# Diffraction by the obstruction of light waves, .i.e, a physical obstacle.

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Diffraction by the obstruction of light waves, .i.e, a physical obstacle.
```Diffraction - Deviation of light from rectilinear propagation caused
by the obstruction of light waves, .i.e, a physical obstacle.
There is no significant physical distinction between interference and
diffraction. However, interference generally involves a superposition
of only a few waves whereas diffraction involves a large number of
waves.
Huygens-Fresnel Principle: Every unobstructed point of a wavefront
serves as a source of spherical secondary wavelets. The amplitude of
the optical field at any point is the superposition of these wavelets.
Suppose light strikes a screen containing an aperture. The effects of
diffraction can be understood by considering the phasor addition of
the electric field for an array of point sources emitting E-M waves
that are unobstructed by the aperture.
The principles of diffraction can be understood by considering the
interference of water waves whose wavelength is made to vary in
comparison to the size of the aperture.
To illustrate diffraction consider water waves in
a ripple tank: OPD max   max  AP  BP  AB
Destructive
interference of
most phasors
for large
angles; N large
Effective number
of point sources
N for phasor
construction
(a)    max
Intermediate
region; N
medium, mixed
interference
N=5
(b)    max
Constructive
interference of
most phasors for
all angles; N
small  Approx.
spherical wave
(c)    max
Light waves striking an aperture
a
Fresnel Diffraction (near-field
diffraction) when incoming and
outgoing waves are both nonplanar.
R
Source S and Screen C are moved
to a large distance R, resulting in
Fraunhofer diffraction (Far-field
diffraction). As a rule-of-thumb:
R > a2/ for Fraunhofer diffraction.
Both incoming and outgoing
wave contributions are
considered as plane waves,
and  = const.
Fraunhofer diffraction conditions
produced by lenses, leaving source
S and screen C in their original
position. Note that the lenses act to
filter non-plane wave components
from striking point P on the
observation screen C.
Consider a serious of diffraction measurements
on a screen whose distance (R) from the slit
varies from close (bottom) to far (top).
R6 > R5 > R4 > R3 ….
R6
R5
R4
R3
R2
R1
a
The diffraction pattern in the near-field
(bottom) is sensitive to variations in R whereas
the shape is independent of distance for R
large (top).
The sharp structure shown in the near-field
pattern is a result of rapidly changing phasor
(E-field) orientations whose summations at
each point on the screen are sensitive to
distance, slit geometry, and angular spread of
the waves. As shown, small changes in R in
the near-field can cause large changes in the
distribution on the screen. If a >> , the
pattern will resemble a sharp geometric
For large distances (e.g., R6 > a2/), the parallel
nature of the plane waves will result in phasor
irradiance, whose shape is independent of R.
As a bridge between interference and diffraction, consider a linear array of
equally spaced N coherent point oscillators. We examine the superposition of the
fields from each source at a point P sufficiently far such that rays are nearly
parallel.
The field of each wave is equal
so that E0(r1) = E0(r2) = … =
E0(rN) = E0(r).
In order to evaluate the field, we
employ a phasor sum, as follows:
~
E  E0 (r )ei ( kr1 t )  E0 (r )ei ( kr2 t )
 ...E0 (r )ei ( krN t )
 E0 (r )ei ( kr1 t ) [1  eik ( r2  r1 )  eik ( r3  r1 )
 ...eik ( rN  r1 ) ]
Due to a difference in OPL, there is a phase difference
between adjacent sources of  = k0, where  = ndsin,
and  = kdsin. From the figure,  = k(r2-r1), 2 = k(r3-r1),
3 = k(r4-r1), …
Therefore, at point P, the phasor sum is obtained using a geometric series
(similar to what was done for multiple beam interference) and yields:

~
E  E0 (r )e  it e ikr1 1  e i  e 2i  e 3i  ...  e ( N 1) i
e
e
iN
 E0 ( r ) e
 E0 ( r ) e
it ikr1 i ( N 1) / 2
e e
i

iN / 2
iN / 2
iN / 2
 1
e
e

e
 E0 (r )e it e ikr1 i / 2
 1
e
e i / 2  e  i  / 2
it ikr1
e

 sin  N / 2  
 sin  / 2  


~ ~*
EE
sin 2  N / 2 
Therefore, I 
 I0
2
sin 2  / 2 



Note that as   0, I  N 2I0,
as expected for N coherent
sources completely in-phase.
