λ as z increases, the Fresnel number decreases and the true physical distance distance represented on the xX= λ z and y = Yzλ axis are increased. Figure shows the normalized intensity distribution along the xy axis (0)= for various normalized distances from the aperture as represented by fresnel number.

As zN→→∞→∞0()0,, αβ and F becomes very large and Uxy( , ) approaches the product of a delta function and e jkz and shape of the doffraction pattern approachs the shape of the aperture. Limit of the process is the geometrical optics prediction of the complex field:

jkz jkz ⎛⎞xy ⎛⎞ Spring 2010 U(, x y )== e U (, x y ,0)Eradat e rect Physics⎜⎟ Dept. rect SJSU ⎜⎟ ⎝⎠22ww ⎝⎠ 49 Fresnel diffraction patterns at different distances from a square aperture.

As NF->0 diffraction pattern becomes wide and smooth approaching Fraunhofer diffraction

As NF approaches infinity diffraction pattern becomes sharp and narrow Spring 2010 approaching the geometricalEradat Physics Dept. SJSU shadow of the aperture 50 4.5 2 Fresnel diffraction by a thin sinusoidal amplitude grating-Talbot images I Goal: calculated the Intensity distrubution of the diffraction by a thin sinusoidal amplitude grating using the Fresnel diffraction formulation. For si mp lic ity we neg lec t the fin ite ex tent of th e grati ng. The field transmotted by the grating has a periodic nature or we limit the attention to the central region of the pattern. The amplitude transmittance function η y is modeled as: ξ x 1 tmL(,)ξ ηπξ=+[] 1 cos(2 / A 2 Where L is the period of the lines paralle l to the ax is η. Illumination: a unit amplitude normally

incident plane wave. z The field immediately behind Grating structure the grating is tA . Spring 2010 Eradat Physics Dept. SJSU 51 4.5 2 Fresnel diffraction by a thin sinusoidal amplitude grating-Talbot images I We will use the convolution form of the Fresnel diffraction equation

⎧⎫k jkz ∞ jx⎡⎤()()22 y e ⎨⎬⎣⎦−+−ξη Uxy( ,(,)) = ∫∫ U ξξη η e⎩⎭2z dd jzλ −∞ or the Fourier transform of the equation

k 22 j (ξη+ ) Fourier transform of the Ue()(ξη, ) 2 z which is complex field λ just to the right of aperture multipliedξη by a quadratic phase factor

jkz kk22 ∞ 22 2π e jxy()+ ⎪⎪⎧⎫ j()+ −+jxy() Uxy(),(,)= e22zz⎨⎬ Uξη e eddλz ξηξη jz ∫∫⎪ ⎪ −∞∞ ⎩⎭ Seond form of the Fresnel diffraction integral λ xy--ξη Where r >>,1,1, < < or observation is in the 01 zz near field of the aperture or Fresnel diffraction region and scalar theory approximation are assumed OthtfftiOr we can use the transfer function approach:

22 jkz −+jzfπλ ()XY f HffFXY(,)= ee Spring 2010 Eradat Physics Dept. SJSU 52 4.5 2 Fresnel diffraction by a thin sinusoidal amplitude grating-Talbot images II jkz By omitting the constat term eH, F is: 22 −+jzfπλ ()XY f HffFXY(,)= e In any problem that deals with a purely periodic structure, the transfer functio n approach yeilds the simplest calcultions. 1) find the spatial frequency spectrum of the field transmitted by the aperture: 1 tLt ()(ξξξ,ηπ)1= [ 1+ mcos(2 (2/ξ / L] A 2 11mm F {} δtfffffffff(,)ξη δ δ =+++−=() ,() ,() , ; with A24 XYXYXY000 4 L 111mm⎛⎞⎛⎞ F {}tffffffAXYXYXY(,)ξ ηδ=+++−() δ , δ⎜⎟⎜⎟ , , 24⎝⎠⎝⎠LL 40 ⎛⎞1 The transfer function evaluated at ()ffXY,,0=±⎜⎟ becomes ⎝⎠L πλz − j ⎛⎞1 2 He⎜⎟±=,0 L and it is unity at the origin. So after propagation ⎝⎠L of a distance z the Fourier transform of the field becomes: πλλλzzπλ −−jj 1 mmLL22⎛⎞11 ⎛ ⎞ F {}Uxy(), = δ ()ffXY,,+++− eδδ⎜⎟ f X f Y e ⎜ f X , f Y ⎟ 2 44⎝⎠LL ⎝0 ⎠ λπλ πλ zz22xx −−jj 1 mm22jj− Uxy(),,==++FF−1 {} Uxy () eeLLππ ee L L Spring 201024 Eradat Physics 4 Dept. SJSU 53 4.5 2 Fresnel diffraction by a thin sinusoidal amplitude grating-Talbot images III πλ z − j 12⎡⎤2 ⎛⎞π x Uxy(),1=+⎢⎥ me L cos⎜⎟ 2 ⎣⎦⎝⎠L

