Download Lectures 18-20: Diffraction

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
-0.002
-0.001
0.001
Argument
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
Dx j
E j = Eo ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ei
D
kzo -wt
e-i
kx j sinJ
îEtotal =
j
Dx j
Eo ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ei
D
kzo -wt
e-i
kx j sinJ
where the wave at the center of the aperture is represented by Eo ei kzo -wt 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
I E-0.5IADwidth
I E0.5IA Dwidth
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ - ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
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
Etotal = Eo ei
kzo -wt
kDsinJ
sin ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
2
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
îItotal = Eo ei
kDsinJ
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
2
kzo -wt
kDsinJ
sin ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
2
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
kDsinJ
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
2
2
É
The absolute square then simply leaves us with the highly nonlinear expression for the diffracted intensity from a
long, narrow slit as
kDsinJ
sin ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
2
Itotal = E2o ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
kDsinJ
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
2
2
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
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
-0.002
-0.001
Ö Graphics Ö
0.001
Argument
0.002
^2,
Diffraction.nb
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.
5
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 :
2p
i kzo -wt ij◊Df
Df= ÄÄÄÄÄÄÄÄÄÄ
e
l nd = kndî E j = Eo 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
ET =
N-1
Eo ei
kzo -wt
eij◊Df = Eo ei
kzo -wt
j=0
eij◊Df = Eo ei
j=0
kzo -wt
1 - eiN◊Df
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
1 - 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
Eo ei
eiN◊Df 2 e-iN◊Df 2 - eiN◊Df 2
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
=
ei◊Df 2
e-i◊Df 2 - ei◊Df 2
-iN◊
Df
2
iN◊
Df
2
e
-e
ei N-1 ◊Df 2 ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
= Eo ei
-i◊
Df
2
e
- ei◊Df 2
kzo -wt
kzo -wt
kzo -wt
ei
N-1 ◊Df 2
sin N ◊Df 2
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
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 N ◊Df 2
IT = E2o ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
sin Df 2
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
x, -0.06, 0.06 , PlotStyle -> RGBColor 1, 0, 0 ,
PlotRange -> All
10
6. ¥ 10-7
0.00003
0.000018
100
80
60
40
20
-0.06
-0.04
Ö Graphics Ö
-0.02
0.02
0.04
0.06
^2,
8
Diffraction.nb
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 the
definition of the phase shift to figure this one out, but it is not difficult.) These secondary maxima are actually
the principal maxima of the multiple-slit interference pattern, and are generally referred to as the "maxima of
order n," where n=0, ±1, ±2, .... The plus and minus signs refer to the right and left sides of the principal
maximum of order 0.
2. Modify the plotting program above to display (in separate colors), the diffraction pattern and the interference
pattern as well as their product. How is the diffraction envelope and the position of the maxima of various
orders affected by changes in the slit width and separation?
3. The diffraction grating
A practical application of the principles we have been discussing is the diffraction grating, which consists of
multiple apertures or even multiple lines ruled very close together on a transmissive or reflective surface. To see
what the effect of many apertures close together is on a source with more than one color, carry out the following
homework exercises.
Homework Exercises
1. Modify the program for multiple-slit diffraction with finite apertures so that you can plot the diffraction
pattern for three different colors. You can use the previous program as a template. To begin, use wavelengths of
l1 =400 nm, l2 =500 nm and l3 =600 nm, and choose the slit widths and separations to be multiples of the
center 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 are
the 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 the
diffraction 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 diffraction
grating would you have to locate the detector plane to achieve a resolution in wavelength of 3 nm between
pixels?
3. In a diffraction grating, the ability to separate various wavelengths is referred to as the resolving power of the
grating, and it is defined as the ratio l/Dl. Calculate the diffraction pattern for red, green and blue light for a
grating with 100 lines out to fourth order. Looking at the resolving power as a function of wavelength, what can
you say about the advantages and disadvantages of designing a grating spectrometer to work in first order or in
fourth order? Why do you not work in zeroth order in spite of the greater intensity of the signals?
4. Diffraction from a rectangular aperture
Fraunhofer diffraction from a rectangular aperture is a straightforward extension of Fraunhofer diffraction for a
long, narrow slit. One simply integrates over two dimensions instead of one. The variables used in the
following Mathematica calculation are taken from the diagram below. Note the sizes of the slits used!
Diffraction.nb
9
Sinc z_ := Sin z
z
Distance = 1.0
Lambda = 550.0 * 10^ -9
SlitX = 0.00000075
SlitY = 0.0000005
PhaseX = 2.0 * Pi * SlitX
2.0 * Lambda
PhaseY = 2.0 * Pi * SlitY
2.0 * Lambda
Plot3D
Sinc PhaseX * x Distance * Sinc PhaseY * y Distance
x, -2 Pi, 2 Pi , y, -2 Pi, 2 Pi , PlotRange -> 0, 0.005 ,
PlotPoints -> 40
—
General::spell : Possible spelling error: new symbol name "Sinc" is
similar to existing symbols Sin, Sinh .
1.
5.5 ¥ 10-7
7.5 ¥ 10-7
—
General::spell1 : Possible spelling error: new symbol name "SlitY" is
similar to existing symbol "SlitX".
5. ¥ 10-7
4.28399
—
General::spell1 : Possible spelling error: new symbol name "PhaseY" is
similar to existing symbol "PhaseX".
2.85599
0.005
0.004
0.003
0.002
0.001
0
5
0
-5
0
-5
5
^2 ,
10
Diffraction.nb
Ö SurfaceGraphics Ö
Homework Problems
1. If you change the PlotRange command in the program to PlotRange->All, what is the peak amplitude of the
diffraction pattern? Which of the secondary maxima can you see then? Is that consistent with the amplitude of
the 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 to
the size of the short side a by: b=a, b=2a, b=5a, b=10a. Comment on any changes you observe. Do they agree
with your intuition?