Also note that for N = 2, I = 4I0cos2(/2) using sin(2 ) = 2sin cos and
again  = kdsin .
sin 2 N / 2
2
This is true for   0 and for  = 2m, m = 0, 1, 2, …
lim

N
 0 sin 2  / 2
For conditions of constructive interference with N sources,  = kdsin = 2m
 (2/)dsinm = 2m which gives dsinm = m and Im = N 2I0, conditions
which are identical to two beam interference when there is complete
constructive interference.
Consider now a line source of oscillators, where each point
emits a spherical wavelet as shown in the figure. The electric
field for a spherical wavelet emitted from a point is

E  0 sin t  kr  where 0 is the source strength for a
r
point emitter.
(90 -  )
y
Y
Y
sin  
R
Let L = source strength per unit length. Then for
each differential segment of source length dy, the
contribution of the spherical wavelet at point P is
 dy
dE  L sin t  kr  From geometry,
r
Width of line source = D
r 2  y 2  R 2  2 yR cos90   
 y 2  R 2  2 yR sin  ,
from the law of cosines.
 y sin 

y2
2 y sin 
y2
 rR
1
 R 1 

 ...
2
2
R
R
R
2R


 R  y sin   ... where R >>D and D > y.
Note that in order to fulfill Fraunhofer (far-field) diffraction conditions, we also
require R > D 2/. Thus r  R - ysin is the resulting approximation that we will
use. We can calculate the total field appearing on the screen at angle , by
summing or integrating all contributions of dE on the linear source:
D/2
E
D/2
 dE  
D / 2
D / 2
dy
L
r
sin t  kr 
D/2

D /2
dy
L
R
sin t  k ( R  y sin  )
Note that we are also ignoring the presence of ysin in the denominator since
its presence in the phase of the sine in the numerator has a much greater
contribution to the integral.
We can further separate terms in the sine by using the following identity:
sin t  kR  ky sin    sin t  kRcosky sin    cost  kRsin ky sin  
Since sin(kysin) is an odd function in y, the last term cancels in the integral.
Therefore, the integral is as follows:
L
D/2
2 L




E
dy
sin

t

kR
cos
ky
sin


R  D/ 2
R

D/2
 dy cosky sin  sin t  kR
0
2 L sin (kD / 2) sin  
 D sin (kD / 2) sin  
sin t  kR  L
sin t  kR
R
k sin 
R
(kD / 2) sin 
So let   (kD / 2) sin 
 E
 L D  sin  

 sin t  kR,
R   
and the time  averaged irradiance is
1 D
I ( )   L 
2 R 
2
2
 sin  
 sin 

  I (0)
  
 



2
We can now solve the single-slit diffraction problem by considering Figs. 10.6
and 10.7 on the next slides where the variable D  b and  = (kb/2)sin ,
where b is the slit width (typically a few hundred ) and l is the slit height
(typically ~ cm).
Clearly, I() have minima when sin = 0 and  = m, m = 1, 2, 3,…
Note   (kb / 2) sin   m 
2 b
sin   m  b sin   m
 2
(minima)
Note that the condition for diffraction minima, bsin = m, should
not be confused with a similar expression for two-slit interference
maxima, asin = m.
Fig. 10.6 (a) Single-slit Fraunhofer diffraction.
(b) Diffraction pattern of a single vertical slit
under point-source illumination.
