Chapter 8. Fourier Transform
Diffraction
Sample problems
8-S1
A thin and long rectangular aperture is illuminated by a collimated beam of coherent light (wave
length λ). The slit is located at the distance f of a convergent thin lens, where f is the focal
distance of the lens. Provide an expression for the irradiance of the diffraction pattern of the
rectangular slit.
Solution to 8-S1
According to section 7.5.2 in the back focal plane of the lens one observes the FT of the field
distribution at the aperture. The amplitude transmittance of rectangular aperture is given by
    
(8.42), Figure 8.6 t (, )  rect     , a and b are the half widths of the rectangular
 2a   2 b 
aperture. The amplitude of the incident field is E0. The field at the aperture
    
is E 0 t (, )  E 0 rect     . The scalar theory of light propagation will be utilized which in
 2a   2 b 
essence means that the interfering wave fronts due to the diffraction effect have parallel vector
fields and are coherent. Applying the FT property of the lens in the image plane of coordinates
x,y, the field will be (7.34), Section 7.5.1.3The Fraunhoffer approximation,
 i ( f 2 f y 2 )
i e x
E( x, y )  

f
FTE(, )fx  x ,fy  y
f
f
.The field in the focal plane is the FT of the field in
x
y
where the preceding
, fy 
f
f
relationships are the scale of frequencies of the focal plane of the lens. The FT of the rectangular
field amplitude is (8.44)
the aperture evaluated at the spatial frequencies f x 
 i ( f 2 f
2
i4ab E 0 e x y
sin c2af x  sin c 2bf y .
E ( x, y )  

f
But 4ab=A is the area of the slit. The corresponding irradiance is,
)
2
2
2
2
E 20 A 2 
x  
y 
x  
y 

I( x, y )  E( x, y )  2 2  sin c2a   sin c 2b   I 0  sin c2a   sin c 2b 
f 
f  
f 
f  
f 

Figure 7.15 gives the diffraction pattern of a square aperture illuminated by a helium-neon laser.
Figure 8.7 shows the irradiance of a narrow slit such that b>>a plotted as a function of the

2ax
normalized coordinates x 
.
f
2
8-S2
Assuming that a uniform field of amplitude E0 impinges in the double slit shown in figure P8.1
derive the expression of the irradiance of the double slit.
Solution to 8-S2
To get the irradiance of the double slit we can use the properties of the delta function to obtain
the expression of the transmittance of the screen shown in Figure P8.1.
Figure P8.1. Double slit arrangement
The transmittance of a single slit was defined in sample problem 8-S1 as,
    
t (, )  rect     .
 2 b   2a 
To position the slit in the positions  d the property of the delta function,
  ( x  a, y  b) f ( xy)dxdy  f (a, b) or the shift property in two dimensions can be utilized.
Consequently the transmittance function of the double slit can be expressed in the following way
    
t (, )  (  d, )  (  d, ) rect    
 2a   2 b 
Indeed expanding the above expression
   d   
   d   
t (, )  rect 
    rect 
 
 2 b   2a 
 2 b   2a 
represents the transmittance of the double slit. If the screen that contains the double slit is
illuminated by a uniform plane wave front of amplitude E0, the field emerging from the double
slit is,
   d   
   d   
E 0 t (, )  E()  E 0 rect 
    rect 
 
 2 b   2a 
 2 b   2a 
To find the Fraunhofer pattern produced by the double slit in the focal plane of the lens, let us
first compute the FT of the field distribution. This computation will be carried out for each of the
expressions involved in the final expression.

     
FT(  d, )  (  d, )  2 cos 2df x and FTrect      A sin c2bf x sin c2af y 
 2b   2a 

The convolution theorem can be used to write,
E 0 t (, )  2 AE 0 sin c2af x sin c2bf y cos 2df x .
With the above result it is possible to apply (8.44) and get
 i ( f 2 f y 2 )
i2  4ab E 0 e x
E( x, y )  

f
sin c2af x  sin c 2bf y cos 2df x;
Where as in sample problem (8-S1) f x 
x
y
, then the final expression of the irradiance
, fy 
f
f
is,
2
2
4E 20 A 2 
x  
y 
2dx
I( x, y )  E( x, y )  2 2  sin c2a   sin c 2b  cos 2
f 
f  
f 
f
2
8-S10
In problem 8.9 the case of two slits has been solved. Let us now consider the case of a screen
with N narrow parallel slits illuminated by normally incident plane wavefronts
Solution to 8-S10
In sample problem 8-S2 we have considered two slits, let us analyze now the problem of N
identical slits. Following the model of problem 8.1we can generate the N slits by utilizing the
comb function. Figure 8.14 illustrates the process of generating a pattern of N slits of separation
2d and rectangular size given by 2b  2a and contained in square area of 2L×2a. The
transmittance in the x-direction is the function given by,
rect
  Nd
2b
 rect
  n  1d
2b
 rect
  Nd
where n=-N…+N.
 rect
2b
  n  2d
2b
 ....  rect

