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Section 1.5 Phasor Diagrams for Multiple Source Interference

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Sometimes, more than two waves will interfere at a single point. In these cases, the total amplitude of the combined oscillation will be the superposition of all of the incident waves. The phasor addition technique can be used to calculate the total amplitude in these complicated cases. Consider, for example, three oscillations, all with the same angular frequency ω, but with different amplitudes and phases:

y1(t)=(A1cos⁡(ωt+ϕ1),y2(t)=(A2cos⁡(ωt+ϕ2), andy3(t)=(A3cos⁡(ωt+ϕ3).

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These three oscillations can be expressed graphically, as shown in Figure 1.22, and their superposition, ytot=y1+y2+y3, can be determined from vector addition of the three phasors.

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Figure 1.22. Phasors for the three oscillations y1, y2, and y3.

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One of the most common multi-source interference problems is an extension of the two-slit problem considered in class. Imagine a light source passing through a screen containing not two but three slits, each separated from its neighbor by a distance d, as shown in Figure 1.23. The interference pattern for this arrangement of slits will look different from the interference pattern for a two-slit setup.

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Figure 1.23. Three closely spaced slits, illuminated straight-on from the right. The interference pattern is observed on a distant screen to the right.
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Figure 1.24. Light rays exiting the three slits parallel to the normal.

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If the interference pattern illuminates a screen a distance L from the slits, where L≫d, then we can assume that the light rays from each of the slits to any particular point on the screen are parallel. (This is the same approximation we made in the two-slit case.) For example, light passing straight through the slits and continuing straight to the screen produce the central maximum in the interference pattern, as in Figure 1.24. These rays combine constructively because there's no difference in the path length for the three light rays, i.e., Δr=0 for all three rays.

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Now consider the light rays traveling at an angle θ to the normal, as in Figure 1.25. Just as in the two-slit case, these rays do not travel the same distance to reach the screen. Light from the bottom slit travels the shortest distance, while light from the middle slit travels an extra distance Δr21, and light from the top slit travels the same amount of extra distance (Δr32=Δr21) compared to light from the middle slit (Here, I've implicitly labeled the slits, #1, 2, and 3 for bottom, middle, and top.).

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Figure 1.25. Light rays exiting at an angle θ to the normal.
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Figure 1.26. Geometry of path-length difference between rays from three slits. As for the double slit, the path length difference for adjacent slits is dsin⁡θ if the rays are nearly parallel.

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From the geometry of the problem (shown in Figure 1.26), it's clear that Δr32=Δr21=dsin⁡θ. We can then calculate the phase difference between light from adjacent slits,

(1.22)Δϕ=2πΔrλ=2πdsin⁡θλ.

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With knowledge of the phase difference, we can draw a phasor diagram and plot phasors for all three light rays, as shown in Figure 1.27. We've assumed that the amplitudes of all three phasors is the same, and connected the phasors head-to-tail instead of placing all three with tails at the origin. Note that Δϕ is the phase difference between adjacent phasors. The total amplitude of the combined rays is just the vector sum of the phasors:

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Figure 1.27. Three phasors, with adjacent phase difference Δϕadj..
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Figure 1.28. The resultant phasor is the vector sum of the three phasors corresponding to waves from the three slits.

Light of wavelength λ=478nm passes through a three-slit system where the slits are separated by a distance d=1.5×10−6m, and produces an interference pattern on a distant screen. What is the amplitude of the light directed at an angle θ=1.2∘ from the direction to the central maximum in the interference pattern? Assume that the amplitude of the light from each slit individually is A.

Solution.

This is a direct application of the method described above. The phase difference between adjacent slits is

Δϕ=(2πΔrλ=(2πdsin⁡θλ=(2π1.5×10−6×sin⁡(1.2∘)478×10−9=(0.41rad.

Therefore, our phasor diagram looks like Figure 1.30. To calculate the total amplitude, we again add the phasors as vectors:

Figure 1.30. Phasor diagram for Example 1.29.
   Phasor      real part (horizontal)       imaginary part (vertical)   
1 Acos⁡(0)=A Asin⁡(0)=0
2 Acos⁡(0.41)=0.92A Asin⁡(0.41)=0.40A
3 Acos⁡(0.82)=0.68A Asin⁡(0.82)=0.73A
Total 2.6A 1.13A

So the amplitude of the resultant phasor is

(1.23)Atot=(2.6A)2+(1.13A)2=2.83A.

For the same setup as in the previous example, at what angle θ would you expect to find the first minimum?

Solution.

A minimum in the interference pattern will occur when the amplitude of the resultant phasor is zero, so we need to arrange the three phasors such that they add to zero. When we had only two phasors, this was quite easy — we chose Δϕ=π. In this case, however, we need to make three phasors add to zero, and Δϕ=π won't work (what would a phasor diagram with three phasors, each differing in phase from its neighbor by Δϕ=π look like?)

Figure 1.32. Phasor diagram at the first minimum of the three-slit interference pattern.

To get three phasors to add to zero, we need a diagram like Figure 1.32, where the phase difference between adjacent phasors is Δϕ=2π3rad. Make sure you understand why this arrangement yields a total amplitude of zero.

Now that we know the required value for Δϕ, we can relate this to the path length difference between adjacent slits:

(1.24)Δr=Δϕ2πλ.

Since Δr=dsin⁡θ, we can solve for sin⁡θ:

sin⁡θ=(Δrd=(Δϕλ2πd=(2π(478×10−9)6π(1.5×10−6)=(0.106

and therefore θ=sin−1⁡(0.106)=6.1∘.

Figure 1.33. Phasor diagram at second minimum of the three-slit interference pattern.

There's actually another way to make three phasors add to zero, and the phasor diagram looks like Figure 1.33: In this case, Δϕ=4π3, and you can show that this minimum will occur for θ=12.3∘.