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Fourier Optics Phys 322 Chapter 11 Lecture 31 Fourier Optics Introduction Fourier Transform: definition and properties Fourier optics Introduction Motivation: 1) To understand how an image is formed. 2) To evaluate the quality of an optical system. Optical system Star (Telescope) Film Delta function Fourier transform Inverse Fourier and filtering transform The quality of the optical system: • The image is not as sharp as the object because some high frequencies are lost. • The filtering function determines the quality of the optical system. Fourier transform: 1D case Any function can be represented as a Fourier integral: angular spatial frequency k 2 / 1 f x Ak cos kx dk B k sin kx dk 00 Fourier sine transform of f(x) Fourier cosine transform of f(x) Ak f x cos kx dx Bk f x sin kx dx Alternatively, it could be represented in complex form: 1 f x Fk eikxdk complex 2 functions! ikx Fk f x e dx Fk A k iBk Fkei k Fourier transform 1 F(k) is Fourier transform of f(x): f x Fk eikxdk 2 Ffk Y x Fk f x eikxdx f(x) is inverse Fourier transform of F(k): -1 fx Y Fk Application to waves: 1 If f(t) is shape of a wave in time: f t F eitd 2 F f t eitdt Reminder: the Fourier Transform of a rectangle function: rect(t) 1/2 1 Fitdtit( ) exp( ) [exp( )]1/2 1/2 1/2 i 1 [exp(ii / 2) exp( i exp(ii / 2) exp( 2i sin( F() Imaginary F(sinc( Component = 0 Sinc(x) and why it's important Sinc(x/2) is the Fourier transform of a rectangle function. Sinc2(x/2) is the Fourier transform of a triangle function. Sinc2(ax) is the diffraction pattern from a slit. It just crops up everywhere... Transform of the Gaussian function 2 f x Ceax Amplitude: C a e1/ 2 a 0.607 a Standard deviation : f()=Ce-1/2 x 1 2a Fourier transform: 2 Fk f x eikxdx Ceax eikxdx 2 0.607 Fk ek 4a k 2a For a Gaussian function Fourier transform is also a Gaussian and x k 1 Properties of FT The Fourier Transform of a sum of two functions f(t) F() t g(t) G() Y {()af t bgt ()} t aftbgtYY{ ( )} { ( )} F() + f(t)+g(t) G() Also, constants factor out. t Properties of FT : The Fourier Transform of the complex conjugate of a function ** Y ft() F ( ) Proof: ** Y f ()t f ()exp(titdt ) * ft( )exp( itdt ) * f (titdt )exp( [ ] ) F * ( ) Derivative Theorem The Fourier transform of a derivative of a function, f’(t): Y {'()}ft iF () Proof: Y {f '()}tftitdt '()exp( ) Integrate by parts: f ()[(ti )exp( itdt )] Remember that the function must be zero at ±∞, so the other term, [f(t) exp(-it)] +∞ -∞ iftitdt( ) exp( ) vanishes. iF( if t 0 The Dirac delta function ()t 0 if t 0 It’s useful to think of the delta function as the limit of a series of peaked continuous functions. 2 fm(t) = m exp[-(mt) ]/√ (t) f3(t) f2(t) f1(t) t The (Dirac) delta function 0 x 0 (x) x f(x) x 0 x dx 1 x For arbitrary function f(x): 0 x x0 x x0 x x0 f x x dx f 0 f x x x dx 0 f x0 Sifting property - function can extract only one value of f Delta function: Fourier transform 1 Fourier transform of function: f x Fk eikxdk 2 Fk x eikxdx ikx Fk f x e dx Fk eik0 =1 function can be represented as a sum of an infinite number of harmonic components with amplitudes 1: 1 1 x eikxdk eikxdk 2 2 -1 x Y 1 Inverse Fourier transform of unity Delta function: displacement and phase (x) What happens if function is shifted? F k x x eikxdx 0 x0 x amplitude: 1 Fk eikx0 phase: kx0 -1 ixk 0 xx0 Y e Fourier transform of a function that is displaced in space (or time) is the transform of the undisplaced function multiplied by an exponential that is linear in phase ixk 0 YY f xx0 fxe Delta function: inverse Fourier transform 1 f x Fk eikxdk 2 Fk k k 0 Fk f x eikxdx 1 f x k k eikxdk 0 2 1 f x eik0x 2 Superposition of delta functions d d f x x x 2 2 FT of sum of functions is a sum of FT of individual functions: idkk/2 id /2 Y fx e e 2coskd / 2 1 Fk A k iBk Fkei k f x Ak cos kx dk B k sin kx dk 00 Even function: only cosines! If function f(x) is real and even its FT is also real and even..
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