Physical Optics

14.1  Interference

14.1.1  Interference due to Differing Distances

Before the 1800's, the view that light had a particle nature rather than the wave nature described by Huygens, held sway over many opinions. The difficulties with Huygens model are rather obvious. The light being emitted into directions other than the wavefront were deemed "to weak to see", the difficulty of believing that waves could propagate across the space between the earth and the sun without an intervening medium, and so on. The first experimental "proof" that light behaves as a wave was provided by Thomas Young in 1801. That experiment shows that light undergoes interference effects. Interference is a classic hallmark of the behavior of waves and is a phenomenon not seen to be intrinsic to matter until the advent of quantum mechanics in the 20th century.

Before considering Young's experiment, let's note a relatively simple example of interference with mechanical waves. Since mechanical waves experience interference (consider that we introduced the notions of complete constructive and destructive interference with mechanical waves), it is necessary that electromagnetic waves should as well. For the simple example, we consider two "clappers" that hit the water in a shallow tank at periodic intervals. Since the frequency of the two "clappers" is the same and they have a fixed phase relationship, we say that they serve as coherent wave sources. Coherence simply implies same frequency/fixed phase relationship.

To see this phenomenon with water waves, see figure 14.1.


Figure 14.1: Two strikers hit the water a certain distance apart and create transverse water waves in phase, i.e. at the same time. The distinctive interference pattern of the two waves is shown.

The cause of this pattern is easy to determine if we remember how constructive/destructive interference can come about. To review what happens when waves overlap, look again at figure  11.5 or at this Java applet.

In the case of the clappers, we note that they are different distances away from most positions in the tank of water. We see for one dark area and one light area how the distances from one clapper (r1 and r1¢) differ from the corresponding distances of the other clapper (r2 and r2¢). The different distances traveled mean that even if the waves started out in phase from the two clappers, they will be out of phase by the time they reach the same destination.


Figure 14.2: The two waves from the clappers are in phase, but become out of phase as they travel different distances to get to the same destination. The waves superpose at every position in the tank, but those phase differences due to distance differences to a point which differ by an integral wavelength lead to a pileup of water at that position (and hence a dark spot) while those points which lead to distance differences which are a non-even integer of half-wavelengths lead to a low spot of water (and hence a brighter spot).

We can quantitatively describe this effect by stating that constructive interference occurs when the path difference, r2 - r1, satisfies the following relationship:
r2 - r1 = ml   (m = 0, ±1, ±2, ±3, ...)
(14.1.1.1)
i.e. an integer number of wavelengths for the path difference has the two waves looking "back" in phase and completely constructive interference occurs.

If the path difference is an odd integer of half-integral wavelengths, then completely destructive interference occurs since the phase difference is an odd integer multiple of p/2 and so troughs for one wave correspond to crests of the other and vice versa. The formula for this is
r2 - r1 = æ
ç
è 		m + 1
2
 ö
÷
ø 		l   (m = 0, ±1, ±2, ±3, ...).
(14.1.1.2)

14.1.2  Young's Experiment

We can now understand Young's experiment in that we can see that any two coherent sources can generate phase differences over a space due to distance differences. Young produced his coherent sources starting with an incoherent light source by letting monochromatic (i.e. one frequency) light fall on a slit. This is necessary since light, even monochromatic light, from a source is generally not synchronized in phase from different parts of the light source. Allowing the light to fall on a slit means that only the light from one small region of the light source gets through. This light is now allowed to fall onto two other slits. Light from these slits forms two coherent sources and interference from these two can be seen by allowing the light from both slits to fall onto a screen. We can see the result in figure 14.3 and in this Java applet.


Figure 14.3: Young's experiment showing constructive and destructive interference for light. Constructive interference corresponds to the regions where the bright colored bars appear. The dark areas in between are the regions of destructive interference. The graph at the far right end of the picture shows the intensity of the light.

