Lectures 18-20: DiffractionReading: Kurt Möller, Optics, Chapter 3

Matt Young, Lasers and Optics, Section 5.5Lipson, Lipson and Tannhauser, Optical Physics, Chapter 7

Many experimental situations in optics feature a coherent superposition of waves from an extended source whichhave slight differences in phase introduced as a result of travelling slightly different optical paths. We havediscussed this situation for the problem of interference, in which the width of the slits was assumed to play norole at all. We must now consider the effect of slits which are finite. We shall find that the same notions ofinterference will serve us well as guides to what happens in this case, except that we have to integrate over theextended aperture instead of summing over point-like apertures. We shall also calculate some cases which areregularly encountered in optical practice.

1. Diffraction from a single rectangular slitWe have learned that interference from multiple slits produces a unique intensity distribution at a distance fromthe slits. However, up to now we have not examined the effects of varying the spacing between the slits. Theproblem of the interference from neighboring elements of a finite aperture is called Fraunhofer diffraction, afterthe man who first described it. (And also, incidentally, was the first to measure the solar spectrum using a prismspectrometer — 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 patternoccupies a total width of 1 mm.

Wavelength = 500.0* 10^ -9Distance = 1.0Nslits = 40Separation = 50* WavelengthPhase = Pi*Sin x * Separation WavelengthPlot 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"

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You will note that the secondary maxima in this case seem to be of quite different amplitudes, and the width ofthe 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 wecalculated earlier? Where does the next principal maximum fall in this case? [Hint: Try extending the limits ofthe argument to plot much greater distances.]

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Fraunhofer's solution to the diffraction problem is depicted in the drawing above this paragraph. We imagine theslit to be long and narrow, with a width d, so that the solution in a plane vertical to the screen is the same atevery point. We subdivide the slit into small elements of area, and add up the contributions to the electric fieldat the observation angle J from the entire linear extent of the slit. This procedure yields the following for a lineelement dx of the slit

Ej = Eo Dx jÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄD

ei kzo -wt e-i kx j sinJ îEtotal =j

Eo DxjÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄD

ei kzo -wt e-i kxj sinJ

where the wave at the center of the aperture is represented by Eo ei kzo -wt and the phase shift of each element oflength with respect to that central plane wave is given by e-i kxj sinJ . But we know from integral calculus whatto do with this sort of thing when we see it: We take the limit as the number of line elements gets larger andtheir 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 auxiliaryvariable HWidth which is just half the slit width D. We also set a constant A=k·sinJ. Now ask Mathematica tooblige us with the definite integral:

Hwidth=0.5*DwidthIntegrate[Exp[I*A*x]/Dwidth,{x,-Hwidth,Hwidth}]

0.5 Dwidth

I E-0.5IADwidth

ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄA Dwidth

-I E0.5IA Dwidth

ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄA Dwidth

But this, if we remember our complex variables, is simply related to the sine function. If we multiply thenumerator and denominator of this expression by (-2i), we get

Diffraction.nb 3

Etotal = Eo ei kzo -wt sin kDsinJÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ2ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ

kDsinJÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ2

îItotal = Eo ei kzo -wt sin kDsinJÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ2ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ

kDsinJÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ2

É2

The absolute square then simply leaves us with the highly nonlinear expression for the diffracted intensity from along, narrow slit as

Itotal = Eo2

sin kDsinJÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ2ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄkDsinJÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ2

2

Now consider this exact analytical solution as it was discovered by Fraunhofer. We take as an example a narrowaperture which is 1 mm wide and very tall. We know from experience that the main effect we should expectfrom shining light through a large aperture is that a spot of light approximately the size of the aperture shouldappear on a screen just opposite it. The exact solution is plotted below for this case.

Wavelength = 550.0* 10^ -9Distance = 1.0SlitWidth = 0.001Wavek = 2* Pi WavelengthPlot 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"

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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 withcoherence 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 ordenominator? Where are the zeros and secondary maxima of the sinc function? Do they agree with what yousee plotted in the graph?

