Physlets run in a Java-enabled browser, except Chrome, on the latest Windows & Mac operating systems. If Physlets do not run, click here for help updating Java & setting Java security.
Section 5.4: Wave Diffraction and Interference
Please wait for the animation to completely load.
When light passes through one slit or two slits, the interference of waves accounts for the observed pattern. Restart. First look at light from two distinct sources separated sufficiently far apart to be considered a double source. What type of pattern do you observe as you drag the detector slowly across the light shown on the screen? (position is given in mm and intensity is in arbitrary units, scaled so that the maximum intensity is 1) How does it change with wavelength (color of light)? How does it change with distance to the screen? Compare this with the single-slit and double-slit cases. Figure 5.6 shows a schematic of a double slit experiment.
Fig 5.4b. A schematic of the double-slit experiment.
To qualitatively understand the pattern on the screen in each of the cases, you need to think about the superposition of waves. If two waves match up or are in phase then the waves constructively interfere and this results in a bright spot on the screen. Light that has traveled the exact same distance from two source points will be in phase and constructively interfere. Constructive interference also occurs when the two light waves travel distances that differ by one, two, or other integer multiples of the wavelength. Similarly, if the difference between the two paths is a half-integer multiple of the wavelength of the light, the waves will be out of phase and will destructively interfere (a dark spot). Where are the positions on the screen that correspond to destructive interference?
The quantitative explanation requires further examination of each case. We will begin with the double-source case and consider the light on the screen from the first source. It has an electric field given by the real part of E1 = E0 ei(kz-ωt), where k is the wave number (2π/λ) and ω the angular frequency. The set up is as shown in Fig 5.4b.
Fig 5.4c. Geometry for interference from two sources and single slit diffraction. The electric field from the second source takes almost the same form, E2 = E0 ei(kz' − ωt) = E0 ei(kz -ωt + δ),
except with a different path length described by the phase shift, δ = k (z' - z). Thus the total electric field at our point is:
E = E1 + E2 = E0 ei(kz − ωt) (1 + eiδ) = E0 ei(kz - ωt) eiδ (e−iδ/2 + eiδ/2) ,
E = E0 ei(kz − ωt)eiδ [2 cos(δ/2)].
The intensity of the wave is proportional to <E2>, which is averaged over time. Using the fact that the time average of sin2(ωt + φ/2) and cos2(ωt + φ/2) over one period is 1/2, gives:
I = 2E02 cos2(δ/2) or Iavg = Imaxcos2(f/2).
When the path length difference is one wavelength, for instance, what is the phase difference between the waves? Since the phase difference, δ, is equal to 2π when the path length difference is one wavelength, λ:
δ = (2π/λ)*[path length difference] = (2π/λ)[d sin(θ)]
where θ is the angle the point of interest on the screen is from the center as measured from the source (see diagram). So for small angles, I = Imaxcos2(πyd/λL), where y is the distance along the screen measured from the center and L is the distance between the sources and the screen.
Now, consider a single slit. We can use the same analysis for one slit, by thinking of the slit as made of multiple sources and the interference between waves from these different tiny sources. In this case, the field due to any small portion of the slit, ds, (see Fig 5.4c) is given by dE = A ei(kz − ωt) ds, where z depends on the value of s. From the diagram (using the law of cosines),
z/R = (1 + 2(s/R)sin(θ) + (s/R)2)1/2,
where assuming R >> s, yields z = R − s sin(θ). Substituting in this expression for z into dE and integrating from −D/2 to D/2 where D is the slit width,
E = ∫dE = A ei(kR − ωt) ∫e−iks sin(θ) ds ,
so the intensity, I = <E2> is
I = I0sin2[(kD/2)sin(θ)] / [(kD/2) sin(θ)]2 ,and for small angles, sin(θ) ≈ y/L and substituting in k = 2π/λ yields
I = I0sin2(πDy/λL)/(πDy/λL)2, (5.4)
Finally, then, the double slit combines these two descriptions as the cos2(α) interference term (where α = πyd/λL) falls within an envelope given by the diffraction term of sin2(β)/β2 where β = πDy/λL. Remember that d is the distance between the slits (double-source term) while D is the width of the slit itself (single-slit diffraction term).
Section by Andrew Duffy and Anne J. Cox