Fresnel and Fraunhofer Diffraction

Fresnel and Fraunhofer Diffraction . 4.1.1 The intensity of a wave field 4.1.2 The Huygens-Fresnel principle in rectangular coordinates 4.2 The Fresnel approximation . . . 4.2.2 Accuracy of the Fresnel approximation 4.2.3 The Fresnel approximation and the angular spectrum 4.2.4 Fresnel diffraction between confocal spherical surfaces . 4.4 Examples of Fraunhofer diffraction patterns 4.4.1 Rectangular aperture 4.4.2 Circular aperture . . 4.4.4 Thin sinusoidal phase grating 4.5 Examples of Fresnel diffraction calculations 4.5.1 Fresnel diffraction by a square aperture . -

4.1 Background •

In cha chapt pter er 3 we we deal dealtt with with mos mostt gene genera rall form form of of the the diff diffra ract ctio ion n theory.

•

n c ap er er

we w

ea w

– Intensity of a wave wave field – Huygens-Fresnel principle – Certain approximations to reduce reduce the problem to a simpler simpler mathematical form. These approximations are: • Fresnel • Fraunhofer – We consider the wave propagation phenomenon phenomenon as a system. . – Preparation for the calculations related related to the approximations

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Eradat Ph sics De t. t. SJSU

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4.1.1 The intensity of a wave field Intensity is is the physically physically measurable attribute of an optical wavefield Intensit and ower densit are not the same but ro ortional Intensity of a scalar monochromatic wave at point P I ( P) =| U ( P ) |2

For a narrow-band (not perfectly monochromatic) intensity is given by 2

I ( P) = | u ( P, t ) | and I ( P , t ) = | u (P , t ) |   

An infinite time average

2

  

Instantaneous Intensity

In calculating a diffraction diffraction pattern, we are looking for the intensity of e pa ern.

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4.1.2 The Huygens-Fresnel principle in rectan ular coordinates According to the th e first Rayleigh-Somefeld Rayleigh-Somef eld solution the diffracted field I

0

Huygens-Fresnel principle can be written as: U I P0

,

= =

jkr 01

jλ

Σ

U ( P1 )

z jλ

where r01 =

Σ

z

2

,

r01 e

 

cos θ ds w ere cos θ = cos( n, r01 ) =

jkr 01

r 012

,

+ ( x − ξ ) + ( y − η ) 2

2

η

y x

Two approximations are used in this result:

P0

1) inherent approximation in the scalar theory Σ

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r 01


z

P1

θ

4

4.2 The Fresnel approximation I Goal: to reduce the H-F principle to a simple and usable expression s y approx ma ons or r 01 :

e ac ev e

Binomial Binom ial expansion: ( x + y ) n

= x n + nx n−1 y +

n( n − 1)

2!

x

n −2

y

2

+ ... + y n , n = 1, 2, 3, ...

(1 + b) ±1/ 2 = 1 ± b ∓ b 2 ± ... for − 1 < b ≤ 1 : the higher order terms are negligable 2 8

⎛ x − ξ ⎞ ⎝ z ⎠

01

2

⎛ y −η ⎞ ⎝ z ⎠

⎡

2

⎣⎢

2

1 ⎛ x −ξ ⎞ 2⎝ z ⎠

2

1 ⎛ y −η ⎞ 2⎝ z ⎠

⎤

...

⎦⎥

W here do we cut the series? series? W e will use r 01 in the diffracted diffracted field equation U x, y ) =

jkr 01

jλ

Σ

U (ξ ,η )

d ξ dη

2

r 01

The term e jkr 01 is very sensitive sensitive to the values of r 01 specially specially since it is multiplied

=

y a very arge num er

. n

π

evs

eo

e or er

.

e eep wo

terms for the exponent. For r012 error introduced by dropping all terms but z is small. x, y

=

z jλ

Σ

,η

e

jkr 01

2 r01

The integration limit is let to S ri rin 20 2 010

η

=

e jkz jλ z

∞

,η e

⎧ j ⎨

k

⎡( x −ξ ) 2 + ( y −η ) 2 ⎤ ⎫⎬ ⎣ ⎦

η

−∞

∞ using the usual boundary conditions. Eradat Ph sics De t. t. SJSU

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4.2 The Fresnel approximation II U ( x, y ) =

e

∞

jkz

U (ξ ,η )e ∫ ∫ jλ z

2 2 ⎨ j ⎡⎣ ( x −ξ ) + ( y −η ) ⎤⎦ ⎬ z 2 ⎩ ⎭

d ξ dη this looks like a convolution

−∞

∞

U ( x, y ) =

e

kz

h ( x, y ) = e ∫ ∫ U (ξ ,η )h( x − ξ , y −η )d ξ dη where jλ z

⎧ k ⎩ 2 z

−

2

−

2

⎫ ⎭

−∞  First form of the Fresnel diffraction integral

Another form of the Fresnel diffraction integral is expressed as the following j

k

2 z

(ξ 2 +η 2 )

Fourier transform of the U (ξ ,η ) e which is complex field just to the right of aperture multiplied by a quadratic phase factor

  ∞

2π k j ξ +η ) ⎫ − j ⎧ ( xξ + yη ) ( λ z 2 z 2z U ( x, y ) = e d ξ dη ⎨U (ξ ,η )e ⎬e ∫ ∫ jλ z ⎭ −∞ ⎩   

e jkz

j

k

( x2 + y 2 )

2

2

Observation in the near field of the aperture or Fresnel diffraction region x - ξ

y - η

z

z

and scalar theory approximation are assumed S rin 2010

Eradat Ph sics De t. SJSU

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4.2.1 Positive vs. negative phases Goal: to understand the meaning of the signs of the

y

phase exponentials: in the s herical wave e jkr 01 and its e uivalent in the quadratic approximation e

j

k

2 z

( x2 + y 2 )

z

(for z > 0)

Sign convention: our phaso rs rotate in the clockwise direction (the angle becomes more negative as time goes) and their time dependence is e − j 2πν t We move in space in such a way that we encounter portions of the wavefield that were emitted earlier in time. The phase must become more positive since the

Wavefront

Wavefront

later

earlier

y

phasor had not have time to rotate as far in clockwise. We move in space in such a way that we encounter portions of the wavefield that were emitted later in time.

