PHY 102: Waves & Quanta
Topic 8
Diffraction II
John Cockburn (j.cockburn@... Room E15)
•Single slit Intensity distribution
•Double slit intensity distribution
•Resolving powers
•X-ray diffraction
Single Slit Diffraction
•From diagram, can see that for slit of width A, we will get destructive
interference (dark band on screen) at angles  which satisfy…..:
a

sin  
2
2
a

sin  
4
2
sin  
2
sin  
a

a
Choice of a/2 and a/4 in diagram is entirely arbitrary, so in general we have a
dark band whenever;
m
sin  
a
(m=±1, ±2, ±3………..)
Position of dark fringes in single-slit diffraction
m
sin  
a
If, like the 2-slit treatment we assume small angles, sin ≈ tan  =ymin/R, then
ymin
Rm 

a
Positions of intensity
MINIMA of diffraction
pattern on screen,
measured from central
position.
Very similar to expression derived for 2-slit experiment:
n
ym  R
d
But remember, in this case ym are positions of MAXIMA
In interference pattern
Width of central maximum
•We can define the width of the central maximum to be the distance
between the m = +1 minimum and the m=-1 minimum:
R
R 2 R
y 


a
a
a
Intensity
distribution
image of diffraction
pattern
Ie, the narrower the slit,
the more the diffraction
pattern “spreads out”
Single-slit diffraction: intensity distribution
To calculate this, we treat the slit as a continuous array of infinitesimal sources:
Can be done algebraically, but more nicely with phasors………………..
Single-slit diffraction: intensity distribution
E0 is E-field amplitude at
central maximum
ETOT
 sin( / 2) 
 E0 


/
2


 sin(  / 2) 
I  I0 


/
2


2
 = total phase difference
for “wavelets” from top
and bottom of slit
Single-slit diffraction: intensity distribution
 sin(  / 2) 
I  I0 


/
2


2
How is  related to our slit/screen setup?
Path difference between light rays from top and bottom of slit is
x  a sin 
phase difference
path difference

2


d sin 

2

From earlier (2-slit)

2a sin 

Single-slit diffraction: intensity distribution
 sin(  / 2) 
I  I0 


/
2



2
2a sin 

 sin(a (sin  ) /  ) 
II 

 a (sin  ) /  
0
2
“real” intensity distribution for double slits
The distribution derived earlier is for the idealised case of infinitely narrow
slits:
 dy 

I TOT  I 0 cos 2 

 R 
“real” intensity distribution for double slits
If we allow the slits (spacing d) to have finite width a, then the TOTAL
Intensity distribution is the product of the 2-slit intensity distribution
I TOT
 dy 
 I 0 cos 

 R 

2
and the intensity distribution for a single slit of width a:
 sin(a (sin  ) /  ) 
II 


a
(sin

)
/



0
2
“real” intensity distribution for double slits
    sin(  / 2) 
I  I cos   

2

/
2
 

2
2
0

2d sin 


2a sin 

“real” intensity distribution for double slits
“ideal” pattern
Single slit pattern
“real” pattern
Missing Orders
Rm 
y 
a
min
Positions of zero intensity
In diffraction pattern
n
y R
d
max
Positions of maximum intensity
In interference pattern
So when:
Rm Rn
md

n
a
d
a
Every nth order is missing in the 2-slit intensity distribution…….
Example Calculation
•An interference pattern is produced by 2 identical parallel slits of
width a and separation (between centres) d = 3a. Which interference
maxima will be missing in the observed pattern?
Circular apertures and resolving powers
Diffraction effects can, of course be observed with any shape of aperture
(see 2nd year optics course).
Take circular aperture, for example:
Angular radius of bright
central spot (Airy disc):
sin   1.22
1

D
Rayleigh Criterion
•Two point objects can just be resolved when the first minimum in the
diffraction pattern from one overlaps with the centre of the Airy disc
(central bright spot) in the diffraction pattern of the other. ie limiting
angle of resolution given by:
1.22
 
D
increase resolution by going to shorter wavelengths (eg electron microscopes)
X-ray diffraction
X-rays: electromagnetic radiation with wavelengths ~10-10m………..
Expect to see interference/diffraction effects when interacting with objects
on this length scale………
Atomic spacing in crystals/complex molecules ~ 10-10m…….X- ray crystallography
X-ray diffraction
Angle of incidence =
Angle of scattering
Constructive interference from adjacent planes of atoms, spacing d:
n  2d sin 
“Bragg condition”
(NB factor of 2)