2. The Fraunhofer diffraction pattern derived here is for an aperture which is finite in one dimension andinfinitely long in the other. What do you think the diffracted intensity distribution would look like for arectangular aperture with widths A and B? Can you sketch the shape of this function without resorting to thecomputer? Can you plot this function in Mathematica?

2. Interference by multiple finite slitsIf diffraction occurs for any finite aperture, we must revisit our ideas about multiple-slit interference. If we havea set of N slits with finite apertures of width a, spaced a distance b apart, we expect the resultant intensitydistribution to be a superposition of the amplitudes from N slits each one modified by the diffraction pattern dueto the finite slit width. Let us revisit the multiple-slit interference problem with this in mind, using the diagramshown below.

Diffraction.nb 5

The phase shift Df between any two successive slits is found in the usual way, by computing the optical pathdifference d. This gives the following :

Df= 2 pÄÄÄÄÄÄÄÄÄÄl nd = kndî E j = Eo ei kzo -wt eij◊Df

The total electric-field amplitude ET due to the N slits is then found by summing over all of them using theformula for summing a geometric series to yield:

ET =j=0

N-1

Eo ei kzo -wt eij◊Df = Eo ei kzo -wt j=0

N-1

eij◊Df = Eo ei kzo -wt 1 - eiN◊Df

ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ1 - ei◊Df

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.

ET = Eo ei kzo -wt eiN◊Df 2

ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄei◊Df 2

e-iN◊Df 2 - eiN◊Df 2

ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄe-i◊Df 2 - ei◊Df 2

=

Eo ei kzo -wt ei N-1 ◊Df 2 e-iN◊Df 2 - eiN◊Df 2

ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄe-i◊Df 2 - ei◊Df 2

= Eo ei kzo -wt ei N-1 ◊Df 2 sin N ◊Df 2ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ

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 thecomplex conjugate squared of this expression, in which all the imaginary exponentials multiply out to 1, yielding

6 Diffraction.nb

IT = Eo2

sin N ◊Df 2ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ

sin Df 2

2

To obtain the multiple slit pattern including the effects of diffraction, we have to multiply the amplitude for themultiple-slit interference by the sinc function. Notice that we have defined a phase factor for the multiple slitinterference and the single-slit diffraction separately; both have to be divided by the wavelength in order to getthe phase shift. To make the programming more transparent, we have used the standard Mathematica way ofdefining 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 zNSlits = 10Lambda1 = 600.0*10^ -9a = 50.0*Lambda1d = 30.0*Lambda1Int z_ := Pi* a*zDiff z_ := Pi* d*zPlot 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

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Diffraction.nb 7

Homework Exercises

1. At what values of the angle J do the secondary maxima appear in this plot? (You will have to go back to thedefinition of the phase shift to figure this one out, but it is not difficult.) These secondary maxima are actuallythe principal maxima of the multiple-slit interference pattern, and are generally referred to as the "maxima oforder n," where n=0, ±1, ±2, .... The plus and minus signs refer to the right and left sides of the principalmaximum of order 0.

2. Modify the plotting program above to display (in separate colors), the diffraction pattern and the interferencepattern as well as their product. How is the diffraction envelope and the position of the maxima of variousorders affected by changes in the slit width and separation?

3. The diffraction gratingA practical application of the principles we have been discussing is the diffraction grating, which consists ofmultiple apertures or even multiple lines ruled very close together on a transmissive or reflective surface. To seewhat the effect of many apertures close together is on a source with more than one color, carry out the followinghomework exercises.

Homework Exercises

1. Modify the program for multiple-slit diffraction with finite apertures so that you can plot the diffractionpattern for three different colors. You can use the previous program as a template. To begin, use wavelengths ofl1 =400 nm, l2 =500 nm and l3 =600 nm, and choose the slit widths and separations to be multiples of thecenter wavelength, d=10l2 and a=20l2 . Plot a few examples to get a feeling for how slit width, number of slits,and slit separation affect the maxima for the various colors.

