Lecture 20. Fraunhofer Diffraction - Transforms

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Bronisława Popławska
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1 Lecture 20 Fraunhofer Diffraction - Transforms

2 Fraunhofer Diffraction U P ' &iku 0 2 zz ) e ik PS m A exp ' &iku 0 2 zz ) e ik PS exp ik 2z a [x 2 m % y 2 m ]. m A exp ik 2z a & x m ) 2 % (y 0 & y m ) 2 ds ik z a x m % y 0 y m dx 0 dy 0 This is effectively saying that BOTH the source and the observation point are far from the aperture

3 Fraunhofer Diffraction If we now generalise this by replacing the plane wave from the source (U 0 /z ) by a general distribution across the aperture U ) Recognise that in a paraxial approximation a long way from the aperture it is the ANGLES that matter, not the positions u = kx m /z a, v = ky m /z a U P ' &iku 0 2 zz ) e ik PS exp ik 2z a [x 2 m % y 2 m ]. m A exp ik z a x m % y 0 y m dx 0 dy 0 ' MOC&FF ma U 0 )exp(&i [ux 0 % vy 0 ])dx 0 dy 0 In case you don t recognise it - the final expression is a 2-

4 D Fourier Transform relation! Fraunhofer Diffraction U P '. &ik 2 z e ik PS exp iz a 2k [u 2 % v 2 ] m A U 0 )exp(&i [ux 0 % vy 0 ])dx 0 dy 0 relates x 0, y 0 space to u, v space e.g. for a square aperture and a plane wave - U P ' &ik 2 z iz exp 2k [u 2 % v 2 ] U a R x R y sinc R x u 2 sinc R y v 2 the intensity is the square of this function For a circular function it is almost the same except that it s a Bessel function, not a sinc function

5 Fraunhofer Diffraction U P '. &ik 2 z e ik PS exp iz a 2k [u 2 % v 2 ] m A U 0 )exp(&i [ux 0 % vy 0 ])dx 0 dy 0 If this is so then we have a powerful technique Applies for large distances Applies for paraxial approximation Maybe could relieve those with thought BUT any U 0 ) can be used We have computers - Have FFT - will compute!

6 Slits and Gratings Consider a slit of dimension a,b at origin in x 0, y 0 plane with light incident at angle Wavefront is exp(ikx 0 sin )rect /a)rect(y 0 /b) Also called the aperture function Now consider a whole set of slits (a grating) use a convolution Wavefront is exp(ikx 0 sin )rect /a)rect(y 0 /b)*comb / ) But that s infinite - need it finite exp(ikx 0 sin )rect /a)rect(y 0 /b)* [comb / )rect /(n ))] For the angular distribution - just need the Fourier Transform Assume b infinite

7 Slits and Gratings Û{exp(ikx 0 sin )rect /a) * [comb / )rect /(n ))]} Û{exp(ikx 0 sin )rect /a) Û{[comb / )rect /(n ))]} Use shift property + six transforms on first term a sinc( a[u + ksin ]/2 ) Û{[comb / )rect /(n ))]} Second term is a finite sum of function Û (N&1)/2 j n'&(n&1)/2 (N&1)/2 & n ) ' j n'&(n&1)/2 exp(&inu ) and the total result becomes a sinc [sin % sin i ] a/ sin N [sin % sin i ] sin [sin % sin i ]

8 Grating Resolving Power Peaks are from sin ratio (squared for intensity) Envelope is from the sinc function For normal incidence close to axis i =0 then width of peak is of order ( N / )cos = Maxima occur at ( / )sin = m cos = m combining these / = mn at 500 slits/mm and 50mm grating resolving power is 2.5 x 10 4 if m = 1

9 Grating Energy Split a sinc [sin % sin i ] a/ sin N [sin % sin i ] sin [sin % sin i ] This is the product of the FT of the slit (broad) the repetitive structure (narrow) The slit peaks the energy in the forward direction m=0 where there is no dispersion (resolving power)!!! - bad idea!! Playing with the slit might permit us to move the peak energy from straight on (or straight back for reflection) to another order

10 Grating Energy Split Suppose we use a (reflection) grating for which the slit height is h = h 0 x 0 tan - ie a sawtooth Phase delay is composed of two parts phase delay of tilted plane wave - kx 0 sin i phase delay due to sawtooth - hcos i Total phase delay is the sum of these and therefore result is

11 Grating Energy Split sinc [sin % sin i % tan cos i ] / sin N [sin % sin i ] sin [sin % sin i ] Notice that if = 0 - grating disappears!! if m / + sin i + cos i tan = 0 the power peaks in order m

12 A Lens as an Aperture If we compute the phase delay across a lens we can make it an aperture function! Take a plane before the lens and a plane after Can be shown that in the paraxial approximation the thickness of the material is... ) ' 0 & x 2 0 %y /R 1 & 1/R 2 and the phase delay is given by ' kn 0 & k 2f x 2 0 % y 2 0

13 A Lens as an Aperture Take the pupil function (ie physical extent of the lens) as P ) - almost always a cylinder function Take the incoming wavefront as t ) - e.g. plane wave Take the phase manipulation of the lens Take the z dimension to be the focal length - f Ignore phase factors and quadratic terms %4 mm&4 P ) t )exp& 2 i f x 0 x % y 0 y dx 0 dy 0 Its a Fourier transform again!! Focal plane image is FT of aperture function in front of lens

14 %4 mm&4 A Lens as an Aperture P ) t )exp& 2 i f x 0 x % y 0 y dx 0 dy 0 if input wave is plane and P is a cylinder function cyl(r/r) Image plane is an airy function of size 2 ( 1.22 f/(2r)) Or using f/(2r) as the f-number f * So if = 500nm and f * of order 1 image size is of order 1µm Generally f * > 1 so spot size is > 1µm

15 A Lens as an Aperture In the formula above low spatial frequencies in incoming wave turn up near centre high spatial frequencies turn up away from the centre By putting an aperture at the focal point we can clean up a mucky plane wave - only allow low frequency components through

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