θ z

The phasor will have advanced in the clockwise direction, therefore the phase must become more negative. S rin 2010


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4.2.2 Accuracy of the Fresnel approximation I parabolic wavefronts in the Huygens-Fresnel principle x, y

=

z jλ

,

Σ

e

jkr 01

e

2

jkz

jλ z

r01



∞

, −∞

⎧ j ⎨

⎫

k

⎡( x −ξ ) 2 + ( y −η ) 2 ⎤ ⎬ ⎦ 2 ⎣

e Parabolic wavelets

Spherical wavelets

order terms in binomial expansion. A sufficient condition for accuracy is:

⎡

⎤

2 2 2 ⎢ 1 ⎛ x − ξ ⎞ 2 1 ⎛ y − η ⎞2 ⎥ 1 ⎧⎪⎛ x − ξ ⎞ ⎛ y − η ⎞ ⎫⎪ r01 ≈ z ⎢1 + ⎜ + ⎜ + +⎜ ⎨⎜ ⎬ ...⎥ ⎟ ⎟ ⎟ ⎟ 2 2 8 ⎝ z ⎠ ⎝ z ⎠ ⎝ z ⎠ ⎭⎪ ⎢ ⎥ ⎩⎪⎝ z ⎠    ⎢ ⎥ b /8 term must be much less than one radian ⎣ ⎦ 2 2 2 ⎧ 2 1 − − π ⎪⎛ x ξ ⎞ ⎛ y η ⎞ ⎫ ⎪ j kr → Δφ (O 2 ) = e ⎨⎜ ⎟ +⎜ ⎟ ⎬ << 1 2

φ



01

Max

z

3

>>

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π

(x −ξ ) { 4λ

2

+ ( y − η )

2 2

}

Max


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

1cm with a light of λ = 0.5μ m. (Answere: z >> 25 cm ) z

3

>>

π

2

( x − ξ ) + ( y − η )

2 2

have their hint: x − ξ and y − η should

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.

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4.2.3 The Fresnel approximation and the an ular s ectrum I Goal: understand the implications of the Fresnel approximations from . 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

 

H RS ( f X , f Y ) = ⎨

⎪ ⎪⎩0

h x

=

e

jkz

jλ z

e

e

⎛ j 2π z 1− λ f 2 − ( λ f ) 2 ⎞ ( X ) Y ⎜ λ ⎟ ⎝ ⎠

f X

2

RS theory + f Y 2 < 1 / λ ←

otherwise

⎧ j k ⎡( x −ξ ) 2 + ( y −η )2 ⎤ ⎫ ⎨ ⎦ ⎬⎭ ⎩ 2 z ⎣

A constant phase delay due to traveling

 

⎯⎯→ H FT

 Fresnel approximation impulse response

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=

e

jkz



Qadratic phase dispersion

  − jπλ z ( f X 2 + f Y 2 )

e

All plane waves Different plane-wave suff er equally components suffer different phase delays

10

4.2.3 The Fresnel approximation and the an ular s ectrum II ⎧ ⎛⎜⎝ j 2π λ z ⎪ H RS ( f X , f Y ) = ⎨e ⎪0 H F ( f X , fY ) = e jkz e

⎞ ⎠

2

1−( λ f X ) − ( λ f Y )2 ⎟

f X

2

+

f Y

2

< 1 / λ ← RS theory

otherwise

− jπλ z ( f X 2 + f Y 2 )

We can see that H F ( f X , fY ) is an approximation to the H RS ( f X , f Y )

Applying the binomial expansion to the H RS ( f X , f Y ) we get: 2

1 − ( λ f X ) − (λ fY ) ≈ 1 − 2

( λ f X )

2

2

⎛ 2π z ⎡ ( λ f ) ( λ f ) ⎜j ⎢1− X − Y ⎜ λ ⎢ 2 2 ⎣ ⎝ 2

e

⎛ j 2π z 1− λ f 2 − ( λ f ) 2 ⎞ ( X ) Y ⎜ λ ⎟ ⎝ ⎠

≈e

−

( λ f Y ) 2 2

⎤⎞ ⎥⎟ ⎥⎦ ⎠⎟

2

if ( λ f X )

=e

jkz

e

2

 1 and

− jπλ z ( f X 2 + f Y 2 )

( λ f Y )

2

 1

= H F ( f X , fY )

Conclusion:

⎛ k ⎞ 2 H RS ( f X , fY ) ≈ H F ( f X , fY ) When the conditions: ( λ f X ) = α = ⎜ x ⎟ ⎝ | k | ⎠ 2

2

( λ f Y ) = β 2 = ⎜

k Y

⎟

⎝ | k | ⎠

 1 are

2

 1,

satisfied. So Fresnel approxilation is equivalent

to the paraxial approximation that is limitted to sma ll propagation angles.

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4.2.4 Fresnel diffraction between confocal s herical 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 z

= 0 and z = z. r 01 is the distance between two spherical caps.

We write the e uations for both surfaces and find the distance between them. Make paraxial approximation by using the binomial expansion. Assuming the extend of the spherical caps about the z axis is small x −

, . . r01

=

z

2

+ ( x − ξ )2 + ( y − η )2 →

U ( x, y ) =

z

j λ ∫∫

Σ

jkz

U x, y

  

Field on the right hand spherical cap

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=

j λ z

U (ξ ,η )

e

r01

≈ z-

xξ z

-

yη

y − η ,

η

y

z

x

r 01

2 r 01

∞

U (ξ ,η )e

d ξ d η , ⎧ 2π ξ η ⎫ ⎨− j [ x + y ]⎬ z

r 01

d ξ dη

−∞    Fourier transform of the Field on the left hand spherical cap


z 12

4.2.4 Fresnel diffraction between confocal s herical surfaces II

η

y

ξ r 01

Comparing the two form: U ( x, y ) 

=

e jkz

∞

∫ ∫ U (ξ ,η )e

⎧ 2π ⎫ ⎨− j [ xξ + yη ]⎬ ⎩ λ z ⎭

d ξ dη

−∞   

Field on the right hand spherical cap

z

Fourier transform of the Field on the left hand spherical cap

Compared with the Fresnel diffractionintegral: k

2+ 2

Fourier transform of the U (ξ ,η ) e 2 z which is complex field just to the right of aperture multi plied by a quadratic phase factor

 

k j ( ξ +η ) ⎪ ⎫ − j 2λ π z ( xξ + yη ) ⎪⎧ 2z U ( x, y ) = e d ξ dη ⎬e ∫ −∞∫ ⎪⎩⎨U (ξ ,η )e j λ z ⎪⎭ 

e

jkz

j

k

2 z

( x 2 + y2 )

∞

2

2

Seond form of the Fresnel diffraction integral

We see that by replacing the two plates with spherical caps, the quadratic factor in ( x, y ), e

j

k

2 z

( x2 + y2 )

, and (ξ ,η ), e

j

k

2z

(ξ 2 +η 2 )

, 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.