Fraunhofer diffraction using lenses so that the
source and fringe pattern can both be at
convenient distances from the aperture.
We can determine the conditions for maxima in the diffraction pattern by
setting the derivative of the Irradiance to zero:
 sin 
I  I 0 
 



2
 sin 
dI
 I 0 2
d
 


 1  1
 sin    2   cos  

   

  cos   sin  
 2 I 0 sin  
0
3



 (i ) sin   0 or (ii )  cos   sin   tan   
The first condition (i) is just the condition
for minima that we have already seen (i.e.
 = m, m = 1, 2, 3, …)
The second condition (ii) results in a
transcendental equation whose graphical
solutions can be observed on the left. The
solutions can easily be found numerically:
  1.43, 2.46, 3.47…which
approaches (m+1/2) for large .
Principal or Central
Maximum
Subsidiary Maximum (SM)
Minima (I = 0)
1st SM
2nd SM 3rd SM
For the figure on the far
left, the conditions for
minima of m = bsin are
understood by dividing the
phasors of rays in two and
four equal segments in (b)
and (c), respectively. In (b),
for each ray in the top half,
a phasor 180 out-of-phase
can be found in the bottom
half, resulting in zero Efield. Likewise, in (c), each
successive quarter contains
phasors that are 180 outof-phase, resulting in zero
Phasor constructions of the E-field for
(a) the central maximum, (b) a
direction slightly displaced from the
central maximum, (c) the first
minimum, and (d) the first maximum
beyond the central maximum for
single-slit diffraction. The example
corresponds to N = 18 phasors.
Consider now the set-up for a double-slit diffraction measurement. The
rectangular geometry of the slit is shown such that l >> b, where l and b are the
height and width of each slit. The screen on which the diffraction pattern is
observed is located a distance R from the slits, where R >> a and R >> b2/.
The point P is located on the screen. We will need to calculate the resulting Efield contribution at point P from all points (sources) located in both slits, using
essentially the same procedure performed for the single-slit case.
Figure 10.13 (a) Double-slit
geometry. Point P on the screen
 is essentially infinitely far
away. (b) A double-slit
diffraction pattern for a =3b.
The E-field is obtained, as before, by integration over all sources in each slit:
b/2
E C
 F ( z)dz  C
b / 2
a b / 2
 F ( z)dz where C 
a b / 2
L
R
and
F ( z )  sin t  k R  z sin  
Integration procedure is similar to that of the single slit and results in:
 sin  
sin t  kR  sin t  kR  2  where   ka / 2sin  and   kb / 2sin 
E  bC 
  
Note that the definition for  is precisely as before for single-slit diffraction.
Using the identity: sin x + sin (x + 2y) = 2cos y sin(x + y), we can express E as
 sin  
 cos  sin t  kR   
E  2bC 
  
After squaring and time-averaging, we obtain the irradiance as:
 sin 2   2
in which  =  = 0 represents  = 0 and I(0) = 4I0


I    4 I 0 
cos 
2

since E0 = 2bC and twice that relative to a single slit.
  
2
2
Note that I0= (bC) = (E0/2) is the contribution from
one slit.
Examine some limits:
If kb << 1, (sin / )  1 and we get I() = 4I0 cos2 = 4I0 cos2(/2), which
yields the expected result for Young’s two slit interference when the slit
width b is very small.
If a = 0,  = 0, then I() =4I0 (sin / )2 = I0’ (sin / )2 , which is just the
situation of two slits coalescing into a single slit.
Note that the expression I() =4I0 (sin / )2 cos2 is that of an interference
term modulated by a diffraction effect.
Note that in general for two slits, minima occur when  = m, m = 1, 2,
3, … and  = (2m+1)/2, m = 0, 1, 2, 3, …
As a general rule-of-thumb, for a = mb, we observe 2m bright fringes within
the central diffraction peak, including fractional fringes. So, in Fig. 10.13,
for a =3b, we have 5 + 2(1/2) = 6 bright fringes within the central
maximum.
