- This must be due to a bending of the light around the edge of S 1 . - If this hole is made small enough, the light will not propagate as a narrow beam but as a spherical wave from the centre of the hole (see Figure 4.2). - This simple principle has proved to be very fruitful and constitutes the foundation of the classical diffraction theory.. - It is easily shown that the diffraction pattern from an opaque strip will be the same as from a slit of the same width.. - where denotes the open aperture of the screen, ds is the differential area, u(P 0 ) is the field incident on the screen and is the angle between the incident and the diffracted rays at point P 0 . - Figure 4.5 Diffraction from a square wave grating. - The function t (x) is called the complex amplitude transmittance of the grating. - on a photographic film) of the interference between a plane wave and a spherical wave (see Figure 4.7(a. - In Section 1.10 (Equation (1.20)) we have shown that a plane wave in the xz-plane with propagation direction an angle θ to the optical axis (the z-axis) will focus to a point in the focal plane of the lens at a distance x f from the z-axis given by. - Figure 4.10 is a reproduction of Figure 4.9 apart from a rescaling of the ordinate axis from x f of dimension length to f x = x f /λf of dimension inverse length, i.e. - Figure 4.9 Diffraction patterns from (a) square wave grating and (b) sinusoidal grating. - (a) The transmittance function of the grating. - (b) The constant term and the first harmonic of the Fourier series. - and (e) The sum of the four first terms of the series. - The transmittance function of the grating is shown dashed. - In that way we get a direct representation of the frequency content or plane- wave content of the gratings. - This would be a proof of the fact that a square-wave grating can be represented by a sum of sinusoidal (cosinusoidal) gratings of frequencies which are integer multiples of the basic frequency f 0 , in other words a Fourier series. - This is further evidenced in Figure 4.11 where, in Figure 4.11(e) we see that the approximation to a square-wave grating is already quite good by adding the four first terms of the series. - To improve the reproduction of the edges of the square wave grating, one has to include the higher-order terms of the series. - t (x, y)u i (x, y) and the field in the x f -plane becomes. - For example, the calculation of the frequency spectrum of a sinusoidal grating given by. - The last equality follows from the definition of the delta function given in Equation (B.11) in Appendix B.2. - f 0 (see Figure 4.10(b)).. - Figure 4.12 shows a point source (1) placed in the focal plane of a lens (2) resulting in a plane wave falling onto a square wave grating (3) which lies in the object plane. - Figure 4.12 Optical filtering process. - lens (4) placed a distance a from the object plane, images the square wave grating on to the image plane where the intensity distribution (5) of the image of the grating can be observed.. - We can therefore, in the same way as in Figure 4.8, observe the spectrum of the grating in the back focal plane of the lens.. - The distance between the focal points now becomes 2λff 0 and the basic frequency of the imaged grating will be 2f i. - Such a manipulation of the spectrum is called optical filtering and the back focal plane of lens (4) is called the filter plane. - nλ (4.21) This is illustrated in Figure 4.13.. - Figure 4.14 shows the same set-up as in Figure 4.8 apart from the grating being moved to the other side of the lens a distance s from the back focal plane. - Figure 4.13. - Figure 4.14. - Figure 4.15. - in the focal plane independent of x, i.e. - Figure 4.16. - The conclusions of the above considerations are collected in Figure 4.16. - Here, P i is the image point of the point source P 0 on the optical axis. - The important point to note is that independent of the positioning of the grating (or another transparent object) in the light path between P 0 and P i , the spectrum will be found in the x f -plane. - (1) to the right of the lens at a distance s b from P i. - (2) to the left of the lens at a distance s a from P 0. - The object in the x-plane is imaged to the x i -plane while the filter plane is in the x f -plane, the image plane of the point source P 0. - The film is placed just to the right of the lens a distance s b from the filter plane. - When the hole in the filter plane is made wide enough to let through this modulated side order, the image of the object on the film will be covered by moir´e or speckle Fourier fringes. - In Section 4.5 we have shown that by placing a transparent diffracting object anywhere in the light path between P 0 and P i of Figure 4.16, we obtain the spectrum (i.e the Fourier transform) of the object in the x f -plane. - where D is the diameter of the lens, J 1 is the first-order Bessel function and where we also have made use of the similarity theorem, Equation (B.2b), Appendix B.1. - The intensity distribution in the x f , y f -plane in Figure 4.16 with a point source at P 0. - The first minimum of J 1 (π x) occurs for x = 1.22 which gives for the radius of the so-called Airy disc. - Figure 4.19 The Airy pattern. - (a) Intensity distribution and (b) Photograph of the pattern. - With this model of the imaging system, it is possible to associate all diffraction lim- itations with either the light propagation from object to