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4.3 The Fraunhofer approximation I Goal: applying another more stringent approximation to the Fresnel diffraction integral to simplify the calculations for valid cases. Fourier transform of the quadratic phase function, U (ξ ,η ) e aper ure

U ( x, y ) =

e jkz jλ z

e

j

k

2 z

( ,η ) , mu p e

s r u on

k

2 z

(ξ 2 +η 2 ) , which is the

y a qua ra c p ase ac or e

j

k z

ξ 2 +η 2

  ∞

k j ⎧ (ξ 2z ξ η U ( , ) e ∫∫⎨

( x2 + y 2 )

j

2

+η 2 )

⎫ − j 2λ π z ( xξ + yη ) dξ dη ⎬e

−∞    Seond form of the Fresnel diffraction integral

Applying the Fraunhofer approximation: z  phase factor e U ( x, y ) =

e jλ z

j

e

k

ξ 2 +η 2 ) ( 2 z

j

(

2 2 k ξ + η

k

2 z

(x +y )

the quadratic

η

≈1 ∞

∫∫

)max

U (ξ ,η )e

− j

2π λ z

xξ + yη

dξ dη

−∞    Fourier transform of the aperture distribution evaluated at f X =

λ z

and f Y =

y P0

Σ

z

θ

λ z

Spring 2010

Eradat Physics Dept. SJSU

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4.3 The Fraunhofer approximation II Fraunhofer approximation: z  z

Since k =

2π

k ξ 2 + η 2

x-

max

2 2

+

or

zλ

π

 aperture

size,

y -η 2

is a large number Fraunhofer approximation is much

λ stringent than Fresnel approximation

At optical frequencies: λ

= 0.6μ m;

aperture width = 2.5 cm; z  1600m

"

"

for an aperture with linear dimension of D, the Fraunhofer approximation

2 D 2 λ The Frauhofer diffraction pattern will form at very far distances but we

. Will see in the problems. S rin 2010


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Bessel functions I The Bessel functions or cylinder functions or cylinderical harmonics o

e rs

n ,

n

x , are e ne as 2

Bessel differential equation: x

2

d y dx 2

e so u ons o

+x

dy dx

e

+ ( x 2 − n2 ) y = 0 .

⎧∞ (-1)l 2 l +|m| ⎪∑ 22l +|m| l !(| m | +l )! x ⎪ l =0 J m ( x ) ⎨

2

sin x

⎪ π x ⎪ 2 cos x ⎪ x π ⎩ J − m ( x ) = (−1) m J m ( x )

m=

1 2

1 2

m=-

1

m = 0,1, 2, 3, ... d dx

An integral identity:

| m |≠

∫

u

m m

=

m m −1

u ' J 0 (u ') du ' = uJ1 (u )

0

∞ n

∑

∞

J ( x ) = 1; k = - ∞ k

e

m =−∞

n ) + 2∑Dept. j JSJSU = J 0 ( zPhysics n ( z ) cos(nθ ) Eradat −∞

iz cosθ

∞

m

n− m

Spring 2010

There are more of these identities. Check you favorite math handbook.

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

Bessel functions of first kind 1 J1 J2 J3 0.5

J

- .

0

5

10

15

u

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Bessel functions II J n ( z ) = J n ( z ) = J n ( z ) = J 0 (a ) = J n ( z ) =

1

π ∫

π

0

−n

cos( z sinθ − nθ )d θ

π ∫

π

0

1

eiz 2π

π

1 2π

∫

2π

0

cosθ

eiz

e dθ → with n = 0

eia cos dθ θ

π ( 2n − 1)!!

1 J n ( x) = 2π i

∫

γ

cos( nθ ) d θ

cosθ inθ

z n

2

∫

x z −

e2

z

Bessel's first integral

z

or

π / 2

J0 (z) =

∑ k = 0

( −1) k

(

2 Z / 4

k !

)

k

2

sin 2 n u cos( z cos u) du for n = 1, 2,...

0

− n −1

dz for n > -

1 2

The Bessel functions are normaized: 2

∞ 0

J n x dx = 1 for n = 0 1 2 ...

∞ ⎡ J ( x) ⎤ 4 ⎡ J ( x) ⎤ Integrals involving J1 ( x ) : ∫ ⎢ 1 ⎥ dx = and ∫ ⎢ 1 ⎥ 0 0 3π ⎣ x ⎦ ⎣ x ⎦ 2010 Eradat Ph sics De t. SJSU ∞

S rin

∞

2

xdx =

1 2

18

Fourier transform of a circularly symmetric function I Most apertures and lenses have circular symmetry for example 1

x 2 + y 2 ≤ a

0

x

g ( x, y ) = {

2

+ y2 >a

expresses a circular aperture with radius of a.

The circular symmetry justifies usage of cylindical coordin ates. x = r cos θ ; y = r sin θ ;

r

=

x2

+ y2 ;

= ρ cos φ ; f y = ρ sin φ; ρ = f X2 + dxdy = rdrdθ ; df X dfY = ρ d ρ d φ ; f x

θ = tan −1 ( y / x) φ = tan -1 ( fY / f X )

fY2 ;

∞

} = G ( f X , fY ) = ∫ ∫−∞ g ( x, y ) e − j 2π ( f

{

F g ( x, y )

X x

+ fY y )

dxdy

Now apply change of varaibles:

{

2π

} = G0 ( ρ , φ ) = ∫0

F g (r , θ )

dθ

∫

∞

0

− j 2π ( ρ cos φ r cos θ + ρ sin φ r sin θ )

g (r , θ )e

rdr

For circularly symmetric functions g is only function of r . So we write: g ( r , θ ) = g R (r ) G0 ( ρ , φ ) = S rin 2010

∫

2π

0

∫

dθ

∞

0

g R (r )e

− j 2π r ρ cos(θ −φ )

rdr

∞

= ∫0


g R (r )rdr

∫

2π

0

e


d θ 19

Fourier transform of a circularly symmetric function II G0 ( ρ , φ ) =

∫

∞

0

g R ( r ) rdr

∫

2π

0

e


dθ

this relation is correct for any value of φ including φ = 0,

1 Value of the integral 2π

∫

2π

0

e

− ja cos(θ )

dθ

= J 0 (a) is own known as the

zeroth order Bessel function of the first kind. With substituting a = 2π r ρ and φ = 0 we get:

( ρ ) = G0 ( ρ ) = 2π B

∫

∞

0

rg R ( r ) J 0 (2π r ρ ) dr ←

Fourier-Bessel transform,

, or B

The inverse Fourier-Bessel transform is then: −1

B g ( r , θ )

= g R (r ) = 2π

∞

ρ G0 ( ρ ) J 0 (2π r ρ ) d ρ

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. S rin 2010


20

Fourier transform of a circularly symmetric function III Following the Fourier integral theorem. and simmilarity theorem, we get: BB

−1

r )} = g R ( r ) ← when g R ( r ) { g R (r )} = B −1B { g R (r )} = BB {g R (

{

}=

g R (ar ) B

for B

a2

G0 ⎜

is continuous.