2b
 ...  rect
  n  2d
2b
 rect
  n  1d
2b
This function has a plot similar to the one shown in Figure 8.17 where the pulse is not centered at
the origin of coordinates. The transmittance in the x direction can be the expressed then,
n N

t ()   (  nd )rect
2b
n N
The transmission function is,
 n N

 x   y  
t (, )    ( x  nd )rect     
 2b   2a  

n  N
The field at the multiple slit apertures is,
 n N

 x   y  
E( x, y)  E0 t (, )  E0   ( x  nd )rect     
 2b   2a  

n  N
Following the same process applied in Problem 9.1, the FT of
n N
j
 n 
 1 j 
FT   (  nd )    f x   
ei 2 f x nd , this simplification applies if the size of the

d
d

j 
n N
 n  

aperture in the x-direction is L and d<<L. We can now utilize the identity,
n N
1  e N
n
 e  1  e with   i2fxd , hence we have,
n N
n N
 ei 2fxnd 
n N
1  e N i 2fxd
. This expression can be transformed utilizing another identity,
1  ei 2 fxd
1  e N i 2 f xd sin Ndf x

1  ei 2 f x d
sin df x
With the above expression,
 n 
 sin Ndf x
FT  (  nd ) 
n 
 sin df x
Returning to,
 i ( f 2 f
2
)
i e x y
E( x, y )  
FTE(, )fx  x ,fy  y and recalling that
f
f

f

     
FT rect      A sin c2bf x sin c2af y 
 2 b   2a  

Utilizing the convolution theorem,
x
i4ab E 0 e
x 
y 

f .The irradiance is,
E ( x, y )  
 sin c2b   sin c 2a 
x

f
f  
f  sin d

f
2
x 

2
2 sin Nd
2
2
E A 
x  
y  
2
f 
I( x, y )  E( x, y )  02 2  sin c2b   sin c 2a  

f 
f  
f   sin d x 
f 

sin Nd
 i ( f x2  f y 2 )
2
y 

Assuming that a>>b, then  sin c 2a  the contribution of this term is equal to 1 and one can
f 

write
x

2
sin Nd
E 20 A 2 
x  
f
I( x, y )  2 2  sin c2b  
x
f 
f   sin d
f

following way:
x
sin c2b

f
2


 , the arguments of the functions can be put in the


2bx
2bx
sin
f 
f and the argument can be define as   kb x ,and   kdx
2bx
2bx
f
f
f
f
sin
I(. )  I 0 sin c 
2
 sin N 
 sin  
2
8-S11
Get expressions that give the position of the maximum of the field flux and the corresponding
irradiance in the focal plane of the lens for the multiple slit problem.
Solution to 8-S11
sin N
is a maximum. To obtain this maximum one can follow
sin 
the classical calculus procedure of getting the derivative equal to zero to obtain extremes of the
function. This leads to an indeterminate expression. To solve this problem the L'Hôpital's rule is
applied. In calculus, L'Hôpital's rule utilizes the derivatives of a function to obtain limits of
indeterminate expressions. Applying this rule the maximum of this function occurs for    n ,
 sin N 
 N ; it occurs at the origin of coordinates of the
n=0,1,2,3,.. and the maximum value is 
 sin  
image plane, Simultaneously the value of the sinc β function is equal to one, then I(0,0)=N2 I0;
one has the total energy of the N slits.
The maximum flux occurs when
8-S12
Get the expressions that give the position of the extremes of the field flux and the corresponding
irradiances in the focal plane of the lens for the multiple slit problem.
Solution to 8-S12
Taking into consideration the expression derived in sample problem 8-S10,
2
x 

