Lectures 18-20: Diffraction Reading: Kurt Möller, Optics, Chapter 3 Matt Young, Lasers and Optics, Section 5.5 Lipson, Lipson and Tannhauser, Optical Physics, Chapter 7 Many experimental situations in optics feature a coherent superposition of waves from an extended source which have slight differences in phase introduced as a result of travelling slightly different optical paths. We have discussed this situation for the problem of interference, in which the width of the slits was assumed to play no role at all. We must now consider the effect of slits which are finite. We shall find that the same notions of interference will serve us well as guides to what happens in this case, except that we have to integrate over the extended aperture instead of summing over point-like apertures. We shall also calculate some cases which are regularly encountered in optical practice. 1. Diffraction from a single rectangular slit We have learned that interference from multiple slits produces a unique intensity distribution at a distance from the slits. However, up to now we have not examined the effects of varying the spacing between the slits. The problem of the interference from neighboring elements of a finite aperture is called Fraunhofer diffraction, after the man who first described it. (And also, incidentally, was the first to measure the solar spectrum using a prism spectrometer — the spectrum which eventually led to the discovery of blackbody radiation.) We start by calculating the interference pattern from 40 slits, spaced 50l apart; this means that the slit pattern occupies a total width of 1 mm. 2 Diffraction.nb Wavelength = 500.0 * 10^ -9 Distance = 1.0 Nslits = 40 Separation = 50 * Wavelength Phase = Pi * Sin x * Separation Wavelength Plot 4 * Sum Cos j - 1 * Phase , j, 1, Nslits ^2, x, -0.002, 0.002 , PlotRange -> All, PlotStyle -> RGBColor 1, 0, 0 , AxesLabel -> "Argument", "Amplitude" 5. ¥ 10-7 1. 4 0.00025 1570.8 Sin x Amplitude 60 50 40 30 20 10 Argument -0.002 -0.001 0.001 0.002 Ö Graphics Ö You will note that the secondary maxima in this case seem to be of quite different amplitudes, and the width of the central intensity maximum is aproximately 1 mm, as you can read off from the Argument (units in meters). Thought Question: Why does this pattern look so different from that of the multiple-slit interference pattern we calculated earlier? Where does the next principal maximum fall in this case? [Hint: Try extending the limits of the argument to plot much greater distances.] Ö Graphics Ö Diffraction.nb 3 Fraunhofer's solution to the diffraction problem is depicted in the drawing above this paragraph. We imagine the slit to be long and narrow, with a width d, so that the solution in a plane vertical to the screen is the same at every point. We subdivide the slit into small elements of area, and add up the contributions to the electric field at the observation angle J from the entire linear extent of the slit. This procedure yields the following for a line element dx of the slit Dxj Dxj i kzo -wt -i kxj sinJ i kzo -wt -i kxj sinJ Ej = Eo ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ e e îEtotal = Eo ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ e e D j D i kz -wt where the wave at the center of the aperture is represented by Eo e o and the phase shift of each element of length with respect to that central plane wave is given by e-i kxj sinJ . But we know from integral calculus what to do with this sort of thing when we see it: We take the limit as the number of line elements gets larger and their individual size gets smaller. So now we use Mathematica to help us do the definite integral without pain. To avoid confusing Mathematica about the variables of integration, it turns out to be useful to define an auxiliary variable HWidth which is just half the slit width D. We also set a constant A=k·sinJ. Now ask Mathematica to oblige us with the definite integral: Hwidth=0.5*Dwidth Integrate[Exp[I*A*x]/Dwidth,{x,-Hwidth,Hwidth}] 0.5 Dwidth -0.5IADwidth 0.5IA Dwidth ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄI E - ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄI E A Dwidth A Dwidth But this, if we remember our complex variables, is simply related to the sine function. If we multiply the numerator and denominator of this expression by (-2i), we get 4 Diffraction.nb 2 sin ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄkDsinJ sin ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄkDsinJ E = E ei kzo -wt ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ2 î I = E ei kzo -wt ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ2 É total o kDsinJ total o kDsinJ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ2 ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ2 The absolute square then