ρ

⎟ ⎝a⎠

Fourier-Bessel transform.

case of the general two-dimensional Fourier transforms.

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Fourier transform of a circular aperture with radius a ⎧⎪1 g ( x, y ) = ⎨ ⎪⎩0

x 2 + y 2

≤a

⎧1 → g R (r ) = ⎨ ⎩0 >a

x 2 + y 2

≤a r>a r

Substituting g R (r ) in G0 ( ρ , φ ) = G0 ( ρ ) = 2π G0 ( ρ ) = 2π

∫

a

0

∫

∞

0

rg R ( r ) J 0 (2π r ρ ) dr

rJ 0 (2π r ρ ) dr

Usin the the inte ral identit : r ' = 2π r ρ

1

a

=0→r'=0

0

u ' J u ' du '

and r

=a

= uJ

u

r ' = 2π a ρ

1

2π a ρ

= ' 0 ' ' 0 2πρ 2 0 2πρ 2 0 1 J1 (2π a ρ ) 2 J1 (2π a ρ ) = G0 ( ρ ) = π a ρ J π a ρ a π a = with kα = 2 πρ 2 ) 2 1 (2 2 2π 2π a 0

=

r

u

⎡ J (k a ) ⎤ G0 ( kα ) = F ( kα ) = 2π a 2 ⎢ 1 α ⎥ where J1 is a first order Bessl function. ⎣ kα a ⎦

S rin 2010


22

Circular aperture with Bessel functions ⎧⎪1 g ( x, y ) = ⎨ ⎪⎩0

x=(-15.1:0.5:14.9); y=(-15.1:0.5:14.9); A=y.'*x; _ = for i=-15.1:0.5:14.9 j_index=0; i_index=i_index+1; for j=-15.1:0.5:14.9 j_index=j_index+1; ^ ^ if r <=5 A(i_index,j_index)=1; else A(i_index,j_index)=0; end end en 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'); axis([-15.1 14.9 -15.1 14.9 -1 4]);Eradat title('Fourier Bessel of Circular Aperture')

2

+ y2 ≤ a

2

+ y2

x x

G0 (kα ) = F ( kα ) = 2π a

Physics Dept. SJSU

2

⎧1 → g R ( r ) = ⎨ ⎩0 >a ⎢ ⎣

1

α

kα a

≤a r>a r

⎥ ⎦

Spring 2010 23

Circular aperture with Bessel functions in MATLAB


Spring 2010 24

Circular aperture with FFT in MATLAB %PHYS 258 spring 07, Nayer 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= - : . : ; A=y.'*x; i_index=0; for i=-2:0.05:2 j_index=0; _ =_ 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); mes x,y, ; po xlabel('x'); ylabel('y'); zlabel('E'); title('Circular aperture'); fft_v=abs(fft2(A)); fft_val=fftshift(fft_v); of spectrum subplot(2,1,2); mesh(x,y,fft_val); Eradat Physics Dept. SJSU xlabel('fx'); ylabel('fy'); zlabel('E'); title('fft of Circular aperture');

Spring 2010 25

4.4 Examples of Fraunhofer diffraction atterns • 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 . • In the following examples we will calculate the intensity distributions across the apertures.

S rin 2010


26

Screen amplitude transmittance function complex field amplitude Screen amplitude transmittance function=

imediately behind the screen

Screen amplitude transmittance for an infinite opaque screen: A

,

⎧1 ⎩0

in the aperture outside the aperture

It is possible to introduce for example plates of various thickness b) Amplitude mask: spatial attenuation by placing an absorbing photographic transparency with real values between 0 ≤ t A

≤1

These two techniques extend all realizable values of t A

.

S rin 2010


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. Aperture amplitude transmittance:

⎛ ξ ⎞ ⎛ η ⎞ rect ⎟ ⎜ 2w ⎟ w 2 ⎝ X ⎠ ⎝ Y ⎠

t A (ξ ,η ) = rect ⎜

where w X and wY 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 ust across the a erture is the transmittance function t A , and the Fraunhofer diffraction pattern is: U x

=

e jkz jλ z

e

j

k

2 z

( x2 + y 2 )

∞

U

e

− j

2π λ z

( xξ + yη )

d d

−∞    Fourier transform of the ape rture distribution evaluated at f X

U ( x, y ) = S rin 2010

e jkz jλ z

e

j

k

2 z

( x2 + y 2 )

=

x

λ z

{ (ξ ,η )}

U F

and f Y =

f X

y

λ z

= x / λ z , fY = y / λ z


28

4.4.1 Rectangular aperture II U ( x, y ) =

e

jkz

jλ z

e

j

⎛

t A (ξ ,η ) = rect ⎜

k

2 z

( x2 + y2 )

ξ

2w X

{

} f

F t A (ξ ,η )

X

= x / λ z , fY = y / λ z

⎞ ⎛ η ⎞ rect ⎟ ⎜ 2w ⎟ Y

} = 2wX sin c ( 2wX f X ) 2wY sin c ( 2wY fY )

{

F t A (ξ ,η )

k

jkz

U x, y x, y

, S rin 2010

= = =

jλ z e

jkz

jλ z

,

2 z

e e

j

k z

2

x 2 + 2

A sin c 2w X f X sin c 2wY fY

( x2 + y2 )

=

with A = 4wX wY

A

s nc

2w x

⎝

2

λ 2 z 2

2

λz

⎠

s nc

⎛ 2w X x ⎞ ⎝ λz ⎠


f X

= x / λ z , f Y = y / λ z

2w

⎝ 2

λ z

⎠

⎛ 2wY y ⎞ ⎝ λ z ⎠ 29

4.4.1 Rectangular aperture III I ( x, y) =| U ( x, y) |

2

=

A2

λ 2 z

⎛ 2w X x ⎞ sin c2 ⎛ 2 wY y ⎞ ⎟ ⎜ λ z ⎟ z λ ⎝ ⎠ ⎝ ⎠

2 sin c ⎜ 2

Exercise: prove that width of the maine lobe or distance between the rs wo zeros s x =

λ z w X

.

Solution: we need to find roots of the I, when y=0, we have

sin c 2 ⎜

⎝

2wY y λ z

⎟ = 1 so we need to require ⎠

⎛ 2w X x ⎞ = 0 → ⎟ ⎝ λ z ⎠

s n π

sin c 2 ⎜

with m = ±1 we get x+ S rin 2010

⎝

π

=

2w x

λ z 2w X x λ z

2 w X

⎠ = 0 → π 2 wX x = mπ → x = mλ z 2w X

λ z

and x−

=−

λz

2 wX


→ Δx =

λ z wX

30

S rin 2010


31

4.4.2 Circular aperture I Goal: calculate the intensity of the Fraunhofer diffraction pattern at a distance z from a circular aperture of radius q located on an infinite .

