Chapter 7. Optical Methods – Interference & Diffraction of Light
Interference
Sample problem
7-S1
From Section 7.3.1 the equation for interference of two wavefronts is given by


I  I1  I 2  E 01 E 02 cos 
The last term is the interference term and it contains the scalar product of the two electric
potential vectors. Analyze this result and compare with the expression arrived at utilizing the
scalar theory of light propagation that neglects the vectorial nature of light. Discuss the
relationship between the two results and indicate the limitation of the scalar theory.
Solution to 7-S1
In the scalar theory of light propagation each propagating beam can be represented by a phasor
of the form indicated by (7.19). The superposition of the phasors yields the interference equation:
I  ( A1  A 2 ) .( A1  A 2 )  A 21  A 22  A1A 2 cos(1 2 )  I1  I 2  I1I 2 cos(1 2 )
Comparing (7.31) it is possible to see that the scalar theory is assuming that the two interfering
beams have parallel states of polarization by not considering the vectorial properties of light
propagation. This of course is not generally a valid assumption and the scalar product can have
any value between one and zero. Orthogonal states of polarization will have scalar products
equal to zero and hence will not produce states of interference. To remediate this problem the
coefficient of visibility is introduced in the preceding equation providing a means of
compensating for the approximation involved in the scalar theory of interference. Thus the
visibility ν in the scalar theory accounts for the effect of the state of polarization and takes into
consideration the coherence of the light
I  I1  I2   I1I2 cos(12 ) where    1
Note: In chapter 7 the derivation of the fundamental equation of interference was limited to a
linear state of polarization. However the phenomenon is not limited to plane polarization. The
Chapter devoted to Photoelasticity will cover more general cases of interference.
7-S2
Two identical partially coherent beams overlap in a region of space. Utilizing the concepts
introduced in Chapters 6 and 7 analyze the possible outcome.
Solution to 7S-2


Each one of the wavefronts is defined by the corresponding k 1 and k 2 vectors. With respect to



these vectors the vectors E1  E1P  E1S can be defined. In this vectorial expression the vectors
representing each of the wavefronts is split into P-polarization and S-polarization beams. In the



same way E 2  E 2 P  E 2S . Since these polarizations are orthogonal forms they will not interfere.


However due to the state of partial polarization E1P and E 2 P are coherent and interfere. In the


same way E1S and E 2S will also interfere. Hence interference fringes will be observed as a result
of the partially coherent beams.
Note: More specific examples will be presented in the chapter of Photoelasticity.
7-S3
Compute the intensity distribution resulting from the interference of two beams of equal intensity
that have parallel field vectors.
Solution to 7-S3


Applying I  I1  I 2  E 01 E 02 cos  and considering that the modulus of the two vectors E01 and
Eo2 are equal one gets,

I  2I0 (1  cos ) , but taking into consideration that 2 cos 2  1  cos  one obtains
2

I  4I0 cos 2
2
7-S4
Assume that one has two coherent point sources that have parallel states of polarization.
Determine the loci of the interference points in the space
Figure P7.1 Interference of two coherence point sources in space.
Solution to 7-S4
Figure P7.1 shows a cross section in space of the pattern formed by two point sources. It is
known from sample problem 6-S11 that the emerging wavefronts are spherical wavefronts
defined as,



 

 
E
E
E1 ( r, t )  0 cos( k  r  t  1 ) and E 2 ( r, t )  0 cos( k  r  t   2 ) .
r
r


Let us consider a point in space (Figure P7.1) and characterized by the vectors r1 and r 2 . The
fields at the point will be,








