Coherent Sources
Wavefront splitting Interferometer
Young’s Double Slit Experiment
Young’s double slit
© SPK
Path difference:
SP  S P
 D   x  d 2  D   x  d 2
2
2
D  x, d
2
2
  x  d 2
 D 1 
2
D

For 1  y
1  y 
n
2



12
  x  d 2
 D 1 
2
D

 1  nx
x  d 2   x  d 2


2
2D
  2 x  d  2d  xd D
2
2



12
For a bright fringe,
SP  S P  m
For a dark fringe,
SP  S P   2m  1  2
m: any integer
For two beams of equal irradiance (I0)
 xd
I  4 I 0 cos
D
2
Visibility of the fringes (V)
I max  I min
V
I max  I min
Maximum and adjacent minimum of the fringe system
Photograph of real fringe pattern for Young’s double slit
The two waves travel the same distance
– Therefore, they arrive in phase
S'
S
•The upper wave travels one wavelength farther
–Therefore, the waves arrive in phase
S'
S
•The upper wave travels one-half of a
wavelength farther than the lower wave.
This is destructive interference
S'
S
Uses for Young’s Double Slit
Experiment
•Young’s Double Slit Experiment provides a
method for measuring wavelength of the light
•This experiment gave the wave model of light a
great deal of credibility.
Phase Changes Due To Reflection
• An electromagnetic wave undergoes a phase change of 180°
upon reflection from a medium of higher index of refraction
than the one in which it was traveling
– Analogous to a reflected pulse on a string
μ2
μ1   
1
2
Phase shift
  k0   
Fresnel double mirror
P1
P2
© SPK
Problem
In a Fresnel mirror the angle between the mirrors a=12’. The
distance r= 10 cm and b=130 cm. Find
(a) The fringe width on the screen and the number of possible
maxima.
(b) the shift of the interference pattern on the screen when
the slit S is displaced by l=1 mm along the arc of radius r
about the center O.
(c) The maximum width of the source slit at which the fringe
pattern on the screen can still be observed sufficiently sharp.
Fresnel biprism
© SPK
Lloyd’s mirror
© SPK
Billet’s split lens
© SPK
Wavefront splitting interferometers
•Young’s double slit
•Fresnel double mirror
•Fresnel double prism
•Lloyd’s mirror
Division of Amplitude
Optical beam splitter
Fringes of equal inclination
C
n1
nf
n2
t
t
D
A
C
D
i
t
A
d
B
B
Optical path difference for the first two reflected beams
  n f [AB  BC]  n1 (AD)
AB  BC  d /cos t
nf
AD  AC sin  i  2d tan  t
sin  t
n1
  2n f dcost
Condition for maxima
dn f cos t  (2m  1)
f
4
m  0, 1, 2,...
Condition for minima
dn f cos t  2m
f
4
m  0, 1, 2,...
Fringes of equal thickness
Constant height contour of a topographial map
Wedge between two plates
1 2
glass
glass
t
x
Path difference
= 2t
Phase difference  = 2kt - 
D
air
(phase change for 2, but not for 1)
Maxima 2t = (m + ½) o/n
Minima 2t = mo/n
Fizeau Fringes
Newton’s Ring
• Ray 1 undergoes a phase change of 180 on
reflection, whereas ray 2 undergoes no phase
change
R= radius of curvature of lens
r=radius of Newton’s ring
d  R  R2  r 2
 1  r 2

 R  R 1     ...
 2  R 

1 r2

2 R
For bright ring
1
2 d   ( n  )
2
2
1r
1
2
  ( n  )
2 R
2
1 
1
rbright  (n  ) R  (n  ) R , n  0,1, 2...
2 
2
For dark ring
2d  n
rdark  nR , n  0,1, 2 ...
Reflected Newton’s Ring
Newton’s Ring
Types of localization of fringes
Interference fringes
Real
Virtual
Localized
Non-localized
Localized fringe
Observed over particular surface
Result of extended source
Non-localized fringe
Exists everywhere
Result of point/line source
POHL’S INTERFEROMETER
Real
Non-localized
Refer Hecht for details
Virtual
Localized
Problem
The width of a certain spectral line at 500 nm is 2×10-2 nm.
Approximately what is the largest path difference for which
the interference fringes produces by the light are clearly
visible?