entrance pupil or from the exit pupil to the image. - Figure 4.12). - h(x i − mx o , y i − mx o )u o (x o , y o ) dx o dy o (4.36) h(x i − mx o , y i − my o ) is termed the impulse response of the imaging system since it gives the response of an impulse in spatial coordinates, i.e. - Such a lens is said to be diffraction limited since its performance is limited only by the size of the exit pupil.. - Figure 4.20 shows a plot of the coherent transfer function for a diffraction limited lens with a circular aperture along the f x -axis. - which is called the cut-off frequency of the coherent transfer function.. - results in the symmetrical expression. - The denominator simply normalizes this area of overlap by the total area of the pupil. - Figure 4.21 Geometry of the area of overlap when calculating the optical transfer function of a diffraction-limited system with a square pupil. - (4.50) where is the triangle function (see Equation (B.15), Appendix B.2) and f 0 is the cut-off frequency of the same system used with coherent illumination. - We therefore have that the cut-off frequency for the incoherent transfer function is twice that of the coherent transfer function.. - H (f x , f y ) is commonly known as the optical transfer function (OTF) of the system.. - Figure 4.22 shows | H (f x. - 2 δ(f x + f g ) (4.54) The spectrum of the ‘output’ image becomes. - This means that the modulation (or the visibility, see Section 3.3) of the input sinusoidal grating has changed by a factor | H (f g. - the value of the MTF at frequency f g. - From the definition of the autocorrelation integral (see Equation (B.2f), Appendix B.1) we therefore can write the frequency spectra of the image intensity in the two cases as. - Another aspect of the imaging properties of a lens is the depth of focus. - Consider Figure 4.23 where the rays passing through the edge of the lens aperture from an on-axis point source are drawn.. - Let us take the depth of focus to be the distance b when the radius of the light spot on the image. - Figure 4.23. - plane is equal to the radius of the Airy disc r f . - which by inserting the value of the radius of the Airy disc, Equation (4.33), gives b = 2.44λ. - From this expression we see that the depth of focus is inversely proportional to the square of the lens aperture or proportional to the square of the aperture number.. - The spectrum of such a grating is indicated in Figure 4.24(b) which should be compared with the spectrum of the grating given in Equation (4.61) in Figure 4.24(a). - We see that the side orders have become broader because of the slowly varying function ψ (x). - Figure 4.24 The spectrum of (a) a sinusoidal grating of period d and (b) the same grating phase- modulated. - After development, I (x) becomes equal to the transmittance function of the film. - In the filter plane we place a screen with a hole a distance λf/d from the optical axis and with an opening wide enough to transmit the full width of one of the modulated side orders given in Figure 4.24(b). - Figure 4.25 The spectrum of the gratings in Figure 4.24(a, b) multiplied. - How big a light spot would be produced on the sur- face of the moon a distance of km away? Neglect any effects of the Earth’s atmosphere.. - 4.3 Imagine an opaque screen containing thirty randomly located circular holes of the same diameter. - 4.4 Determine the Fourier transform of the function U (x). - 4.8 Make a sketch of the resulting function arising from the convolution of the two functions depicted in Figure P4.3.. - where t = 2πρr 0 and r 0 is the radius of the aperture. - By comparison with Equation (4.29) we obtain another representation of the Bessel function J 1 (t):. - (a) Find the intensity distribution in the Fraunhofer diffraction pattern of the double- slit aperture shown in Figure P4.4.. - (b) Sketch the intensity distribution along the x f -axis of the observation plane. - 4.18 Find an expression for the intensity distribution in the Fraunhofer diffraction pattern of the aperture shown in Figure P4.5. - Assuming L = 10 mm and f 0 = 10 lines/mm, sketch the intensity distribution across the x f -axis of the focal plane. - Indicate the numerical values of the distance between the diffracted components and the width (between first zeros) of the individual components.. - A square stop, of width D/2, is placed at the centre of the exit pupil, as shown in Figure P4.5.. - (b) Sketch the limiting form of the optical transfer function as the size of the stop approaches the size of the exit pupil.. - (a) Find the Fourier spectrum of the field-amplitude distribution transmitted by the object screen.. - on the lens axis in the focal plane. - Sketch the resulting intensity distribution in the image plane.. - 4.24 The so-called central dark ground method for observing phase objects is achieved by placing a small opaque stop in the back focal plane of the imaging lens to block the undiffracted light. - 4.25 According to the so-called Rayleigh criterion of resolution, two incoherent point sources are just resolved by a diffraction-limited system when the centre of the Airy pattern generated by one source falls on the first zero of the Airy pattern generated by the second. - (a) Find s in terms of the exit pupil diameter D, the image distance b and the wavelength. - (b) Calculate the ratio between the intensity of the central dip and the maximum.
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