⎛q⎞ ⎟ ⎝ w⎠

t A (q ) = circ ⎜

Circular symmetry suggests using the cylinderical coordinates and the Fourier-Bessel transformation. The Fraunhofer diffraction pattern is: U ( x, y ) =

e

k

k

jλ z

2 z

e

x

∞

+y

∫ ∫ U (ξ ,η )e

−

2π λ z

x

+ yη

dξ dη

−∞    Fourier-Bessel transform of the aperture distribution X

U ( x, y ) =

e

jkz

jλ z

e

j

k 2 r

2 z

{ ( q )}

U B

=

x

λ z

ρ = r / λ z

Y

=

y

λ z

where r =

x

2

+ y 2 is the radius in the

aperture plane and ρ = f X2 + f Y 2 is the radius in the spatial frequency plane. S rin 2010


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 t U (r) =

⎧ ⎩

e

jkz

jλ z

e

j

k 2 r 2 z

{

t A B

( q )}

ρ = r / λ z

=

e

jkz

jλ z

e

j

k 2 r 2 z

⎧ ⎩

⎛ q ⎞⎫ ⎟⎬ ⎝ w ⎠⎭

B ⎨circ ⎜

= r / λ z

⎛ q ⎞ ⎫ = A J1 (2π wρ ) where A = π w2 . With ρ = r ; 2π wρ = kwr ⎬ ⎟ z π wρ λ z ⎝ w ⎠⎭

B ⎨circ ⎜

U (r) = r

=

A jλ z

jkz

e e

A

⎝ λ z ⎠

2

⎢⎣

j

2

2 z

r

⎢2

J1 (kwr / z )

⎣

kwr / z

J kwr / z kwr / z

⎥⎦

2

⎥⎦

←

e

ry pa ern.

Width of the central lobe measured along the x and y axis: d = 1.22

S rin 2010


λ z

33

4.4.2 Circular aperture III xerc se: rove

a w

o

e cen ra o e measure a ong

on the Airy pattern is: d = 1.22

e x an y ax s

λ z w 2

2

⎛ ⎞ ⎡ 2 J1 (kwr / z ) ⎤ = 0 for the roots*. we start from the Airy pattern: I ( r ) = ⎜ ⎟ ⎝ λ z ⎠ ⎢⎣ kwr / z ⎥⎦ A

= 0 for r ≠ 0 so

kwr / z r = 1.2203

z

=

λ z

= 3.8317 → r =

3.14

w

λ z

Using the table we an calculate the other zeros * 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

14.9309 S rin 52010

16.4706

17.9598

19.4094

20.8269 Eradat Ph22.2178 sics De t. SJSU

34

The grating equation on

on or

e cons ruc ve n er erence or a g

pass ng

through a transmission grating: n2 Λ sin θ 2

− n1Λ sin θ1 = mλ

The grating equation n2 sin θ 2

= n1 sin θ1 + mλ

λ

Λ

←→ θ 2 > θ 1 A "nagative" diffraction order (m<0) ←→ θ 2 < θ 1 A "positive" diffraction order (m>0)

θ1

1

2

θ2

Grating

> θ 1 corresponds to the zeroth order

For a reflection grating bothe incident and reflected

θ2

rays are on the same side so n1

= n2 = n n1

S rin 2010


n2 35

4.4.3 Thin sinusoidal amplitude grating I Goal: calculate the intensity of the Fraunhofer diffraction pattern at a distance z from a thin sinusoidal amplitude grating. The amplitude trnsmittance function is:

⎡ 1 + m cos(2π f ξ ) ⎤ rect ⎛ ξ ⎞ rect ⎛ η ⎞ 0 ⎜ 2w ⎟ ⎜ 2w ⎟ ⎥⎦ ⎣2 2 ⎝ ⎠ ⎝ ⎠

t A (ξ ,η ) = ⎢

We have assumed that the grating structure is bounded by a square t A

aperture of width 2 w. m is the eak-to- eak chan e of am litude

transmittance across the screen. f 0 is the sptial f reuency of the grating.

amplitude transmittance (no effect on the phase). Illumination: a unit-amplitude plane wave t A :

e e

s r u on across

e aper ure.

Figure: cross section of the grating amplitude transmittance function. S rin 2010


x 36

4.4.3 Thin sinusoidal amplitude grating II The Fraunhofer diffraction pattern is the Fourier transfor of t A : U ( x, y ) =

e jkz

e

j

k

2 z

( x2 + y 2 )

F

{U (ξ ,η )}

,

=

e jkz

e

j

k

2z

( x2 + y2 )

{

}

F t A (ξ ,η )

,

but first:

⎧ 1 + m cos(2π f ξ ) ⎫ = 1 δ f , f + m δ f + f , f + m δ f − f , f ( X Y ) ( X 0 Y) ( X 0 Y ) ⎬ 0

F ⎨

2

⎧ ⎩

2

2

4

⎛ ξ ⎞ rect ⎛ η ⎞ ⎫ = ⎟ ⎜ 2w ⎟⎬ ⎝ 2w ⎠ ⎝ ⎠⎭

F ⎨rect ⎜

4

Asnic ( 2wf X ) snic ( 2wfY ) where A = 4 w2 is the

area of aperture bounding the grating.

}=

{

F t A (ξ ,η )

With f X

2

snic ( 2wfY

= x / λ z and

U ( x, y ) =

S rin 2010

A

A j λ z

jkz

e e

j

k

2 z

fY

m m ⎧ ⎫ ) ⎨snic ( 2wf X ) + snic (2w( f X + f 0 ) + snic (2w( f X − f 0 ) ⎬ 2 2 ⎩ ⎭

= y / λ z

( x 2 + y2 )

⎛ 2wy ⎞ ⎧ snic ⎛ 2wx ⎞ + m sin c ( 2 w ( f + X ⎟⎨ ⎜ λz ⎟ 2 λz ⎝ λz ⎠⎩ ⎝ ⎠

snic ⎜


f λz) +

m

2

sin c (

2w λz

⎫ ⎭

( f X − f 0 λ z )⎬

37

4.4.3 Thin sinusoidal amplitude grating III n

na y 2

⎡ A ⎤ ⎡ jkz j 2k z ( x + y ) ⎤ 2 ⎛ 2 wy ⎞ I ( x, y ) = ⎢ e e snic ⎜ ⎢ ⎥ ⎟ ⎥ ⎝ λ z ⎠ ⎣ jλ z ⎦ ⎣ ⎦ 2

2 wx

2

2w

m

2

2w

m

⎨ snic ⎜ ⎟ + sin c( λ z ( f X + f0 λ z) + 2 sin c( λ z ( f X − f0 λ z) ⎬ ⎝ λ z ⎠ 2 ⎩ ⎭ 2 ⎡ A ⎤ ⎛ 2 wx ⎞ + m sin c( 2 w ( f + f λ z ) + m sin c( 2 w ( f − 2 ⎛ 2 wy ⎞ ⎧ I ( x, y ) = ⎢ snic snic ⎨ X X 0 ⎜ ⎟ ⎜ ⎟ ⎥ For f 0

 1/ w

⎫ f0 λ z) ⎬

2

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. I x, y

A

2

≈⎢ ⎥ ⎣ jλ z ⎦

snic

⎜ ⎝

2w λz

⎟ snic ⎜ ⎠⎩ ⎝

2 wx λz

m

2

2w

⎟ + 4 sin c ( λ z ( f X + ⎠

f0 λ z) +

2

m

4

sin c (

2w λz

( f X − f0 λ z)

⎭

η = Diffraction efficiency = fraction of the power in a single order of the Fraunhofer diff. patteren.