2
2
2
sin
N

d
E A 
x  
f 
I( x, y )  02 2  sin c2b  

f 
f   sin d x 
f 

Extracting the square root of this expression the field can be represented by,
x
2bx 

f
E( x, y )  I 0  sin c

f  sin d x

f
A change of variables can be introduced, x= nλ and the angular variable is defined as sinθ= nλ/f,
n=1,2,3…and call this variable ρ= nλ/f . The following notation is introduced p/2=d. The
argument of the second function to the right of the equal sign, that can be called interference
function, because it represents the interference effect of the different slits can be written,
kp
.
' 
2
Likewise for the sinc function calling s=2b the width of the slit, the argument becomes
ks
.
' 
2
With this notation,
sin N'
E(' , ' )  I 0 sin c ' 
sin '
the flux of the field is given by the product of two functions. The first function represents the
diffraction of one single slit. The second function is the interference function. The sinc function
modulates the interference function. Each function has its own maxima and minima and the
overall intensity maxima and minima are the result of the superposition of the functions.
Considering the equation that represents the interference term, the plot is shown in the
normalized pattern of Figure 8.7. Considering this function the maxima are reproduced each time
that the sine function becomes equal to 1.
kp
This occurs every time that,
 m , and replacing k and ρ,
2

kp 2p sin 
,
 m or sin   m .
2
2
p
sin Nd
Hence if this result is plotted as a function of ρ, this plot agrees with the result of the diffraction
produced by a grating equation (8.73). Consequently the maxima correspond to the diffraction of
the grating obtained by the derivation performed in (8.7). These maxima are surrounded by
sin N'
 N , this condition
subsidiary minima. Now considering this function has maxima when
sin '
occurs when '  m where m=0,1,2,..
sin N'
0.
The minima are given by
sin '
 2 3 ( N  1) N  1
This value of the function corresponds to '  , , ,...
,
N N N
N
N
.

N 2 N 3N
If one uses the notation m , the values of m that are multiples of N such as
,
,
N
N N
N
are not included because these values correspond to maxima. The above considerations describe
the behavior of the interference function.
FigureP8.2.Plot of the flux component that depends on the sin function as a function of ρ
To understand the effect of the sinc function that describes the diffraction of a single slit on the
intensity distribution the relationship of the angular variables β’ and α’ need to be analyzed. To
simplify the derivations let us assume that p=ns where n is an integer n=2,3,4…and relate the
total number N to the value of n. Let us assume that p=4s, and N=4+2=6. The maxima of the
interference pattern occur as seen before each time that '  m . The sinc function has the zeros
' p
every time that '  m'  , The ratio of both arguments is
  4 . The first minimum of
' s
'   hence '  4 , the second missing maxima corresponds to 8π, and the next to 12π
Figure P8.3. Multiple slit diffraction intensity distribution
Figure P8.3 shows the result of the sinc function on the interference function plotted in Figure
P8.2. It can be seen that the four maximum is suppressed and the amplitude of the other maxima
is reduced. The maximum intensity is observed at the center of coordinates and corresponds to
the zero order of diffraction. It is possible to see that the equation derived in Section 8.7 provides
the angular distribution of the diffraction orders but does not give information concerning the
corresponding intensities. The present analysis indicates that this distribution depends on the
ratio p/s. This is a simple model, it corresponds to the type of intensity function shown in Figure
8.17.
Problems to solve
8.3
Compute the maxima and minima of the diffraction pattern of a single slit along the x-coordinate
8.4
Determine the intensity of the first four subsidiary maxima
8.5
A single slit is illuminated by a solid state laser of λ=635 nm. The slit has a width of 0.1mm and
length of 10 mm. The slit is imaged by a lens of focal length f=50 cm. Plot the two dimensional
pattern of the intensity distribution at the focal plane assuming an arbitrary unit intensity and a
scale to represent relative intensity.
8.6
If you capture the intensity distribution of a single slit, could you determine the value of the
parameter a of problem 8.1?
8.7
Plot the intensity distribution of a double slit along the x-axis. Assume that d=5b assuming an
arbitrary maximum intensity I0.
8.8
The Young’s experiment is closely related to the diffraction pattern of a double slit. The
variables that control the diffraction pattern are the dimensions that were indicated in Figure 8.1,
a, and d. If you want to simulate as close as possible a system of parallel fringes what should be
your selection of these variables? Plot the results of your choice.
8.9
In what way by measuring the intensities of the diffraction pattern could you determine the
values of b and d of Figure P8.1.
8.12
Considering the developments of the problems in this chapter comment about the symbolical
operations illustrated in Figure 8.15.
8.13
For the solution of sample problem 8-S11 assume that λ=635 nm, p= 2s, and s=1.25 microns.
Following the developments from sample problem 8-S11, plot a graph of the relative intensity
distribution in the focal plane of the imaging lens for the first three orders of diffraction.
8.14
If the lens in problem 8.12 has a focal distance of 38 cm plot the position of the first three orders
utilizing a graph similar to the one represented in Figure 8.19. Indicate the relative intensities by
giving the zero order the value 1.