5. Diffraction from a circular aperture
Diffraction from a circular aperture is very much like diffraction from an extended rectangular aperture, except
for the difference in symmetry. The mathematical problem is "out of bounds" for this course, but the solutions
are the Bessel functions of order 1, which can be calculated by Mathematica. (Thank goodness!) As in the case
of the rectangular aperture, we find the intensity distribution by integrating the phase differences over a circular
aperture.
The Bessel function has a series expansion as follows:
g
g
1
J1 g = ÄÄÄÄÄÄ
- ÄÄÄÄÄÄÄÄÄÄÄÄÄ
ÄÄÄÄÄÄ
2
12 ◊2 2
3
g
1
+ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
ÄÄÄÄÄÄ
12 ◊22 ◊3 2
5
- ...
where the argument g is equal to
k
g= ÄÄÄÄÄÄ
2 D ◊sinJ
Homework Problem
1. Plot the Bessel function BesselJ[1,x] using Mathematica and determine where the first few zeros of the
function occur.
In this next section, we will use Mathematica to compute the diffraction pattern from a circular disk, while
making the small-angle approximation sinJ≈J.
Diffraction.nb
11
Wavelength = 500.0 * 10^ -9
Distance = 1.0
ApertureRadius = 0.0005
Wavek = 2.0 * Pi Wavelength 2.0
r = Sqrt x^2 + y^2
Plot3D 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
5. ¥ 10-7
1.
0.0005
6.28319 ¥ 106
x2 + y2
0.015
0.01
0.002
0.005
0.001
0
-0.002
-0.001
0
-0.001
0
0.001
0.002-0.002
Ö SurfaceGraphics Ö
This pattern of circular rings is known as an "Airy pattern," after the Irish physicist who first studied the
diffraction from a circular aperture in some detail. The central disk in this pattern is called the "Airy disk," and it
will play a role in our discussion about limits of resolution.
12
Diffraction.nb
6. Limits of resolution
If two light sources are close together, the far-field intensity distributions of the two sources may overlap and
make it impossible to separate the sources. In this section, we examine this problem quantitatively, and develop
the 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 first
compute 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^ -9
Distance = 1.0
Separation = 0.001
ApertureRadius = 0.0005
Wavek = 2.0 * Pi Wavelength 2.0
r1 = Sqrt x - Separation ^2 + y^2
r2 = Sqrt x + Separation ^2 + y^2
Plot3D 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
5. ¥ 10-7
1.
0.001
0.0005
6.28319 ¥ 106
-0.001 + x
0.001 + x
2
2
+ y2
+ y2
Diffraction.nb
13
0.2
0.002
0.1
0.001
0
-0.002
-0.001
0
-0.001
0
0.001
0.002-0.002
Ö SurfaceGraphics Ö
Now start reducing the separation and see how large it is compared to the slit dimensions when the diffraction
patterns begin to overlap. What happens to the overlap when you change the wavelength? Does this surprise
you? Why or why not?
14
Diffraction.nb
Wavelength = 500.0 * 10^ -9
Distance = 1.0
Separation = 0.00035
ApertureRadius = 0.0005
Wavek = 2.0 * Pi Wavelength 2.0
r1 = Sqrt x - Separation ^2 + y^2
r2 = Sqrt x + Separation ^2 + y^2
Plot3D 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
5. ¥ 10-7
1.
0.00035
0.0005
6.28319 ¥ 106
-0.00035 + x
0.00035 + x
2
2
+ y2
+ y2
Ö SurfaceGraphics Ö
To get an idea of the quantitative relationship between the diffraction pattern and the resolving power, let us take
a section through the Bessel function pattern for two sources separated by a distance S, and look at how their
diffraction patterns compare. We will arbitrarily take the y-coordinate to be equal to zero.
Diffraction.nb
15
Wavelength = 500.0 * 10^ -9
Distance = 1.0
Separation = 0.0005
ApertureRadius = 0.0005
Wavek = 2.0 * Pi Wavelength
r1 = Sqrt x - 0.5 * Separation ^2
r2 = Sqrt x + 0.5 * Separation ^2
Plot
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
Wavek * ApertureRadius * r1 Distance ^2 +
BesselJ 1, Wavek * ApertureRadius * r2 Distance
Wavek * ApertureRadius * r2 Distance ^2 ,
x, -0.003, 0.003 , PlotStyle ->
RGBColor 1, 0, 0 , RGBColor 0, 1, 0 , RGBColor 0, 0, 1
PlotRange -> All
5. ¥ 10-7
1.
0.0005
0.0005
1.25664 ¥ 107
-0.00025 + x
0.00025 + x
2
2
,
16
Diffraction.nb
0.25
0.2
0.15
0.1
0.05
-0.003 -0.002 -0.001
0.001
0.002
0.003
Ö Graphics Ö
For the sake of definiteness, Lord Rayleigh proposed long ago that the criterion for two sources to be just
resolvable would be for the principal maximum from one to fall just on top of the first secondary maximum of
the other. About what slit separation does it take to do this, and what is the argument of the Bessel function
when this criterion is satisfied? Are you surprised that we get the ratio of wavelength to aperture diameter and
not 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 that
the fog lamps are 1.5 meters apart. Using the Rayleigh criterion, determine at what distance your fully
dark-adapted eyes will be able to determine that what you see coming toward you is two headlights rather than
one. If you compute the diffraction pattern using Mathematica, do you see a pattern which agrees with this?
7. Imaging a lattice of apertures