4.4.3 The convolution theorem The convolution theorem says that the Fourier transform of a convolution c (x) of f (x) an

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4 4 3 The Convolution Theorem The Convolution Theorem Says That The Fourier Transform Of A Convolution C X Of F X An 1 (112.14 KiB) Viewed 48 times

Please answer part b only, equation 4.2 is
below:

4 4 3 The Convolution Theorem The Convolution Theorem Says That The Fourier Transform Of A Convolution C X Of F X An 2 (84.04 KiB) Viewed 48 times

4.4.3 The convolution theorem The convolution theorem says that the Fourier transform of a convolution c (x) of f (x) and g(x) is the product of the Fourier transforms of f(x) and g(x). Using the notation F (k) is the Fourier transform of f(x): F(x) = Lt ()exp(-ikx]dx (4.7) we can write C(k)= F(k) G(k). (4.8) The gives us a simple and quick way of obtaining the Fraunhofer diffraction pattern of a series of slits, for example a Young's double slit set-up or a set of multiple slits. [Note that to solve the Optics problem we just need to replace k by ksin (O).) The aperture function of a double slit experiment as convoluted with the sum of two 8-functions: 8(x) = 8(x+d/2)+8(x-d/2). (4.9) Sketch the aperture function of a double slit of slit spacing d and slit width a as a function of x. b) By substituting g(x) from Eq. 4.9 for A (x) in Eq. 4.2 show that the intensity pattern for a dou- ble slit of zero width is given by 7(0) = 4 cos [ sin (O (0) (4.10) 2

= Optics course, the amplitude of the far-field diffrac- tion pattern is the Fourier transform of the transmission function of the diffracting aperture: Ares (0) = LA(x)exp(-ikxsin (O))dr, (4.1) and the observed intensity 1 (0) = |Ares? is given by 1(0) = LA A (x) exp[-ikxsin (0)]dx (4.2) where Ares is the resultant amplitude at an angle e, with e being the angle between the straight through direc- tion and a point on the pattern. Here the aperture func- tion A (x) represents the wave amplitude at a position x along the source, and k = 21/2 for monochromatic light of wavelength 2.