IMIT Swedish original by Leif Ek English verison by Anna Thaning Optics exp eriment Fourier Optics Royal Institute of Technology Department of Micro electronics and Information Technology Optics Section Intro duction In this exp eriment we will as implied by the title use optical systems to study F ourier transforms Fourier analysis describ es how a function may b e divided into a sum or an integral of sines and cosines This is often done in one dimension only but the optical system provides us with them means to easily do it in two functions or images expressed dimensions It is describ ed how twodimensional in the spatial co ordinates x y are transformed The twodimensional Fourier transform is dened as Z Z 1 ff x y g f x y expik x k y dxdy F x y 1 To day twodimensional Fourier transforms can quite easily b e found numerically for example using image pro cessing software you can also nd a function for it t in Matlab But this still requires some time and computational p ower During this exp eriment we will create an instant Fourier transform of a pattern A diractive slit in uniform illumination gives rise to a diraction pattern which turns out to b e the fourier transform of the diractive slit It is placed at innity ie very far from the slit but a fo cusing lens can b e used to move it to the fo cal plane of the lens Similarly a lens may b e used to retransform the image Schematically the i m a g e o f o b j e c t Fourier plane t h e o b j e c t l a s e r transform ing lens re-transform ing lens Figure Schematic illustration of the exp eriment setup is shown in Fig It provides us with the opp ortunity to view the Fourier transform but it also gives us the p ossibility to meddle with it By blo cking some parts of the Fourier transform we remove certain frequencies from the image The eect can then b e studied in the image of the ob ject This pro cess is known as ltering Unfortunately this metho d for viewing the Fourier transforms has some limitations The lenses used are not p erfect they have ab errations which will distort the transform But even with a p erfect lens the true transform will not b e found since the nite size of the lens will cut o the highest frequences The lens will act as a lowpass lter The extent of this ltering dep ends on whether the coherent or the incoherent illumination is used The aim of this exp eriment is to show that Fraunhofer diraction gives the Fourier transform of the diracting ap erture what twodimensional transforms of some simple images lo ok like the dierences that app ear for coherent and incoherent illumination As an extra b onus you also get the Babinets principle Theory Fourier transformation by Fraunhofer diraction Now one would think that the Fourier transform on an ob ject eg a square could b e obtained just by illuminating the ap erture and then study the diraction pattern In reality its a little bit more complicated since the diraction pattern of an ap erture there are changes its shap e dep ending on the distance of observation Generally two cases Fresnel diraction where the diraction pattern close to the ap erture is considered and Fraunhofer diraction where the pattern is studied at innity Fresnel diraction is normally much more complicated so we will stick to Fraunhofer diraction and show that the Fraunhofer diraction pattern is really the Fourier transform of the ob ject Assume that the ob ject consists of an ap erture of any y x r R Figure Diraction from an arbitrary ap erture at P X Y Z shap e in the x y plane We study the diraction pattern in the p oint P of resnel co ordinates X Y Z as illustrated in Fig Accordning to the HuygensF principle the ap erture may b e regarded as the sum of many p oint sources that pro duce spherical waves From the p oint source lo cated in the small surface element dS we obtain a spherical wave which gives rise to the electric eld dE A dE exp ik r tdS r where is the strength of the source ie the incident eld strength at this par A ticular p oint and r is the distance b etween this p oint on the ap erture and the observation p oint P as illustrated in Fig This is the normal expression for a spherical wave on complex form see eg Eq in Hecht To obtain the total eld at P we must sum all the dierent contributions This is done by integration ZZ A E P exp ik r tdS r A where A is the area of the diracting ap erture This is a fairly complicated ex pression and we need to do some approximations According to the geometry see Fig p r X x Y y Z p R X Y Z Since the size of the ob ject is small compared to the distance R we can approximate R in the denominator However due to the large value of k r very small