It can be found from:

⎧ 1 + m cos(2π f ξ ) ⎫ = 1 δ f , f + m δ f + f , f + m δ f − f , ( X Y ) ( X 0 Y) ( X 0 ⎬ 0 4 4 ⎩2 2 ⎭ 2

F ⎨

f Y )

Since the delta functions determine power of the pattern and sinc functions only spread them. 2

η0

= ⎜ ⎟ = 0.25, ⎝2⎠

η 1

m

2

m

=⎜ ⎟ = , ⎝ 4 ⎠ 16

η −1

m

2

m

=⎜ ⎟ = ⎝ 4 ⎠ 16

so η 1 max

=

16

= 6.25% and total power

in 3 orders is 3/8. The rest of the power is lost by absorption of the grating. S rin 2010


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: t A (ξ ,η )

=

⎡ m ⎤ ⎢⎣ j 2 sin(2π f 0ξ ) ⎥⎦

e Sinusoidal phase difference introduced by the grating

⎛ ξ ⎞ rect ⎛ η ⎞ ⎟ ⎜ 2w ⎟ ⎝ 2w ⎠ ⎝ ⎠

rect ⎜

. 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. 0

.

thin means the structure can be modeled by a simple phase transmittance (no effect

on amplitude). um na on: a un -amp u e p ane wave t A : the field distribution across the aperture.

S rin 2010


39

4.4.4 Thin sinusoidal phase grating II t A (ξ ,η ) = e

⎢⎣ j 2 sin(2π f 0ξ ) ⎥⎦

⎟ rect ⎜

k

jλ z

2

k

jkz

2

2 z

e

η

⎟ ⎝ 2w ⎠

⎝ 2w ⎠

jkz

U ( x, y ) =

ξ

rect ⎜ F

{U (ξ ,η )} f

X

, fY

=

jλ z

e

⎡ j m sin(2π f ξ ) ⎤ 0 ⎥ ⎢ 2

2

{

∞ q =−∞

⎧

m ⎞ j 2π qf ξ ⎫ ⎛ Jq e ⊗ ⎩q =−∞ ⎝ 2 ⎠ ⎭

=

t

F t A (ξ ,η )

=

⎡

∞

0

∞

J q =−∞

{

}=

F t A (ξ ,η )

∞

∑ =−

U ( x, y ) = S rin 2010

A jλ z

jkz

}f

F t A (ξ ,η )

2z

=

2

⎛m⎞ q ⎝2⎠

X

, f Y

2π

⎧ ⎛ ξ ⎞ rect ⎛ η ⎞⎫ rect ⎝ 2w ⎠ ⎝ 2 w ⎠⎭ ⎩

⎛ m ⎞ δ ( f − qf , f ) ⎤ ⊗ 0 X Y

A sin c (2wf X ) sin c (2wfY )

⎛ m ⎞ sin c ⎡ 2w f − qf ⎤ sin c 2wf ( Y ) 0 )⎦ ⎟ ⎣ ( X

AJ q ⎜

e e

j

k

∞

⎛ m ⎞ sin c ⎡ 2w x − qf λ z ⎤ sin c ⎛ 2wy ⎞ ( )⎥ ∑ 0 ⎟ ⎜ λ z ⎟ ⎢ 2 λ z ⎝ ⎠ ⎣ ⎦ ⎝ ⎠ =−∞ q Eradat Ph sics De t. SJSU

x2 + y2 ) ( 2 z

Jq ⎜

40

4.4.4 Thin sinusoidal phase grating III t A (ξ ,η ) = e U ( x, y ) =

⎡ m ⎣ 2

⎤ ⎦

0

rect ⎜

⎟ rect ⎜

⎟ ⎝ 2w ⎠ ⎛m⎞ ⎡ 2w x − qf λ z ⎤ sin c ⎛ 2wy ⎞ J q ⎜ ⎟ sin c ( ) 0 ⎜ ⎟

⎝ 2w ⎠

A

jkz

e e

j

k

2 z

( x2 + y2 )

∞

∑

q =−∞

⎛ A jkz j 2k z ( x + y ) ∞ 2w 2wy ⎞ ⎞ m⎞ ⎛ ⎡ ⎤ ⎛ =⎜ e e J q ⎜ ⎟ sin c ⎢ ( x − qf 0λ z )⎥ sin c ⎜ ∑ ⎟⎟ j λ z λ z λ z 2 ⎝ ⎠ ⎣ ⎦ ⎝ ⎠⎠ =−∞ q ⎝ 2

I

η

For f 0

 1/ w

2

2

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.

⎡ I

A ≈ ⎛⎜ ⎞⎟ ⎝ λ z ⎠

2

∞

∑

q =−∞

2

Jq

⎞⎤

⎛

⎛ m ⎞ sin c 2 ⎢ 2w ⎜ x − qf λ z ⎟ ⎥ sin c 2 ⎛ 2wy ⎞ 0 ⎜ ⎟ ⎜ λ z ⎟ ⎢ λ z ⎜ ⎥ ⎟ ⎝2⎠ ⎝ ⎠ ⎢ ⎜ Displacement ⎥ ⎟ of the order fromthe center ⎥ ⎢ 

We see that introduction of the sinusoidal phase grating has deflected power from the zeroth order to the higher ordes. 2

Peak intensity of the qth order = ⎢

J q ⎜

⎟⎥ ⎝ 2 ⎠⎦ It happens when y = 0 and x − qf 0λ z = 0 → x = qf 0 λ z ⎣ λ z

S rin 2010


41

4.4.4 Thin sinusoidal phase grating IIII Displacement of qth order from the center of the pattern

= qf 0 λ z

For q = 0 or zeroth order y

= 0 and x = 0 For q = ±1 or first order y = 0 and x = ± f 0 λ z function of frequency of the grating lining, wavelength, and distance of observation. So for spectroscopy in the blue region we need hight f 0 grating or larger spectrometer. η = Diffraction efficiency = fraction of the power in a single order of the Fraunhofer diff. patteren. ∞

It can be found from:

{

m

} = ⎢ ∑ J q ⎜ ⎟ δ ( f X − qf 0 , fY )⎥ ⊗ [ A sin c (2wf X ) sin c (2wfY ) ] ⎣ q =−∞ ⎝ 2 ⎠ ⎦

F t A (ξ ,η )

Since the delta functions determine power of the pattern and sinc functions only spread them. η 0

m

= J q ⎜ ⎟ ⎝2⎠

Plot η0

m = J q2 ⎛⎜ ⎞⎟ we see that when m/2 is root of J0 then the central lobe wanishes. η 1 max = 33.8% ⎝2⎠

1 = 6.25. 16 There is n o power absorption and sum of the power in all orders is equal to the total incident power. which is much greater than that of the sinusoidal amplitude grating which is

The distribution of power between the lobes varies as m changes.

