Optics

Physics 425 Modern Optics Fourier Optics Lecture Notes W. Andreas Schroeder

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VII. FOURIER OPTICS

The Fourier Plane of a Lens

In diffraction theory, the field at a point ro due to the diffraction of a field ( )riE incident on a diffraction surface is given by the Huygens-Fresnel integral;

( ) ( ) [ ]dsiR

EiE io Rkr

r

= exp

where the double integral reflects the 2-D nature of , == rdrddxdyds , and the geometry is

[NOTE: [ ]Rk

iRi

exp can be interpreted as a Huygens wavelet of unit amplitude.]

y

x

z

P

r

ro

R

ds


2

Let us consider the one-dimensional case of a plane wave normally incident on a diffraction object, O, a distance Z/ from a lens;

If a screen is positioned a distance Z from behind the lens, what is the field pattern on the screen?

The Huygens-Fresnel integral allows the field just in front of the lens, ( )LxE , to be evaluated:

( ) 22 'ZxxR Ls += ( )'2

'

2

Zxx

Z Ls

+

( ) ( ) [ ] ( ) sLssoL dxZxxikikZxfE

Zi

xE

=

'2exp'exp

'

2 (1)

where oE is the amplitude of the incident plane wave, ( )sxf is the transmission function of O, and '' ZR has been assumed for the denominator.

equivalent to 1-D factor

S

plane wave

O

R/ R

P

lens screen

z

Z Z /

sx Lx


3

Now, the thin lens introduces a phase shift to the field as illustrated below;

( )222 ozff += 222 oo zfz =

In the paraxial approximation oz is small, hence

fzo 22

so that the lens introduces a phase change, , given by fkkzo 2

2==

fkxL2

2=

Thus, the field just after the lens, ( )LA xE is

( ) ( )

= f

xikxExE LLLA 2exp

2 (2)

negative since center of wavefront is retarded by lens.

in our case.

z

f

zo

spherical wave

plane wave


4

The field at point P may now be calculated in a similar manor to that used previously;

( ) ( ) [ ] ( ) LLLAp dxZxikikZxE

ZiE

=

2expexp

2

Substitution of equations (1) and (2) into this expression gives

( ) ( )

+

+

++

=

'2exp

'

exp11'

12

exp2

'exp'

222

Zx

ikZZ

xikxfZZ

xikxfdxdx

ZZZik

ZZiE

E ssLL

ssLo

p

In general, this double integral in difficult to solve, but two conditions provide for a simple solution;

i) Assume the lens is infinite in extent (i.e. has a constant, unity transmission). ii) Place the object and screen a distance f from the lens (i.e. Z = f and Z/ = f).

( ) ( ) ( )( )

++

+

=

222

22

exp2

2exp ssLLssLop xxxxfik

xfdxdxffikfiE

E

The integral Lx can now be evaluated by completing the square;

( ) ( ) ( ) ( ) ( )( ) LsLssssop dxxxfikxfxfikxfdxffikfiEE +

+

=

222

2exp2

2exp

22exp

Using [ ]

pi=

duu 22exp with ( )+= sL xxu and fik

2

= gives

( ) [ ] ( ) sssop dxfikx

xfikfikf

fiE

E

pi

= exp2exp2


5

So the field ( )pE at the back focal plane of the lens is proportional to the Fourier transform of the field transmitted by the diffracting object, ( )xE ,i.e.

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

== dxipxxfxEpEE p exp

where sxx = and ffkp

opi

=

=

2

N.B. It can be shown that even if the object distance does not equal f, then the back focal plane of the lens still contains the Fourier transform of the object.

... a consequence of Abbes imaging theory.

I F O

So

Si

f


6

Image processing

The image of an object can be modified by perturbing the light distribution in the Fourier plane of an optical system. This image processing can be used for edge enhancement, object recognition, etc.. A suitable optical system for image processing is shown below;

All image processing methods are based upon the realization that the dc (background) components of the light from the object are imaged near the optical axis in the Fourier plane, F, while the higher spatial frequency components of the object (e.g. those from hard edges) appear far from the optical axis in F. This spread in the objects spatial frequencies allows their amplitudes to be modified in F by the use of masks etc..