When an interference maximum overlaps with a diffraction minimum (as
shown in Fig. 10.13c) it is often referred to as a missing order.
(a)
(b)
Figure 10.14
Single- and double-slit Fraunhofer
patterns. (a) Photographs taken with
monochromatic light. The faint
cross-hatching arises entirely in the
printing process. (b) When the slit
spacing equals b, the two slit
coalesce into one (of width 2b) and
the single-slit pattern appears (i.e.,
that’s the first curve closest to you).
The farthest curve corresponds to
the two slits separated by a =10b.
Notice that the two-slit patterns all
have their first diffraction minimum
at a distance from the central
maximum of Zo. Note how the
10.13(b) as the slit width b gets
smaller in comparison to the
separation a.
For diffraction from many slits (i.e., the N-slit problem), we generalize the
2-slit problem:
( N 1) a  b / 2
b/2
a b / 2
2 a b / 2
E C
 F ( z)dz  C 
b / 2
F ( z )dz  C
a b / 2
where C 

2 a b / 2
L
R
F ( z )dz  ....  C
 F ( z)dz
( N 1) a b / 2
and F ( z )  sin t  k R  z sin  
The integrals can be evaluated as
before for the single- and double-slit
cases using the same identities. The
result is
 sin  
 sin t  kR  2 j 
 E   bC 
j 0
  
 ka 
where R j  R  ja sin  ,     sin 
 2
N 1
  kR  2 j  kRj (see figure for Rj)
The summation can be evaluated most easily using phasors, as before:
 sin   N 1
 sin    i t kR  N 1 i 2 j 
 sin t  kR  2 j   bC 
 Ime
E  bC 
e




j

0
j

0



 

 
Again, the phasor sum is evaluated as a geometric series, as was done
for the case of multiple beam interference. Therefore, we can write
 sin   sin N 
ka

E  bC 
 sin t  kR  ( N  1) ,   sin  and
2
   sin  
kb
  sin 
2
Therefore, the time-averaged irradiance can be expressed as
2
 sin    sin N 
 
I ( )  I 0 


sin



 
Note that I0 is the irradiance in for  = 0 emitted by one slit and I(0) = N2I0,
which is seen by taking the limit for  = 0 of the above expression. Also note
that if kb << 1 (i.e. very narrow slits), we get the same expression for a linear
coherent array of oscillators.
2
Diffraction term
multiple-slit interference term
2
2
 sin    sin N 
I (0)  sin    sin N 
 
 
Therefore, I ( )  I 0 
  2 

N     sin  
    sin  
2
2
The principal maxima are given by the following positions:  = 0, ,
2, 3, … = m (m = 0, 1, 2, 3 ..) and  = (ka/2)sin  a sinm = m
and (sinN/sin)2 = N2 at these positions.
Minima are determined by (sinN/sin)2 = 0   = /N, 2/N, 3/N, …,
(N-1)/N, (N+1)/N,… Therefore, between consecutive principle maxima
we have (N – 1) minima (see figures on next slide).
Subsidiary maxima: Between consecutive principle maxima we have (N-2)
subsidiary maxima when |sinN| 1. This occurs for   3/2N, 5/2N,
7/2N, …For large N, sin2  2 and the intensity of the firs subsidiary peak
is approximately
2
2
2
 sin    2 
I (0)  sin  
   


I ( )  I (0)
22   
    3 
Figure 10.17 multiple-slit pattern with a = 4b, N = 6
Finally, note that for N = 2 slits, we observe that
I (0)  sin  

I ( )  2 
N   
2
2
2
I (0)  sin    2 sin  cos  
 sin N 

 




sin

4

sin





 
2
 sin  
 cos 2 
 I (0) 
where   (ka / 2) sin    / 2
  
because   ka sin   k  k (OPD ).
2
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