The quantitative description of interference for Young's experiment is characterized by the distance d between the two slits, the wavelength of the light, and the angle q between the horizontal and the line from the top slit to the point of interest on the screen. We find experimentally that
dsinq
=
ml   (m = 0, ±1, ±2, ±3, ...)  constructive interference
dsinq
=
æ
ç
è 		m + 1
2
 ö
÷
ø 		l   (m = 0, ±1, ±2, ±3, ...)  destructive interference
(14.1.2.3)

The light and dark bands are referred to as interference fringes. The center of the pattern, corresponding to m = 0, is equidistant from the two slits. All other points on the screen are further away from one of the slits than the other, hence the constructive and destructive interference effect works the same way here as it did for the clappers in the water wave tank. The "extra" distance traveled by light from one of the slits to point P on the screen is easily shown to be about dsinq, as seen in figure 14.3, if we also assume that the distance L from the slits to the screen is much larger than d. This requirement is necessary so that we can assume that q is approximately the same angle that the line from the bottom slit to the point P makes with respect to the horizontal. Then the vertical position of each bright interference fringe is given by
ym = Ltanqm
(14.1.2.4)
where m identifies which bright fringe from the center we are talking about. Notice that ym can be either positive or negative depending on the value of qm. If L >> d, then, for the first few values of m, we have qm as a small angle. In that case,
tanqm @ sinqm Þym = Lsinqm = L ml
d
 .
(14.1.2.5)

Phase Differences and Path Differences

As a general rule, any path length difference between two coherent sources can be expressed as a phase difference with the following association:
f
2p
 = r2 - r1
l
 Þf = 2p
l
 (r2 - r1) = k(r2 - r1)
(14.1.2.6)
where k is the wave number. We can go further: any path difference which results in a different number of wavelengths to go from the source to a point in space in comparison to the number of wavelengths to go from another coherent source to the same point generates a phase difference. Other than path length differences what could result in such a wavelength difference? It turns out that electromagnetic waves can change their wavelength (compared to the wavelength in vacuum) inside a material with an index of refraction other than 1.0. The frequency of electromagnetic waves is unaffected by passage through most materials, so if the speed of light in those materials is smaller than in a vacuum, this must be the result of a reduction in wavelength! Hence even if the distance traveled by the waves from two coherent sources is the same, they may still experience a phase difference as shown in figure  14.4 or in this Java applet.


Figure 14.4: Wavelength changes in a material can cause phase differences even for coherent waves which travel the same physical distance.

This is incorporated into our general scheme for calculating phase changes by letting the wavelength, l, be given by
l = l0
n
 Þ k = nk0
(14.1.2.7)
where l0 is the wavelength in vacuum and n is the refraction index of any material an EM wave goes through. Finally, if the two coherent sources are separated by a distance d, then the combination of their waves at a point (call it P) which is distant from either of the two sources gives a path difference which is
r2 - r1 = dsinq
(14.1.2.8)
with q defined as the angle of the line from one of the slits to the point P compared to a line from the center of the slits to point P. In this case we can define the general phase difference as
f = k(r2 - r1) = kdsinq = 2pd
l
 sinq
(14.1.2.9)
with l and k defined as in equation 7.

Intensity for Phase Difference Superposition

For mechanical waves, we showed that we can use trigonometric identities to find the result of superposing two waves with different phases. Suppose the electric fields for two waves are described as follows:
E1(t)
=
Ecos(wt)
E2(t)
=
Ecos(wt + f)
(14.1.2.10)
Then the sum of these two waves yields a resultant wave which is described by
EP = ê
ê
ê 		2Ecos 1
2
 f ê
ê
ê 		cos æ
ç
è 		wt + 1
2
 f ö
÷
ø
(14.1.2.11)
Since, for electromagnetic waves, we have seen that the intensity at any point and time is proportional to the square of the amplitude, we have
I
=
I0cos2 f
2
I0
=
Sav, max. = EP, max.2
2m0c
 = 1
2
 e0cEP, max2 = 1
2
 e0c(4E2) = 2e0cE2
(14.1.2.12)
where I0 is the maximum intensity and occurs at positions where the phase difference between the two original waves is zero (i.e. f = 0). Hence we can use our expression for phase differences from equation 9 to predict that the intensity far from two sources is given by
I = I0cos2 f
2
 = I0cos2 æ
ç
è 		1
2
 kdsinq ö
÷
ø 		= I0cos2 æ
ç
è 		pd
l
 sinq ö
÷
ø
(14.1.2.13)
and the directions of maximum intensity occur for
pd
l
 sinq = mp   (m = 0,±1, ±2, ...)
(14.1.2.14)

14.1.3  Example Problems

Some examples are probably in order at this point.

