Tài liệu Quantum Physics doc

Chapter XVII
Chapter XVII
Interference of light
Interference of light
§1. Interference of coherent sources of light
§2. Interference in thin films
§3. Interferometer
§1. Interference of light from coherent sources:
We consider an overlap of light that comes from two sources.
A remarkable phenomen takes place, if two sources satisfy some
following conditions:
The sources are
monochromatic.
It means that they emit light
of a single color.
A monochromatic light corresponds to a
sinusoidal electromagnetic
wave
with a single frequency
f
and wave length

Two sources have the same frequency f (the same wave length )
Two souces are permanently
in phase
, or , at least, have any definite
constant phase difference
Then, two sources are called coherent sources.
1.1 Coherent sources of light:
 Recall the formula for a sinusoidal e-m wave:
       
2
, cos cos 2 cos
speed amplitude
2
wavelength wavenumber or wavevector
frequency 2 angular frequency
y x t A x vt A kx ft A kx t
v A
k
f f

 




 
 
     
 
 
 
  
  
 Notes:
Common sources of light do not emit monochromatic light
(single-frequency light)
→ However one can produce approximately monochromatic light:
• by using filters which block all but a very narrow range
of wave length
• by using light from a laser
1.2 Interference
1.2 Interference
of light through narrow
of light through narrow
slits:
slits:
Monochromatic
light source at a
great distance,
or a laser.
Slit pattern
Observation
screen
Young’s experiment on doubledouble slit interferenceslit interference
(Thomas Young performed in 1800)(Thomas Young performed in 1800)
Light (wavelength is incident on a two-slit (two narrow,
rectangular openings) apparatus:
 If either one of the slits is closed, a
diffuse image of the other slit will appear
on the screen. (The image will be
“diffuse” due to diffraction. We will
discuss this effect in more detail
later.)
Monochromatic light
(wavelength )
S
1
S
2
screen
Diffraction
profile
I
1
 If both slits are now open, we see
interference “fringes” (light and dark
bands), corresponding to constructive
and destructive interference of the
electric-field amplitudes from both
slits.
I
S
1
S
2
Dark fringes
Light fringes
 Important quantity: path difference = r
2
- r
1
The light density at the location of observer depends on the
path difference 
S
1
S
2
Observer
Light
d
 r
1
r
2
A path difference corresponds to a phase difference of
two waves at the observer’s point



One has a simple
One has a simple
formula for the path difference,
formula for the path difference,


,
,
whenwhen the observer is far from sourcesthe observer is far from sources
(Assume 2 sources radiating in phase)
= dsin
When observer distance >> slit spacing
(r >> d) :
d sin
Normal to d

d

Observer
r

d

 The corresponding phase difference
at the observer’s point:
= dsin= m
Constructive
Interference
= dsin= (m +
1
/
2
)
Destructive
Interference
m=0
m=1
m=2
m=-1
m=-2

= sin
-1
(md
“lines” of
constructive
interference:
d


I
/d
/d
0
-/d
r
Usually we care about the linear
(as opposed to angular)
displacement y of the pattern
(because our screens are often
flat):
L
y
y = L tan
At observer’s points that satisfy the condition
m m = 0, ±1, ±2, → two waves
are in phase and
reinforce each other,
we say
that there is constructive interference
at these points
If m + 1)→ two waves
cancell each other
→ there is
destructive interference
y m(d)L
y (m +
1
/
2
)(d)L
d

Y L
I
L/d
L/d
0
-L/d
L
 The slit-spacing d is often large compared to , so that is small.
Then we can use the small angle approximations to simplify our results:
y = L tan L (in radians)
m = 0, ±1, ±2,
m(d)
Constructive
Interference:
(m +
1
/
2
)(d)
Destructive
Interference:
For small angles: (<< 1 radian)
sin tan (only in radians!)
1. What is the spacing y between fringe maxima on a screen 2m away?
a. 1 m b. 1 mm c. 1 cm
2. If we increase the spacing between the slits, what will happen to y?
a. decrease b. stay the same c. increase
3. If we instead use a green laser (smaller ), y will?
a. decrease b. stay the same c. increase


Example:
Example:
A laser of wavelength 633 nm is
incident on two slits separated by
0.125 mm.
I
S
1
S
2
y

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