122⎡⎤⎛⎞⎛⎞πλλzxππ22 ⎛⎞ x Ixy(),=+⎢⎥ 1 2 m cos⎜⎟⎜⎟2 cos + m cos ⎜⎟ 4 ⎣⎦⎝⎠⎝⎠LL ⎝⎠ L Now consider 3 special cases for the observation distance: πλλzn2 L2 1) =→=2nzπ where n =±± 0,1,2 L2 2 1 ⎡⎤⎡⎤⎛⎞λ2212ππxx22 ⎛⎞ π ⎛⎞ x Ixy( , )= ⎢⎥⎢⎥1++ 2mm cos⎜⎟ cos ⎜⎟ =+ 1 m cos ⎜⎟ 4 ⎣⎦⎣⎦⎝⎠LL ⎝⎠4 ⎝⎠ L this is perfect image of the grating. These images that are formed without aid of a lens are called "Talbot images" or "self-images". πλznL(2+ 1) 2 2) =+→=(2nz 1) where n =±± 0, 1, 2 L2 π 2 12⎡⎤⎡⎤⎛⎞xx22 ⎛⎞ 212 ⎛⎞ x Ixy( ,12coscos1cos)=−⎢⎥⎢⎥ m⎜⎟+ m ⎜⎟=− m ⎜⎟ 44⎝⎠L ⎝⎠LL ⎝⎠ ⎣⎦⎣⎦λ This is also image of the ππgrating with a 1800 spatial phase π shift or "contrast reversal". These too are called "Talbot images". Spring 2010 Eradat Physics Dept. SJSU 54 4.5 2 Fresnel diffraction by a thin sinusoidal λπamplitudeπλ grating-Talbot images IIII 1 ()nL− 2 λππλ z 3) =−(2nz 1) →=2 where n =±± 0, 1, 2 then L2 2 λ

⎛⎞zx2 ⎛⎞2 1cos4+ ()π x cos⎜⎟2 == 0 Using cos ⎜⎟ ⎝⎠LL ⎝⎠ 2 22 121⎡⎤22⎛⎞ππxmmx⎡⎤⎛⎞ ⎛⎞ 4 Ixy(),1cos1=+⎢⎥ m ⎜⎟ =⎢⎥⎜⎟ + + cos ⎜⎟ 4422⎣⎦⎝⎠LL⎣⎦⎝⎠ ⎝⎠ This is an image with twice frequency of the original grating and has m2 reduced contrast (nisted of 1 and m we have 1 and and the background 2 is now brighter by + m2 /2 . This is called the "Talbot subimage". For example for m = 0.3 we have m2 /2= 0.045

Spring 2010 Eradat Physics Dept. SJSU 55 Locations of Talbot image planes behind the grating Talbot subimages

A guide to the eye

Talbot Talbot Grating Phase Phase image image reverse reverse TlbtTalbot Talbot image image 2L2/λ 2L2/λ

Spring 2010 Eradat Physics Dept. SJSU 56