S rin 2010


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

S rin 2010


43

4.5.1 Fresnel diffraction by a square a erture I Goal: calculate the intensity of the Fresnel diffraction pattern at a distance z from a square aperture of width 2w located on an infinite opaque screen. The amplitude transmittance function:

⎛ ξ ⎞ rect ⎛ η ⎞ ⎟ ⎜ 2w ⎟ ⎝ 2w ⎠ ⎝ ⎠

t A (ξ ,η ) = r ect ⎜

The complex field imediately behind the aperture: U (ξ ,η ) z =0

= rect ⎛⎜

⎞ rect ⎛ η ⎞ ⎟ ⎜ 2w ⎟ 2w ξ

Illumination: a unit-amplitude, normally incident, monochromatic plane wave Using the co nvolution form of the Fresnel diffraction formula: x, y

=

U ( x, y ) = I ( x )

=

j λ z e

1

e

j

π z

⎡( x −ξ )2 −( y −η ) 2 ⎤

η

−w

jkz

I ( x ) I ( y) w

e ∫ λ z −w

S rin 2010

w

e jkz

j

π

⎡( x −ξ )2 ⎤ ⎥⎦

λ z ⎢⎣

where

d ξ dη

and

I ( y)

=

1 λ z


w

∫e

j

π

⎡( y −η )2 ⎤ ⎥⎦

λ z ⎢⎣

d ξ d η

−w

44

4.5.1 Fresnel diffraction by a square aperture II Change of variables: α

=

2 λ z

(ξ − x )

∫

α1

=

2 λ z

(w + x)

(η − y )

and

1 2 j π 2 β 2 e d β I ( y) = 2 − β 1

and

α 2

λ z

β

∫

=

2 1

2

β

α

1 2 j π2 α 2 e dα I ( x ) = 2 −α1

=

and

2 λ z

(w − x)

2 2

λ z

With the Fresnel number: N F

−

λ z

= w2 / λ z and normalized distance variables

in the observation re ion we have: X

=

1

=

β 1

x

λ z

= 2(

S rin 2010

and

y

Y =

λ z 2

F

N F

+ Y ) and

β 2

the limits of the integrals become:

=

= 2(

F

N F

−

− Y )


45

4.5.1 Fresnel diffraction by a square a erture III Using

∫

α2

e

α 1

π j α 2

2

dα

α2

= ∫0

⎛ πt2 ⎞ C ( z ) = ∫ cos ⎜ dt ⎟ 0 ⎝ 2 ⎠ z

e

π j α2

2

dα

α1

− ∫0

e

π j α2

2

dα and the Fresnel integrals:

⎛ π t 2 ⎞ S ( z ) = ∫ sin ⎜ dt we write ⎟ 0 ⎝ 2 ⎠ z

and

I ( x)

=

C (α 2 ) − C (α1 ) ⎦ + j ⎣ S (α 2 ) − S ( α1 ) ⎦ ⎣ 2

I

=

1 ⎡C 2

U ( x, y ) =

⎤ + ⎡S

− C

e jkz

⎡⎣C (α { 2 j

2

and

⎤

− S

) − C (α1 ) ⎤⎦ + j ⎡⎣ S (α 2 ) − S ( α 1 ) ⎤⎦}

× { ⎡⎣C ( β 2 ) − C ( β1 ) ⎤⎦ + j ⎡⎣ S ( β 2 ) − S ( β 1 ) ⎤⎦} ,

=

1 4

2

{

−

2 1

2

−

2 1

2

× ⎡⎣C ( β 2 ) − C ( β1 ) ⎤⎦ + ⎡⎣ S ( β 2 ) − S ( β 1 ) ⎤⎦ S rin 2010


2

}

46

Fresnel integrals Fresnel integrals are defined as: z

j

π

2

2

t

0

C z

=

z

cos

⎛ πt2 ⎞

dt

and

S z


=

z

sin

⎛ π t 2 ⎞

dt we write

Spring 2010 47

The Fresnel integrals S(u) and C(u) are entire functions i.e. they are analytical at all finite points of the complex plane. The Fresnel integrals are tabulated and are available computer programs


Spring 2010 48

4.5.1 Fresnel diffraction by a square a erture III Fresnel integrals: C z

⎛ πt2 ⎞ = 0 cos dt 2 ⎝ ⎠ z

S z

⎛ π t 2 ⎞ = 0 sin dt 2 ⎝ ⎠ z

2 2 2 2 1 ⎡⎣C (α 2 ) − C (α1 )⎤⎦ + ⎡⎣ S (α 2 ) − S (α1 )⎤⎦ × ⎡⎣ C ( β2 ) − C ( β1 )⎤⎦ + ⎡⎣ S ( β 2 ) − S ( β1 )⎤⎦ 4 2 N F = w / λ z for a fixed w and λ , as z increases, the Fresnel number decreases

{

I ( x, y ) =

}{

}

and the true physical distance distance represented on the x = X λ z and

=Y

λ z axis are increased.

Figure shows the normalized intensity distribution along the x axis ( y = 0) for various normalized distance s from the aperture as represented by fresnel number.

,

,

F

,

the product of a delta function and e jkz and shape of the doffraction pattern appr oachs the shape of the aperture. Limit of the process is the geometrical optics pre c on o

e comp ex e :

⎛ x ⎞ rectSJSU ⎛ y ⎞ jkz Physics Dept. = e jkz rect U ( x, y ) = e U ( x, y, 0) Eradat ⎜ 2w ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ 2w ⎠

Spring 2010 49

Fresnel diffraction patterns at different distances from a s uare a erture. As NF->0 diffraction pattern becomes wide and smooth approaching Fraunhofer diffraction

F

pattern becomes sharp and narrow Eradat Physics Dept. SJSU approaching the geometrical shadow of the aperture

Spring 2010 50

4.5 2 Fresnel diffraction by a thin sinusoidal am litude ratin -Talbot ima es I Goal: calculated the Intensity distrubution of the diffraction by a thin sinusoidal amplitude grating using the Fresnel diffraction formulation. or s mp c y we neg ec

e n e ex en o

e gra ng.