simply leaves us with the highly nonlinear expression for the diffracted intensity from a long, narrow slit as kDsinJ 2 sin ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ 2 2 Itotal = Eo ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄkDsinJ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ2 Now consider this exact analytical solution as it was discovered by Fraunhofer. We take as an example a narrow aperture which is 1 mm wide and very tall. We know from experience that the main effect we should expect from shining light through a large aperture is that a spot of light approximately the size of the aperture should appear on a screen just opposite it. The exact solution is plotted below for this case. Wavelength = 550.0 * 10^ -9 Distance = 1.0 SlitWidth = 0.001 Wavek = 2 * Pi Wavelength Plot Sin 0.5 * Wavek * SlitWidth * x 0.5 * Wavek * SlitWidth * x ^2, x, -0.002, 0.002 , PlotRange -> All, PlotStyle -> RGBColor 1, 0, 0 , AxesLabel -> "Argument", "Amplitude" 5.5 ¥ 10-7 1. 0.001 1.1424 ¥ 107 Amplitude 1 0.8 0.6 0.4 0.2 Argument -0.002 -0.001 0.001 0.002 Ö Graphics Ö Diffraction.nb 5 The function which describes this diffraction pattern is so important that it has a special name: the sinc function. We will see it in a number of other applications in optics where we have to deal with coherence effects over a finite slit width. Homework Exercises 1. Examine the sinc function carefully. Which factor is varying faster as a function of x, the numerator or denominator? Where are the zeros and secondary maxima of the sinc function? Do they agree with what you see plotted in the graph? 2. The Fraunhofer diffraction pattern derived here is for an aperture which is finite in one dimension and infinitely long in the other. What do you think the diffracted intensity distribution would look like for a rectangular aperture with widths A and B? Can you sketch the shape of this function without resorting to the computer? Can you plot this function in Mathematica? 2. Interference by multiple finite slits If diffraction occurs for any finite aperture, we must revisit our ideas about multiple-slit interference. If we have a set of N slits with finite apertures of width a, spaced a distance b apart, we expect the resultant intensity distribution to be a superposition of the amplitudes from N slits each one modified by the diffraction pattern due to the finite slit width. Let us revisit the multiple-slit interference problem with this in mind, using the diagram shown below. 6 Diffraction.nb The phase shift Df between any two successive slits is found in the usual way, by computing the optical path difference d. This gives the following : 2 p i kzo -wt ijàDf Df= ÄÄÄÄÄÄÄÄÄÄl nd = kndîEj = Eo e e The total electric-field amplitude ET due to the N slits is then found by summing over all of them using the formula for summing a geometric series to yield: N-1 N-1 1 - eiNàDf E = E ei kzo -wt eijàDf = E ei kzo -wt eijàDf = E ei kzo -wt ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ T o o o 1 - eiàDf j=0 j=0 Homework Exercise: Convince yourself that this expression is correct. Then factor the numerator and denominator to produce the final result for the amplitude and intensity shown in the equation below. eiNàDf 2 e-iNàDf 2 - eiNàDf 2 E = E ei kzo -wt ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = T o eiàDf 2 e-iàDf 2 - eiàDf 2 e-iNàDf 2 - eiNàDf 2 sin NàDf 2 E ei kzo -wt ei N-1 àDf 2 ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = E ei kzo -wt ei N-1 àDf 2 ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ o e-iàDf 2 - eiàDf 2 o sin Df 2 Notice that this is not the same as the sinc function! Finally, then, the intensity due to the N slits is given by the complex conjugate squared of this expression, in which all the imaginary exponentials multiply out to 1, yielding Diffraction.nb 7 sin NàDf 2 2 I = E2 ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ T o sin Df 2 To obtain the multiple slit pattern including the effects of diffraction, we have to multiply the amplitude for the multiple-slit interference by the sinc function. Notice that we have defined a phase factor for the multiple slit interference and the single-slit diffraction separately; both have to be divided by the wavelength in order to get the phase shift. To make the programming more transparent, we have used the standard Mathematica way of defining functions. We have also tacitly assumed that our computer "experiments" are being carried out in air (index of refraction n=1). Sinc z_ := Sin z z NSlits = 10 Lambda1 = 600.0 * 10^ -9 a = 50.0 * Lambda1 d = 30.0 * Lambda1 Int z_ := Pi * a * z Diff z_ := Pi * d * z Plot Sinc Diff x Lambda1 * Sin NSlits * Int x Lambda1 Sin Int x Lambda1 ^2, x, -0.06, 0.06 , PlotStyle -> RGBColor 1, 0, 0 , PlotRange -> All 10 6.