Plane wave

Object plane

Image plane

Fourier plane

optical axis

I F O

f2 f1 f1 f2


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i) Edge enhancement

If the central dc spatial frequency components of the object are masked out, the reconstructed image will only have its edges high-lighted.

Eg. a square aperture.

N.B. The opposite to edge enhancement transmission of the dc component in the Fourier plane simply produces a uniform illumination in the image plane (edges smeared out).

2a

a

2a

x

Eobject 1

edges at 2a

x = enhanced by

removing dc components in Fourier plane with a dc stop.

p

EF.T.

mask

p

EF.T.

2a

2a

x

Eimage

2a

2a

x

Iimage

=

=


8

ii) Phase-contrast microscope

__

Fritz Zernike (Nobel Prize)

The phase-contrast microscope technique allows the visualization of transparent objects that only produce a phase shift (with respect to the background) in the light transmitted by the object; e.g., micro-organisms in the water.

The transmission function for a phase object in the object plane may be written as

( ) ( )[ ] ( )xixixf += 1exp

for a small phase shift ( )x and no absorption.

( )[ ] ( ) ( )[ ] ( ) ( )pipxipxf +=+=

In the phase-contrast microscope, the dc signals are retarded by a phase of

2pi with respect to the higher spatial frequency components in the

Fourier plane; that is, ( )pEFT is now given by

( ) ( ) ( )piippEFT +

pi=2

exp

( ) ( )[ ]ppi =

The inverse transform gives ( )imageE , i.e.

( ) ( ) 1imageE

so a phase-contrast object has been turned into an irradiance-contrast image which is easier to see or detect.


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iii) The Schlieren method

__

another phase-filtering technique in which exactly half the Fourier plane is masked out.

The technique produces an intensity distribution in the image plane that is proportional to the phase change produced by the object. This method is particularly useful in aerodynamics to measure changes in density associated with airflows.

iv) Pattern recognition

Different objects have different Fourier transform patterns. So if a mask, whose transmission mimics the Fourier transform of one particular object, is placed in the Fourier plane, the strength of the radiation transmitted by the mask will give a measure of how closely the object resembles that represented by the mask. In other words, one performs a correlation between the objects Fourier transform and the expected object useful in military applications and sorting machines.

N.B. The position of the object in the object plane does not matter since the Fourier transform is not affected;

( )[ ] ( )pFeaxf ipa=+

the shift theorem.

O I F

f2 f2

mask


10

Diffraction Limited Focusing

In our discussion of diffraction, we showed that the Fourier transform of a circular aperture is an Airy pattern:

The first minimum 1 is given by

=

a

R2

22.11

where a is the radius of the aperture and R is the distance from the aperture to the far-field viewing screen.

Thus, since

,

when viewed in reverse an Airy field pattern produces a uniform circular illumination.

( ) =

pi aJa 1

2

0 a r

1

Io

( )II

1

( ) ( )aa2 1p pi

= JE

( ) ( )2

1o

a

a2

= JII


11

So, if a lens with a diameter aD 2= is illuminated with a uniform plane wave, the irradiance pattern at the focus (the Fourier plane) will be an Airy pattern with R = f. That is

The radius, 1, of the diffraction limited focal spot for a uniformly illuminated lens of diameter D is, therefore, given by

E.g. for f = 10cm, D = 2.5 cm (1 inch) at = 500nm we get 1 = 2.4m (small!).

=

Df22.11

R

E(r)

E()

r

R = f

E(r)

E()

r

D = 2a

a

0


12

N.B. i) The fact that 1 has important implications for optical information

storage; e.g. compact disks,

To avoid cross-talk we require that bits of information be separated by

13~ , so that one bit requires an area of 2110~ .

Memory ( ) ( )

pi

22

222

21

222.0

10 fDabab bits

The 21

dependence is driving the development of shorter wavelength

visible laser diodes; hence, BlueRay etc.

ii) An optical system is often characterized by its Strehl Ratio; which is defined as the ratio of the light energy within a radius 1 of the perfect diffraction limited image (which is 84%).

( )

( )

=

0

0

84.0

1

dI

dI

RatioStrehl

The Strehl ratio is related to the opd between the perfect and not-so-perfect (real) wavefronts that produce the image.

opd 0 16

8

4

Strehl ratio 1.0 0.988 0.952 0.8095

Area = ( )22 ab pi 2100cm b a

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