Problem 1:
Two slits spaced 0.450 mm apart are placed 75.0 cm from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a light of 500 nm?
Solution:
The ordering of the fringes in terms of values of m are shown in figure  14.5.


Figure 14.5: The ordering of bright and dark fringes corresponding to maxima and minima in the interference pattern for a two-slit experiment.

The second and third dark lines are found for m values of 1 and 2 in the formula
dsinq = æ
ç
è 		m + 1
2
 ö
÷
ø 		l.
(14.1.3.15)
The distance between the dark fringes is given by the same formula as the distance between bright fringes, namely
Dym = R Dml
d
(14.1.3.16)
since consecutive dark fringes correspond to changes of one wavelength in the phase difference between waves from the two slits. Hence, the distance we want is
Dym = (0.75 m) (1)(500×10-9 m)
0.450×10-3 m
 = 8.33×10-4 m
(14.1.3.17)

Problem 2:
Two slits spaced 0.260 mm apart are placed 0.700 m from a screen and illuminated by coherent light that has a wavelength of 660 nm. The intensity at the center of the central maximum is I0.
  1. What is the distance on the screen from the center of the central maximum to the first minimum?
  2. What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to I0/2?
Solution:
  1. The distance between maxima and adjacent minima, in terms of angle is given as follows:
    |sinqmaxima -sinqminima|
    =
    ê
    ê
    ê     		m l
    d
    		 - æ
    ç
    è     		m + 1
    2
    		 ö
    ÷
    ø     		l
    d
    		 ê
    ê
    ê     		Þ
    R|sinqmaxima -sinqminima|
    =
    R l
    d
    		 1
    2
    Dym
    =
    R l
    2d
    =
    (0.700 m) 660×10-9 m
    2(0.260×10-3 m)
    Dym
    =
    8.88×10-4 m
    		(14.1.3.18)
  2. We work from the intensity formulation.
    I
    =
    I0cos2 f
    2
    		 Þ
    cos2 f
    2
    =
    I
    I0
    		 = 1
    2
    		 Þ
    f
    =
    2cos-1 1
    Ö2
    		 = p
    2
    p
    2
    =
    2pd
    l
    		 sinqÞ
    sinq
    =
    pl
    4pd
    =
    660×10-9 m
    4(0.260×10-3 m)
    sinq
    =
    6.35×10-4
    		(14.1.3.19)
    We can use this value of sinq to find the vertical distance from the central maximum position.
    y
    =
    Rsinq
    =
    (0.700 m)(6.35×10-4)
    y
    =
    4.44×10-4 m
    		(14.1.3.20)

    Problem 3: A thin flake of mica (n = 1.58) is used to cover one slit of a double-slit arrangement. The central point on the screen is now occupied by what had been the seventh bright fringe (i.e. m = 7) before the mica was introduced. If the wavelength of light illuminating the slits is 550 nm, what is the thickness of the mica?


    Solution: We note that the central spot on the screen is equidistant from both slits. If this spot is now the position of what was formerly the 7th maximum, then the optical pathlength in the mica must account for a phase shift of 7 wavelengths, i.e. to say, the number of wavelengths to the central spot on the screen for light from the uncovered slit is
    Nl, uncovered = L
    l0
    		(14.1.3.21)
    where L is the distance from the slits to the screen and l0 = 550 nm. For the slit covered with mica, the number of wavelengths to the central spot is
    Nl, covered @ L-t
    l0
    		 + nmicat
    l0
    		(14.1.3.22)
    where t is the mica thickness. The phase shift is equal to 7 wavelengths, so
    Nl, coveredl0 -Nl, uncoveredl0
    =
    7l0
    L - t + nmicat - L
    =
    7l0
    (nmica - 1)t
    =
    7l0 Þ
    t
    =
    7l0
    nmica - 1
    =
    7(550×10-9 m)
    1.58 - 1
    t
    =
    6.64×10-6 m = 6.64mm
    		(14.1.3.23)

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