The field transmotted by the grating has a periodic nature or we limit the attention to the central region of the pattern.

η

The amplitude transmittance function

ξ

is modeled as: t A (ξ ,η ) =

2

y x

[1 + m cos(2πξ / L ]

Where L is the period of the lines para e o

e ax s η .

Illumination: a unit amplitude normally incident plane wave. The field immediately behind the grating is t A. S rin 2010

Grating structure


z

51

4.5 2 Fresnel diffraction by a thin sinusoidal am litude ratin -Talbot ima es I We will use the convolution form of the Fresnel diffraction equation U x

=

e

∞

jkz

U

jλ z

e

⎧ j k ⎡( x −ξ ) 2 + ( y −η )2 ⎤ ⎫ ⎨ ⎦ ⎬⎭ ⎩ 2 z ⎣

d d

−∞

or the Fourier transform of the equation j

,

k

2 z

(ξ 2 +η 2 )

just to the right of aperture m ultiplied by a quadratic phase factor

 

2π k − j ( xξ + yη ) j (ξ +η ) ⎪ ⎧ ⎫ ⎪ U ( x, y ) = e 2 z d ξ d η ⎨U (ξ ,η )e 2 z ⎬ e λ z ∫ ∫ jλ z ⎪ −∞ ⎪   

e

jkz

j

k

( x2 + y2 )

∞

2

2

Seond form of the Fresnel diffraction integral

Where r 01

>> λ ,

x - ξ z

< 1,

y - η z

< 1, or observation is in the

near field of the aperture or Fresnel diffraction region and scalar theory approximation are assumed r we can use H F ( f X , fY ) = e e jkz

S rin 2010

e rans er unc on approac : − jπλ z ( f X 2 + f Y 2 )


52

4.5 2 Fresnel diffraction by a thin sinusoidal am litude ratin -Talbot ima es II By omitting the constat term e jkz , H F is: H F ( f X , fY ) = e

− jπλ z ( f X 2 + f Y 2 )

, approach yeilds the simplest calcultions. 1) find the s patial frequency spectrum of the field transmitted by the aperture:

1 + m cos π 2 m m 1 1 F {t A (ξ ,η )} = δ ( f X , fY ) + δ ( f X + f0 , fY ) + δ ( f X − f0 , fY ); with f0 = L 2 4 4 ,η =

t A

⎟ + 4 δ ⎜ f X − L 0 , fY ⎟ ⎠ ⎝ ⎠ ⎛ 1 ⎞ The transfer function evaluated at ( f X , f Y ) = ⎜ ± , 0 ⎟ becomes ⎝ L ⎠ πλ z 1 ⎞ − j L ⎛ H ⎜ ± , 0 ⎟ = e and it is unity at the origin. So after propagation ⎝ L ⎠

{

}=

F t A (ξ ,η )

2

)+

δ ( f X , fY

4

+ ,

δ ⎜ fX

⎝

fY

L

2

of a distance z the Fourier transform of the field becomes: π z

π z

1 1 m − j L2 ⎛ ⎞ + m e− j L2 δ ⎛ f − 1 , f ⎞ U x y δ f f e δ f f , = , + + , F { ( )} 2 ( X Y ) 4 ⎜ X L Y ⎟ 4 ⎜ X L 0 Y ⎟ ⎝ ⎠ ⎝ ⎠ −1

{

1

}= +

U ( x, y ) = F F U ( x, y ) S rin 2010

m

− j

πλ z

2π x

πλ z

2π x

−j m −j L L L eEradat e Ph +sicseDeL2 t.eSJSU 2

j

53

4.5 2 Fresnel diffraction by a thin sinusoidal am litude ratin -Talbot ima es III − j 2 1⎡ ⎛ 2π x ⎞ ⎤ U ( x, y ) = ⎢1 + me L cos ⎜ ⎟⎥ 2⎣ ⎝ L ⎠ ⎦ πλ z

I ( x, y ) =

1 + 2 m cos ⎜ 4 ⎢⎣ ⎝

π z 2

L

⎟ cos ⎜ ⎠ ⎝

πx L

2 2 ⎟ + m cos ⎜ ⎠ ⎝

π x L

⎟⎥ ⎠⎦

Now consider 3 special cases for the observation distance: 1)

π z L2

= 2nπ → z =

nL

λ

where n = 0, ±1, ±2

1⎡ ⎛ 2π x ⎞ + m 2 cos 2 ⎛ 2π x ⎞ ⎤ = 1 ⎡1 + m cos ⎛ 2π x ⎞ ⎤ 1 + 2m cos I ( x, y =

2

this is perfect image of the grating. These images that are formed without aid of a lens are called "Talbot images" or "self-images". 2)

πλ z L2

I x

= (2n + 1)π → z =

(2n + 1) L2 λ

where n = 0, ±1, ±2

1⎡ 2π x ⎞ 2π x ⎞ ⎤ 1 ⎡ 2π x ⎞ ⎤ = 1 − 2m cos ⎛ + m2 cos 2 ⎛ = 1 − m cos ⎛ 4⎣ ⎝ L ⎠ ⎝ L ⎠⎦ 4 ⎣ ⎝ L ⎠⎦

2

This is also image of the grating with a 180 0 spatial phase shift or "contrast reversal". These too are called "Talbot images".

S rin 2010


54

4.5 2 Fresnel diffraction by a thin sinusoidal am litude ratin -Talbot ima es IIII 1 (n − ) L2 πλ z π 2 3) 2 = (2n − 1) → z = where n = 0, ±1, ±2 then

⎛ πλ z ⎞ = 0 Using cos 2 ⎛ 2π x ⎞ = 1 + cos ( 4π x ) ⎜ L ⎟ 2 ⎟ 2 ⎝ L ⎠ ⎝ ⎠ 1⎡ 1 ⎛ m2 ⎞ m2 ⎛ 4π x ⎞ 2 2 ⎛ 2π x ⎞ ⎤ I ( x, y ) = ⎢1 + m cos ⎜ 1 cos = + + ⎢⎜ ⎟ ⎟⎥ ⎜ L ⎟⎥ 4⎣ 2 ⎠ 2 ⎝ L ⎠ ⎦ 4 ⎣⎝ ⎝ ⎠⎦ cos ⎜

reduced contrast (nisted of 1 and m we have 1 and

m2

2

and the background

+ m / 2. This is called the "Talbot subimage". For example for m = 0.3 we have m 2 / 2 = 0.045 is now brighter by

S rin 2010


55

Fresnel and Fraunhofer Diffraction

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