2. For a wavelength of 500 nm, a slit width of d=10l2 and a slit separation of a=20l2 , how far apart in angle arethe maxima of order 0, 1 and 2 if you have 100 slits? If you were to place a detector on a screen 1 m from thediffraction grating, how far apart would these maxima be in distance? Suppose that we turn the problem around,and have a photodiode array with elements spaced 50 mm apart ("50 mm pitch"). How far from the diffractiongrating would you have to locate the detector plane to achieve a resolution in wavelength of 3 nm betweenpixels?

3. In a diffraction grating, the ability to separate various wavelengths is referred to as the resolving power of thegrating, and it is defined as the ratio l/Dl. Calculate the diffraction pattern for red, green and blue light for agrating with 100 lines out to fourth order. Looking at the resolving power as a function of wavelength, what canyou say about the advantages and disadvantages of designing a grating spectrometer to work in first order or infourth order? Why do you not work in zeroth order in spite of the greater intensity of the signals?

4. Diffraction from a rectangular apertureFraunhofer diffraction from a rectangular aperture is a straightforward extension of Fraunhofer diffraction for along, narrow slit. One simply integrates over two dimensions instead of one. The variables used in thefollowing Mathematica calculation are taken from the diagram below. Note the sizes of the slits used!

8 Diffraction.nb

Sinc z_ := Sin z zDistance = 1.0Lambda = 550.0*10^ -9SlitX = 0.00000075SlitY = 0.0000005PhaseX = 2.0*Pi* SlitX 2.0* LambdaPhaseY = 2.0*Pi* SlitY 2.0* LambdaPlot3D Sinc PhaseX*x Distance *Sinc PhaseY* y Distance ^2 ,x, -2 Pi, 2 Pi , y, -2 Pi, 2 Pi , PlotRange -> 0, 0.005 ,PlotPoints -> 40

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Diffraction.nb 9

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Homework Problems

1. If you change the PlotRange command in the program to PlotRange->All, what is the peak amplitude of thediffraction pattern? Which of the secondary maxima can you see then? Is that consistent with the amplitude ofthe secondary maxima you calculated for a one-dimensional slit?

2. Plot the diffraction patterns for rectangular apertures in which the dimension b of the long side is related tothe size of the short side a by: b=a, b=2a, b=5a, b=10a. Comment on any changes you observe. Do they agreewith your intuition?

5. Diffraction from a circular apertureDiffraction from a circular aperture is very much like diffraction from an extended rectangular aperture, exceptfor the difference in symmetry. The mathematical problem is "out of bounds" for this course, but the solutionsare the Bessel functions of order 1, which can be calculated by Mathematica. (Thank goodness!) As in the caseof the rectangular aperture, we find the intensity distribution by integrating the phase differences over a circularaperture.

The Bessel function has a series expansion as follows:

J1 g = gÄÄÄÄÄÄ2 - 1ÄÄÄÄÄÄÄÄÄÄÄÄÄ12 ◊2 gÄÄÄÄÄÄ23 + 1ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ12 ◊22 ◊3 gÄÄÄÄÄÄ2

5 - ...

where the argument g is equal to

g= kÄÄÄÄÄÄ2 D ◊sinJ

Homework Problem

1. Plot the Bessel function BesselJ[1,x] using Mathematica and determine where the first few zeros of thefunction occur.

In this next section, we will use Mathematica to compute the diffraction pattern from a circular disk, whilemaking the small-angle approximation sinJ≈J.

10 Diffraction.nb

Wavelength = 500.0* 10^ -9Distance = 1.0ApertureRadius = 0.0005Wavek = 2.0* Pi Wavelength 2.0r = Sqrt x^2 + y^2Plot3D BesselJ 1, Wavek *ApertureRadius* r Distance

Wavek* ApertureRadius*r Distance ^2,x, -0.002, 0.002 , y, -0.002, 0.002 , PlotPoints -> 40,PlotRange -> 0.0, 0.015

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This pattern of circular rings is known as an "Airy pattern," after the Irish physicist who first studied thediffraction from a circular aperture in some detail. The central disk in this pattern is called the "Airy disk," and itwill play a role in our discussion about limits of resolution.

