Today Ch.36 (Diffraction)
Next week, Dec.6, Review
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Secs.511-515: December 9, Friday: 12:30-2:30 pm
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Lecture 23 (Ch. 35)
Interference
1. Interference and superposition principle
2. Condition of constructive and distractive
interference: phase difference and path difference
4.Thomas Young’s double-slit experiment
5.Interference in thin films
6.Applications
Interference and superposition principle
Let us consider a superposition of two electric fields oscillating in the same direction,
with the same frequency and given initial phase shift.
E1  E0 cost
E2  E0 cos(t   )
 
Et  E1  E2 , cos  cos   2 cos(


2
 
) cos(

2
)

Et  2 E0 cos( ) cos(t  )  Et 0 cos(t  ), Et 0  2 E0 cos( )
2
2
2
2
The same result can be easily obtained from the phaser’s diagram.


Et  Et 0 cos( t  ), Et 0  2 E0 cos( )
2
2
1.  2m (m  0,1,2,...)  Et 0  2 E0
1
2.  2 (m  )  Et 0  0
2
3. 

2
(2m  1)  Et 0  2 E0
Total intensity does depend on 
(in general I ≠I +I ! ).
This phenomenon is called interference.
t
1
2
4 E02 cos2 ( / 2)(cos2 (t   / 2)
It 

 4 I 0 cos2 ( / 2)
0c
0c
E t2
It  4I 0 cos2 ( / 2)  2I 0 (1  cos )
Phase difference:
It
1.  2m  I t  4 I 0 ( Et 0  2 E0 ) constractive int erfernce
4I 0
1
2.   (2m  1)  2 (m  )  I t  0 ( Et 0  0) destruct. int er.
2
2I 0
3.  m 

2

2


2


2
(2m  1)  I t  2 I 0 ( Et 0  2 E0 ) no int er.
If phase is random interference is absent: cos  0  It  2I 0
Monochromatic waves of the same frequency with fixed relative phase are called
coherent waves. Coherence time:
, coherence length
1
c
c ~
lc ~
f
f
General case
E1  E01 cost
E2  E02 cos(t   )
I  I1  I 2  2 I1 I 2 cos
2 I1 I 2 cos is int erefrenceterm
Prove it is using
ei  e i
cos 
2
Path difference
E1  E0 cos(t  kr1 )
r1
E2  E0 cos(t  kr2  0 )
S1
  k (r2  r1 )  0  kr  0
r2
S2
1.  2m  I t  4 I 0 ( Et 0  2 E0 ) constractive int erfernce
1
2.  2 (m  )  I t  0 ( Et 0  0) destructive int erference
2
3. 

2
(2m  1)  I t  2 I 0 ( Et 0  2 E0 ) no int erference
2

Taking into account that
k
In particular case when
0  0
1.  2m  r  m constractive int erfernce
1
1
2.  2 (m  )  r  (m  ) destructive int erference
2
2


1
3.  (2m  1)  r  (m  ) no int erference
2
2
2
In particular case when
0  0
Thomas Young’s double slits
experiment, 1800
Thomas Young (1773 – 1829)
Max and min positions (bright and dark stripes)
If R  d  r  r2  r1  d sin  ,   kr
max :   2m  d sin   m
1
1
min :   2 (m  )  d sin   (m  )
2
2
If  is sm allsin   tan 
Rm
max : ym 
d
R (m  1 / 2)
min : ym 
d
y
R
Example
.
A very thin sheet of plastic (n=1.6) covers one slit of a double-slit apparatus illuminated by
640nm light. The center point of the screen, instead of being a maximum, is dark. What
is the minimum thickness of the plastic sheet?
d
  (k  k0 )d  

1 1
2 (  )d   ,   0
 0
n
2
0
(n  1)d  1  d 
0
2(n  1)

640nm
 533nm
1.2
Example
.
In a two-slits interference experiment, the slits are 0.2mm apart, and a screen is at a
distance of 1m. The third bright fringe (not counting the central bright fringe) is found
to be displaced 9.49mm from the central fringe. Find the wavelength of the light used.
NB:R>>d, R>>y3
Rm
d
ym d (9.49103 m)(0.2 103 m)


 632.7 nm
Rm
3 1m
max : ym 
Intensity distribution
I t  4 I 0 cos2 ( / 2)  2 I 0 (1  cos ),
  kr 
2

d sin  
2d
y
R
4
Max and min positions (bright and dark stripes)
If R  d  r  r2  r1  d sin  ,   kr
max :   2m  d sin   m
1
1
min :   2 (m  )  d sin   (m  )
2
2
If  is sm allsin   tan 
Rm
max : ym 
d
R (m  1 / 2)
min : ym 
d
y
R
Example
A radio station operating at a frequency of 1500kHz has two identical vertical dipole antennas
spaced 400m apart, oscillating in phase. At distances much greater then 400m, in what
directions is the intensity greatest in the resulting radiation pattern? If intensity produced by
each antenna 400km away along y axis is 2mW/m2 what is the total intensity produced by two
antennas at this point?
y
c 3  108 m / s
1.  
 200m
6
f 1.5  10 1 / s
m m(200m) m
max : d sin   m  sin  

 
d
400m
2
m  0    0, m  1    30 , m  2    90
| m | 2  sin   1  im possible. 2.I  4 I 0  8m W / m 2
Phase change in the reflected wave
na  nb
Er 
Ei
na  nb
if na  nb    0
ifna  nb    
Interference in the thin films
In the case of a normal coincidence:
2
  k 2t   , k 
 film