changes r to r will cause large changes to the exp onential term so here we need a more accurate approximation By some manipulation we get r x y xX y Y r R R R The term x y R is small and may b e neglected Then expansion in a p ower series yields r xX y Y xX y Y R r R R R Insertion of Eq into Eq yields ZZ ZZ xX y Y A A E X Y Z exp ik r tdS exp ik r t expik dS r R R A A To simplify the expression we intro duce the ap erture function A exp ik r t A Ax y x y expix y R which describ es the transmittance of the ob ject It contains an amplitude part A x y and a phase part x y R is constant with resp ect to the integration variables x and y Eq may now b e rewritten as ZZ 1 E X Y Z Ax y expik xX y Y R dxdy 1 The integration can b e extended to innity since the ap erture function is zero everywhere but in the ap erture In order to identify Eq as a twodimensional Fourier transform we dene the spatial frequencies k X k Y k k x y R R Inserted into Eq they yield ZZ 1 Ax y expik x k y dxdy E k k x y x y 1 Except for the names of the variables this is identical to Eq Consequen tly we have shown that the electric eld at Fraunhofer diraction is the Fourier transform of the ap erture function ie that E k k F fAx y g x y Since the ap erture function describ es the geometry of the ap erture this means the diraction pattern far away from the ob ject will b e its Fourier transform Tw odimensional images and transforms In order to b etter understand twodimensional images and their transforms well start with the easier onedimensional case Assume we have a function of one vari able f t and its transform F The function can always b e written as a sum of sine and cosine functions that is as a sum of dierent frequency comp onents The Fourier transform of the function will the b e the sum of the Fourier transforms of the trigonometric functions that constitute ft In two dimensions the same thing happ ens An image can b e viewed as a lot of sinusoidal lattices These lattices have dierent spatial frequencies and dierent orientations as illustrated in Fig The a ) b ) Figure Two sinusoidal gratings of dierent orientations Fourier transform of a sinusoidal lattice is three delta functions one in the origin and one to each side as illustrated in Fig The p ositions of the the delta pulses are determined by the spatial frequency of the lattice the higher the frequency the longer the distance to the origin and by its orientation This can easily b e deduced mathematically but we leave this to the interested student a ) b ) Figure The Fourier transforms of the gratings in Fig Task Some simple transforms Transforms of several identical parts If the ob ject consists of several identical parts the sup erp osition principle says the image will b e the sum of the electrical eld from each part Assume there are N 0 0 parts and that each of them gets the lo cal co ordinate system x y where the origin is at x y j N as illustrated in Fig Intro duce the notation j j Figure An ob ject consisting of four identical parts 0 0 Ax y and A x y for the total ap erture function and the ap erture function for one part resp ectively Then the total eld will b e ZZ N 1 0 0 X X x x Y y y j j 0 0 0 0 E X Y A x y exp ik dx dy R 1 j or ZZ N 1 0 0 X X x Y y X x Y y j j 0 0 0 0 ik dx dy exp ik E X Y A x y exp R R 1 j For simplicity intro duce the spatial frequences k and k as dened in Eq x y Then the eld my b e written as ZZ 0 N 1 X 0 0 0 0 0 0 A x y exp ik x k y dx dy exp ik x expik y E k k x y x j y j x y 1 j The sum at the end of the expression is a phase factor only it will not aect the pattern itself Consequently several identical parts give the same Fourier transform as one of the parts except for the extra phase factor This shows that the electrical eld E at the image plane is the Fourier transform of the ap erture function for one part of the pattern A multiplied by the Fourier transform of delta pulses at the origin of each part ie that N X 0 0 E F fAx y g F fA x y g F x x y y j j j Compare to the convolution theorem which says that F fAg F fA g F fg g F fA g g Tee conclusion is that several identical parts of a gure may b e describ ed as the ap erture function of the part convoluted with delta functions at the centre of each part as illustrated in Fig A A 1 g x= x* x Figure Several identical parts can mathematically b e describ ed as a convolution b etween one part and several delta functions Task The transform of two circles a Use two delta functions and a circ function to describ e two circles at a distance a from each other Place one circle at the origin and the other at a A circ function is dened as x y D circx y x y D where D