Diffraction.nb 11

6. Limits of resolutionIf two light sources are close together, the far-field intensity distributions of the two sources may overlap andmake it impossible to separate the sources. In this section, we examine this problem quantitatively, and developthe Rayleigh criterion for resolving power.

We begin our study of the question of resolving power by posing an astronomical problem: Two sources,located a great distance from the earth, are so close that we do not know whether we will see a single bright disk,or two slightly separated circular disks which look to the eye like an ellipse, or two well-resolved disks. We firstcompute the sum of the two diffraction patterns for two sources separated by some distance S.

Thought Question : We are going to plot the incoherent sum of the two diffraction patterns. Is this justified?Can you think of a case when it might not be justified?

Wavelength = 500.0* 10^ -9Distance = 1.0Separation = 0.001ApertureRadius = 0.0005Wavek = 2.0* Pi Wavelength 2.0r1 = Sqrt x - Separation ^2 + y^2r2 = Sqrt x + Separation ^2 + y^2Plot3D BesselJ 1, Wavek *ApertureRadius* r1 Distance

Wavek* ApertureRadius*r1 Distance ^2 +BesselJ 1, Wavek *ApertureRadius* r2 Distance

Wavek* ApertureRadius*r2 Distance ^2,x, -0.002, 0.002 , y, -0.002, 0.002 , PlotPoints -> 40,PlotRange -> All

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Now start reducing the separation and see how large it is compared to the slit dimensions when the diffractionpatterns begin to overlap. What happens to the overlap when you change the wavelength? Does this surpriseyou? Why or why not?

Diffraction.nb 13

Wavelength = 500.0* 10^ -9Distance = 1.0Separation = 0.00035ApertureRadius = 0.0005Wavek = 2.0* Pi Wavelength 2.0r1 = Sqrt x - Separation ^2 + y^2r2 = Sqrt x + Separation ^2 + y^2Plot3D BesselJ 1, Wavek *ApertureRadius* r1 Distance

Wavek* ApertureRadius*r1 Distance ^2 +BesselJ 1, Wavek *ApertureRadius* r2 Distance

Wavek* ApertureRadius*r2 Distance ^2,x, -0.002, 0.002 , y, -0.002, 0.002 , PlotPoints -> 40,PlotRange -> All

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To get an idea of the quantitative relationship between the diffraction pattern and the resolving power, let us takea section through the Bessel function pattern for two sources separated by a distance S, and look at how theirdiffraction patterns compare. We will arbitrarily take the y-coordinate to be equal to zero.

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Wavelength = 500.0* 10^ -9Distance = 1.0Separation = 0.0005ApertureRadius = 0.0005Wavek = 2.0* Pi Wavelengthr1 = Sqrt x - 0.5*Separation ^2r2 = Sqrt x + 0.5*Separation ^2Plot BesselJ 1, Wavek *ApertureRadius* r1 Distance

Wavek* ApertureRadius*r1 Distance ^2,BesselJ 1, Wavek *ApertureRadius* r2 Distance

Wavek* ApertureRadius*r2 Distance ^2,BesselJ 1, Wavek *ApertureRadius* r1 Distance

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For the sake of definiteness, Lord Rayleigh proposed long ago that the criterion for two sources to be justresolvable would be for the principal maximum from one to fall just on top of the first secondary maximum ofthe other. About what slit separation does it take to do this, and what is the argument of the Bessel functionwhen this criterion is satisfied? Are you surprised that we get the ratio of wavelength to aperture diameter andnot something that involves the wavelength alone?

Homework Problems

1. A car is driving through a moderately dense fog with yellow (sodium, l=590 nm) fog lamps on. Assume thatthe fog lamps are 1.5 meters apart. Using the Rayleigh criterion, determine at what distance your fullydark-adapted eyes will be able to determine that what you see coming toward you is two headlights rather thanone. If you compute the diffraction pattern using Mathematica, do you see a pattern which agrees with this?

7. Imaging a lattice of apertures

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