E1 ( r, t )  E01 cos( k  r1  t  1 ) and E2 (r, t )  E02 cos( k  r2  t   2 )
The interference produced by the two sources according to the result of sample problem 7-S1
depends on the difference of phase of the two wavefronts,
  k(r1  r2 )  (1   2 )
Because the vector k is radial, the dot product reduces to the product of the moduli of the
corresponding vectors as indicated in the preceding equation. Then the interference fringes are of
the form,
2
2
 E 02
I  I0  2Iin cos( k( r1  r2 )  (1   2 ) where I0  E 01
and Iin  E01E02
The maxima of the intensity will occur every time that the difference of phase δ=2nπ, n=0, 1, 2
and the minima δ=2(n+1)π. Then the condition for observing maxima and minima are
2n   2  1
k
( 2n  1)   2  1
r1  r2 
k
r1  r2 
The above equations define concentric hyperboloids of revolution that intersected by a plane as
shown in Figure P7.1 produce hyperbolas. In Figure P7.1 the thin lines represent the spherical
wave’s cross- section with the plane of the page. The lines that create a divergent fan of fringes
are the interference fringes. It is possible to see that along planes that pass through the line S1S2
and at long distances of the source the fringes may appear to be a set of parallel lines.
7-S5
Utilizing Figure P7.2 derive the equation of the fringes formed in the Young’s experiment.
Figure P7.2 Young’s experiment.
Solution to 7-S5
In this experiment the interference is produced by wave front splitting. Assuming that there is a
coherent source (i.e. laser) a point source emits coherent light and two points of the original
wave front are utilized as sources. This problem was solved in sample problem 7-S4, the
resulting fringes observed in plane sections are hyperbolas. As such it is possible to derive an
asymptotic solution introducing some assumptions.
The first assumption is that the distance s between the sources and the observation screen is such
that s>>a, then S1P  S2P . Under this assumption the amplitude of the field vector is such that
the two moduli, E1=E2. Under these circumstances the two propagating fields will have the same
amplitude and assuming that the two sources have the same phase







E1 ( r, t )  E0 cos( k1  r1  t )

E1 (r, t )  E0 cos( k 2  r2  t )
The difference of phase of the two wavefronts is then,
     
   k 1  r1    k 2  r2  ,

 

Utilizing the argument that the dot product reduces to the product of the moduli because the k
vectors are radial one arrives at   k(r1  r2 ) . It is necessary to evaluate the difference between
the paths. The difference of the optical path is S1A. To get an expression for the fringe loci a
system of coordinates with center at P0 is introduced, then the point P has a coordinate y. S1A=a
sin θ, but since s>>a is assumed, then sin θ is very close to θ and S1A=a sin θ. From the Figure
P7.2 the angle θ is given by tanθ=y/s, but again when the angels are small tanθ=sin θ=θ, S1A=
ay/s and then of the form
k
ay
2 ay
, and the equation of the fringes is I  I 0  I1 cos
.
s
 s
The period of the fringes depends only on the y coordinates and not in the x coordinates. This is
of course an approximation, the loci of the fringes is not a straight line but a hyperbola with very
small changes of curvature, to all practical effects in a limited extent of x the fringes can be
assumed to be straight fringes; this is then asymptotic solution, the hyperbole are represented by
their asymptotes.
7-S6
Light from a point source falls onto a thin plate of index of refraction n2 (Figure P7.3). The light
impinges on the plate and the phenomenon of interference that takes place is observed at some
distance from the plate. The observer sees interference fringes. Explain the formation of these
fringes.
Solution to 7-S6
Figure P7.3 shows the effect of a thin transparent plate on a coherent beam that impinges on the
plate with an angle θi1 (sample problems (6-S26) and (6-S27). Assuming that successive
reflections are of very low intensity and considering only the first reflection, the optical path
difference between the two beams is,
AD  AB sin i1
Then,
t
S  2 
 n 2  n1 AD .
cos t1
Also and taking into consideration Snell’s law:
AD  AB sin i1
But
AB  2t  tgt1 , then
n
AD  2 t  tgt1  2 sin t1
n1
t
n t
2 t sin 2 t1
 n 2  2t  tgt1  n 2  sin t1  2 2 
cos t1
cos t1
cos t1
t
S  2 
 n 2 (1  n1 sin 2 t1 )  2n 2 t cos t1
cos t1
S  2 
There is an additional difference of phase   , because the beam reflects from an internal face
and the other from an external phase.
Figure P7.3.Interference fringes produced by a thin plate
Expressing the difference of optical path as an angle,
4t
  ks 
cos t1  

Utilizing the angle of incidence:

4 t
( n 22  n12 sin 2 i1 )  

The two beams emerging from the plate can be seen as coming from two virtual sources located
at infinity. Also if the beam impinging at point A comes from a source far away from the plate,
there will be two sources, one real and the other virtual far away from the plate but very close to
each other. Taking into consideration the irradiance of the field one will get,