1
1
1. max :   2m  2t   film (m  )  0 (m  )
2
n film
2
m0
1
2. min :   2 (m  )  2t  m film 
2
n film
 
film
n<nfilm
Example
The walls of a soap bubble have about the same refractive index as a plain water, n=1.33.
In the point where the wall is 120nm thick what colors of incoming white light are the most
strongly reflected?
0
1
max : 2t 
(m  ) 
n film
2
m  0, 0  4tn film  4 120nm 1.33  640nm (orange)
m  1, 0 
640nm
 213nm(UVradiation, nonvisible)
3
 
t
Example
Light with wavelength 648nm in air is incident perpendicularly from air on a film 8.75 μm thick and with
refractive index 1.35. Part of the light is reflected from the first surface of the film and is reflected back
at the second surface. The second surface of the film is again in contact with air.
1. Find the total phase difference in terms of
at the exit of the film between the rays reflected from the
first and second surfaces.
2. Will you see constructive or distructive interference in the reflected light? Explain.
3. Suppose the second surface of the film is in a contact with a plastic material with refractive index 1.85.
How does it change the answers on the previous questions? Explain.

1.  k 2t   
2n2t
0
2  1.35 17.5m
 
   2 (36  1 / 2)    37  2
0.648m
2. Constractive since the total phase shift is multiple to 2
3.
  k 2t  2  75
Reflection from each surface gives the phase shift
. Interference will be destractive.


. Total phase shift now consists of an even number of
NB: if light is produced by a thermal source the typical coherence length is of an
order of 1μm. Hence for films much thicker then 1μm an interference is impossible.
Interference at the thin wedge of air
NB: Interference between the light reflected from
the upper and lower surfaces of the glass plate is
neglected due to the larger thickness of the plate.
 
1
max : 2t  0 (m  )
2
min : 2t  m0

NB: the fringe at the line of contact is dark,
because of
phase shift produced by
reflection from the second plate.
Example
A monochromatic light with wavelength in the
air 500nm is at normal incidence on the top
glass plate (see the figure). What is the
spacing of interference fringes?
x l
tl
  x  , min : 2t  m0
t h
h
m0l
0l (500109 m)0.1m
xm 
, xm1  xm 

 1.25m m
3
2h
2h
2(0.0210 m)
Newton’s rings
NB: Interference between the light reflected from
the upper and lower surfaces of the lens is
neglected due to the larger thickness of the lens.
R
R
rm
1
max : 2t  0 (m  )
2
min : 2t  m0
 
r
NB: the fringe at the line of contact is dark,
because of
phase shift produced by
reflection from the glass plane.

The positions of max or min:
rm  R 2  ( R  t m ) 2
Example: Newton’s rings can be seen when a plano-convex lens is placed on a flat glass surface
(see the figure). The radius of curvature of the convex surface is 95.2cm. The lens is illuminated
from above with red light, λ=580nm.
1) Is the spot in the center bright or dark?
2) Formulate the condition for the constructive
interference.
3) Find the diameter of the first bright ring.
4) Would the first ring be closer or further from the
R
center for the blue light compare to the red light?
R
 

1. The spot at the center is dark, because of
phase shift
produced by reflection from the
glass plane.
1
2. max : 2t  0 ( m  ) , m  0,1,..
r
2
rm
3.t0= λ0/4
r0  R 2  ( R  t0 ) 2  2 Rt0  t0  t0 (2 R  t0 ) 
2

0
4
(2 R  0 / 4) 
0 R
2
 0.54m m, d 0  1.08m m
4. Closer, since λ for blue light is shorter and hence r is smaller.
Nonreflective and reflective coatings
   
Nonreflecting film corresponds to destructive
interference in the light reflected from the upper
and lower surfaces of the film.
Reflecting film corresponds to the constructive
interference in reflection.
NB: Minimum in reflection corresponds to
maximum in light transmitted through the film into
the glass.
Maximum in reflection corresponds to minimum in
light transmitted through the film into the glass.
1
min : 2t   film (m  )
2
 film
min with m  0 : t 
4

0
4n film
Such quarter wavelength films are used to cover lenses in cameras and solar cells to increase an
amount of light through lens and solar cells. They are used also to make planes “invisible” for
radars (which typically use 2cm wavelength).
NB: If nfilm >nglass (or other substrate) then quarter wavelength film is the reflective film (because the
phase shift at the second interface becomes zero).
Example. A plastic film with index of refraction 1.85
is put on the surface of a car window to increase
the reflectivity and thus to keep the interior of the
car cooler . The window glass has index of

refraction 1.52. (a) What minimum thickeness is
required if light with wavelength 550 nm in air
reflected from two sides of the film is to interfere
constructively? (b) It is found to be difficult to
manufacture and instol coatings as thin as
calculated in part (a). What is the next greatest
thickness for which there will be also constructive
interference?
We need constructive interference in reflection:
.
 
1
2t   (m  ) , a)m  0
2

0
b) m  1
550 nm
t 

 74.3 nm.
4 4n 4(1.85)


3 30 3 550 nm
t


 223 nm.
4
4n
4(1.85)
The Michelson interferometer
Compensator plate D is made of the same glass as C and has the same thickness. It induces the
same phase shift in ray 1 as those enquired by ray 2 in C.
1887
V=3x104m/s