All waves diffract, if they pass through or around obstacles, and interfere, if two or more waves arrive at the same place at the same time. When a monochromatic plane wave passes through a single slit of the right width w, we observe a Fraunhofer single slit diffraction pattern a large distance L >> w away from the slit. When the wave passes through multiple regularly-spaced slits with slit-spacing d, we observe a multiple-slit Fraunhofer interference pattern a large distance L >> d away from the slits.
In this lab you will use Excel to produce some plots of diffraction and interference patterns for electromagnetic waves.
Open a Microsoft Word document to keep a log of your procedures. This log will become your lab report. Address the points highlighted in blue. Answer all questions.
Dark fringes in the diffraction pattern of a single slit are found at angles θ for which w sinθ = mλ, where m is an integer, m = 1, 2, 3, ... .
Assume light from a distant source passes through a narrow slit as shown on the figure on the right. Let the polarization be perpendicular to the plane of the figure. What do we observe on a distant screen?
According to the Huygens-Fresnel principle, the total field at a point y on the screen is the superposition of wave fields from an infinite number of point sources in the aperture region. Each point s on the wave front inside the aperture (-a/2 ≤ s ≤ a/2) is the source of a spherical wave. A distance r from the point s the electric field is due to this point sources is
dE = (Asds/r)cos(kr - ωt).
If r0 is the distance from the point s = 0 on the optical axis to a point y on
the screen, then the contribution dE to the total amplitude on the
screen from the point at s = 0 is
dE(y) = (Asds/r0)cos(kr0 - ωt).
Here As/r0 is the amplitude per unit width and ds is the infinitely small width of a point source. For off-axis points for which s ≠ 0, the distance is longer or shorter than r0 by an amount Δ.
The contribution dE(y) to the total amplitude on the screen from an off-axis point (s ≠ 0) is
dE(y) = (Asds/(r0 + Δ(s))) cos(k(r0 + Δ(s)) - ωt).
To find the total amplitude E(y) we have to add up the contributions from all points on the aperture. Because there are an infinite number of points, the sum becomes an integral.
E(y) = ∫-a/2+a/2(A/(r0 + Δ(s)))cos(k(r0 + Δ(s)) - ωt)ds.
We define sinθ = Δ/s. Since r0 >> Δ, we approximate 1/(r0+Δ) with 1/r0. However we cannot drop the Δ inside the cosine function, since kΔ(s) is not necessarily much smaller than 2π. We then have
E(y) = (As/r0)∫-a/2+a/2cos[(ksinθ)s + (kr0 - ωt)]ds.
Using ∫cos(ax + b)dx = (1/a)sin(ax +b) the integration yields
E(y) = (As/r0)cos(kr0 - ωt)(sin(ka(sinθ)/2)/(ka(sinθ)/2).
or, inserting k = 2π/λ,
E(y) = (As/r0)cos(kr0 - ωt)(sin(πa(sinθ)/λ)/(πa(sinθ)/λ).
The function sin(x)/x = sinc(x) is called the sinc function.
In Excel, produce a plot of sin(x)/x vs x for -4π < x < 4π.
(Divide the region from -4π < x < 4π into ~ 200 data points. You can use this spreadsheet to get started.)
Label the axes and past the plot into a Word document.
The intensity is proportional to the square of the field, I(y) ∝ E2(y). Since the square of a cosine function averages to ½, the time-averaged intensity is given by
<I(y)> = <I0>sin2(πa(sinθ)/λ)/(πa(sinθ)/λ)2,
where <I0> is the average intensity at the center.
The time-averaged intensity has a peak in the center with smaller fringes on the sides. For small angles we may approximate sinθ ~ θ. Then the first zeros on the sides of the central peak occur when πasinθ/λ ~ πaθ/λ = π, or θ = λ/a.
In Excel, produce a plot of <I(y)>/<I0> = sin2(πa(sinθ)/λ)/(πa(sinθ)/λ)2
versus πa(sinθ)/λ, for -4π < πa(sinθ)/λ
In other words, set x = πa(sinθ)/λ, and produce a plot of <I(y)>/<I0> = sin2(x)/x2 versus x, for -4π < x < 4π. Label the x-axis with πa(sinθ)/λ instead of x. Paste the plot into a Word document.
Describe the main feature of the Fraunhofer diffraction pattern. If λ = 500 nm, and you want to observe the diffraction patter you have plotted have a width of ~±2 cm a distance L = 1 m away from the slits, what is a reasonable slit width?
Assume light shines on a series of equally spaced slits. The spacing between the slits is d. The diffraction pattern is observed on a screen a distance L away from the slits, L >> d.
If we view the slits as sources of electromagnetic waves, then these sources
the electric fields E(x,t) = Emaxcos(kx - ωt + φ) of all the sources are in phase.
But if we observe the diffraction pattern on the screen a distance z away from the x-axis so that z/L = tanθ, then the electric field of source n is out of phase with the electric field of source 1 by (n - 1)δ, where
δ = k d sinθ = (2πd/λ) sinθ.
The total electric field at z is the sum of the fields due to all of the
sources. The intensity at z is proportional to the square of the amplitude
of the resultant field. The resultant field at z is given by
E = Emax∑n = 0N-1cos(α + nδ),
where α is the phase of the electric field of source one at position z on the screen and N is the number of sources.
The amplitude of this field at z is given by
E2 = Emax2[(∑n = 0N-1cos(nδ))2 + (∑n = 0N-1sin(nδ))2].
The intensity distribution as a function of δ is given by
I(δ) = E2/Emax2 = (∑n = 0N-1cos(nδ))2 + (∑n = 0N-1sin(nδ))2.
In Excel produce a plot of I(δ) versus δ for up to four sources.
Open a Microsoft Excel Spreadsheet.
With a = b = 1 and c = d = 0 the plot shows the intensity distribution of two sources.
Turn on another source by setting c = 1.
Turn on another source by setting d = 1.
Convert your log into a lab report.
Laboratory 9 Report
Save your Word document (your name_lab9.docx), go to Canvas, Assignments, Lab 9, and submit your document.