I  I 0 cos 2
2
And with the substitution:
 1
cos2  1  sin  one arrives to
2 2
4 t


I  I 0  1  cos
( n 22  n12 sin 2 i1 )   



The above expression does not specify where the point P of observation is located, it depends
only on θi1. One can see also that in the above expression the position of the source is not defined
either. Hence the above expression is valid for a point source as well as for an extended source.
The observed fringes are fringes of equal inclination. Hence to relate the fringe spacing observed
in a particular set up to the thickness of the plate, the particular geometry utilized to perform the
observations must be specified.
7-S7
Similar fringes to those observed in sample problem 7-S6 may be obtained with the set up shown
in Figure P 7.4 .Two glass plates limit a thin wedge of air. A collimated beam of light is
projected normal to the lower plate via a beam splitter and is also observed normal to the surface
of the base plate.
Figure P7.4 Illustration of the formation of fringes of equal thickness.
Derive an expression providing the thickness of the air wedge as a function of the distance to the
point of contact of the two plates assumed to be the center of coordinates.
Solution to 7-S7
From sample problem 7-S6 S  2n 2 t cos t1 .The notation t ( x)  t cos t1  x can be introduced,
since the angle α is small the arc is equal to the tangent. Furthermore the reflection on the base
plate is an internal reflection while the reflection in the upper plate is external, according to the
Fresnel relationship the difference of phase is   depending what is considered external or
internal. Intensity maxima will be observed when the two interfering wavefronts are in phase,
2 n
  n2, n  0,1,2,3... Hence  
n 2 t ( x )   . There is a condition of normal illumination and

normal observation therefore sample problem 7-S6 i1  t1  0, cos t1  1 .
Hence maxima of intensity will be observed when,
2 n

1 3 5 2n  1
n 2 t ( n )    2n , or 2n 2 t ( m )   m where m  , , ...

2
2 2 2
2
The quantity n2 t( x) is the optical thickness of the wedge then 2n 2 t ( m ) 

 m
2
this equation yields,
1 

t ( m) 
m   ,

2n 2 
2
choosing the minus sign for the air wedge t ( n ) 
n
,
2n 2
if the wedge is air n2=1 and we obtain the result t ( n ) 
n
with n=1.2.3…
2
We can compute the fringe spacing,

 p  x n 1  x n 
2n 2 
Problems to solve
7.1
A solid state laser of λ=635 nm sends a parallel beam of light that has an angle of inclination
θ=30o with respect to the normal of a glass plate of index of refraction nr=1.52 as shown in
Figure P.7.3. The thickness of the plate t=1.1 mm. Compute the optical path of the beam going
through the thin plate. Compute the phase difference between the two beams coming out of the
plate.
7.2
A setup similar to the one shown in Figure P.7.4 is utilized to observe a thin glass plate. The
source is an extended coherent source. Assume that the thin glass plate lower face is supported
on a non reflecting surface. Show that the interference fringes formed in the image plane are
circular fringes.
7.3
With the set up shown in Figure P.7.4 a pattern of fringes of pitch  p  31.2 microns is observed
formed by a wedge of glass of index of refraction 1.5. The fringes are almost parallel fringes.
What is the angle between the faces of the wedge? The observation is made with a solid state
laser of λ=635 nm.
7.4
In a Michelson interferometer arrangement a set of 30 fringes are observed in field of 22 cm.
What is the deviation angle from π/2 between the fixed and the movable mirrors of Figure 7.11?
The source of illumination is a laser source of λ=635 nm. Make a sketch of the set up.
7.5
In the Mach-Zender interferometer of Figure 7.12 in the field of observation one can see 10
fringes in a field of 2 cm. Compute the rotation angle between the mirrors M1 and M2.
7.6
In the same Mach-Zender two parallel plates of the same thickness 5 mm are inserted in the
horizontal arms keeping them parallel to each other. The instrument has a scale that allows the
determination of the position of minima of the fringes. After the insertion the minima have
experienced a displacement of ¼ of a fringe. If the index of refraction of one of the plate is 1.52,
what is the index of refraction of the other plate?