Transcript Topic 7 - Diffraction I

PHY 102: Waves & Quanta
Topic 7
Diffraction
John Cockburn (j.cockburn@... Room E15)
•Interference re-cap
•Phasors
•Single slit diffraction
•Intensity distribution for single slit
Electromagnetic Waves
E y  E0 sin( kx  t )
Bz  B0 sin( kx  t )
Where E0 and B0 are related by: E0 =
cB0
INTENSITY of an EM wave  E02
NB. we will see later that EM radiation sometimes behaves like a
stream of particles (Photons) rather than a wave………………
Interference
First, consider case for sound waves, emitted by 2 loudspeakers:
Path difference =nλ
Constructive Interference
Path difference =(n+1/2)λ
Destructive Interference
(n = any integer, m = odd integer)
Interference
Young’s Double Slit Experiment
•Demonstrates wave nature of light
•Each slit S1 and S2 acts as a separate source of coherent light (like the
loudspeakers for sound waves)
Young’s Double Slit Experiment
Constructive interference:
d sin   n
Destructive interference:
1

d sin    n  
2

Young’s Double Slit Experiment
Y-position of bright fringe on screen: ym = Rtanm
Small , ie r1, r2 ≈ R, so tan ≈ sin
So, get bright fringe when:
n
ym  R
d
(small  only)
Young’s Double Slit Experiment:
Intensity Distribution
For some general point P, the 2
arriving waves will have a path
difference which is some
fraction of a wavelength.
This corresponds to a difference
 in the phases of the electric
field oscillations arriving at P:
E1  E0 sin t 
E2  E0 sin t   
Young’s Double Slit Experiment:
Intensity Distribution
Total Electric field at point P:
ETOT  E1  E2  E0 sin t   E0 sin t   
Trig. Identity:
1
 1

sin   sin   2 cos      sin     
2
 2

With  = (t + ), = t, get:
 


ETOT  2 E0 cos  sin  t  
2 
2

 


ETOT  2 E0 cos  sin  t  
2 
2



2
E
cos
So, ETOT has an “oscillating” amplitude:  0
2 

Since intensity is proportional to amplitude squared:
 
2
I TOT  4 E0 cos 2  
2
Or, since I0E02, and proportionality constant the same in both cases:
 
I TOT  4 I 0 cos 2  
2
phase difference
path difference

2


d sin 

2

 
I TOT  4 I 0 cos 2  
2
I TOT
 d sin  
 4 I 0 cos 

  
2
For the case where y<<R, sin ≈ y/R:
 dy 
I TOT  4 I 0 cos 2 

R



Young’s Double Slit Experiment:
Intensity Distribution
I TOT
 dy 
 I 0 cos 

 R 

2
2-slit intensity distribution: “phasor” treatment
•Remember from Lecture 1, harmonic oscillation with amplitude A and
angular frequency  can be represented as projection on x or y axis
of a rotating vector (phasor) of magnitude (length) A rotating about
origin.
Light
•We can use this concept to add oscillations
with the same frequency, but different
phase constant  by “freezing” this rotation
in time and treating the 2 oscillations as
fixed vectors……
•So called “phasor method”
2-slit intensity distribution: “phasor” treatment
Use phasor diagram to do the
addition E1 + E2
E1  E0 cost 
E2  E0 cost   
Using cosine rule:
2
ETOT
 E02  E02  2 E02 cos   
 E02  E02  2 E02 cos 
 2 E02 1  cos 
2
TOT
E
 2  
 4 E  cos   
 2 

2
0
Another way: Complex exponentials
i
Ae  A(cos   i sin  )
E0 e
E0 e
i (t  )
it
 E0 (cos t  i sin t )
 E0 (cos(t   )  i sin( t   ))

it

i (t  )
E0 cos(t )  Re E0 e
E0 cos(t   )  Re E0 e


Another way: Complex exponentials
Single Slit Diffraction
“geometrical” picture breaks down when slit width becomes comparable
with wavelength
Single Slit Diffraction
observed for all types of wave motion
eg water waves in ripple tank
Single Slit Diffraction
Single Slit Diffraction
•Explain/analyse by treating the single slit as a linear array of
coherent point sources that interfere with one another (Huygen’s
principle)…………………….
All “straight ahead” wavelets
in phase → central bright
maximum
Fraunhofer (“far-field”)
case
Destructive interference of
light from sources within slit
for certain angles
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  I0 


a
(sin

)
/



2