Double Beam Wire Diffraction - Explanation

****In order to preform this calculation, you must input your parameters into the dark orange “Input Parameters” sections. There are two sections entitled “Input Parameters” (in dark orange print) and contain a total of 6 variables that need to be inputted in the set of lines after the lines in which the variables are defined. After inputting your parameters, you must drag your cursor over to the right side of the mathematica file (where the brackets are), click on the bracket all the way to the right that runs the entire length of the document, and press the shift + enter keys at the same time. It may take a little while to load but be patient until all of the brackets on the right side of the screen turn from dark black (which means they are calculating) to their normal blue color****

Theory Section:

Franhoufer diffraction. Two beam interference. At center of dark fringes thin wires are placed.
Using Babinet’s principle we study diffraction by thin slits instead of thin wires.

Arbitrary Single Slit Diffraction, where the slit is placed at the center of the dark fringe and the diffraction is (Subscript[D, SingleSlit]).

(Subscript[D, SingleSlit])=(-1)^j*(∫_(j*d-b/2)^(j*d+b/2))(z-j*d)*Sin[F+a*z]\[DifferentialD]z

This can be applied to diffraction due to multiple slits. Input the number of slits in the experiment below as (Subscript[n, slits])

Input Parameter

(Subscript[n, slits])= the number of wires in your experiment

(Subscript[n, slits])=1

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For a setup with (Subscript[n, slits]), the diffraction due to (Subscript[n, slits])is found by taking the sum when j=0 of (Subscript[n, slits])-1 (which we can call (Subscript[n, sum])) and can be found by:

(Subscript[n, Sum])=(Subscript[n, slits])-1

(Subscript[D, NSlits])=(∑_(j=0)^(Subscript[n, Sum]))-(((((-1))^j) Cos[F+a d j] (a b Cos[((a b)/2)]-2 Sin[((a b)/2)]))/(a^2))

In order to find the time average of the square of this function, integrate it on F from 0 to 2π and multiply by the fraction (1/(2π)).

(Subscript[D, NSlitsSquared])=(Subscript[D, NSlits])^2

(Subscript[I, D])=(1/(2π))*Integrate[((Subscript[D, NSlitsSquared])),{F,0,2π}]

The intensity of diffracted light (which we call (Subscript[I, Diffracted])) is defined as:

(Subscript[I, Diffracted])[a_,b_,d_]:=(Subscript[I, D])

Calculations Section:

Input Parameters

You must input values for:
α= 1/2  angle between the two beams
b= width of the wires (or slits)
λ= wavelength of the beam
R= distance from the grating to the screen
W=side length of the square beam required to get proper decrease in number of photons at detector in front of the beam

α=0.001485;
b=0.000017;
λ=632*(10^(-9));
R=2.521;
W=0.001;

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d= distance between the centers of two consecutive wires (or slits)

k=((2π)/λ);
a=k×Sin[θ]
d=(λ/(2α))

Diffracted Light

With slits of different thickness, the first peak remains nearly identical except its height changes.

Manipulate[Plot[(Subscript[I, Diffracted])[a,t,d],{θ,0.0,0.0015}],{t,0,32.*10^(-6)},PlotPoints->200,PlotRange->All]

The Diffraction Pattern

Plot[10^18*(Subscript[I, Diffracted])[a,b,d],{θ,-.06,.06},PlotPoints->500,PlotRange->All,AxesLabel -> {θ, Intensity }]

The First Two Peaks Corresponding to Detector Position

Plot[10^18*(Subscript[I, Diffracted])[a,b,d],{θ,-.0015,.0015},PlotPoints->500,PlotRange->All,AxesLabel -> {θ, Intensity }]

Sum of All the Diffracted Light

(Subscript[L, DiffractedSum])=2*NIntegrate[(Subscript[I, Diffracted])[a,b,d],{θ,0.0,0.1,.17,.23,.34,.41,.49,.5,.6,.69,.78,.88,.99,1.1,1.2,1.4,π/2}]

Wire-Grid Region Photon Count

Number of Photons that Cross the Wire-Grid Region (without wire-grid in place)

(Subscript[P, WireGridRegion])=W*Integrate[(1/2)(((2×Cos[k×0.0009999998333×y]))^2),{y,-W/2,W/2}]

Sum of Two Undisturbed Beams

(Subscript[P, TwoUndisturbedBeams])=2*W*Integrate[(1/2)(((1))^2),{y,-W/2,W/2}]

Number of Photons Stopped by the Wires Located at the Minima 

Approximated by a linear function:

(Subscript[P, WireStopApproximation])=(Subscript[n, slits])*W*Integrate[(1/2)(((2×k×0.0009999998333×y))^2),{y,-b/2,b/2}]

Exact Calculation:

(Subscript[P, WireStopExact])=(Subscript[n, slits])*W*Integrate[(1/2)(((2×Sin[k×0.0009999998333×y]))^2),{y,-b/2,b/2}]

Exact fraction stopped by the wires:

(Subscript[F, StoppedByWires])=((Subscript[P, WireStopExact])/(Subscript[P, WireGridRegion]))

Fraction of Photons that go to Detector 1

Detector 1 is centered at α
Starts at (Subscript[β, 1])=α+W/(2R)
Ends at (Subscript[β, 2])=α-W/2R

Plot[10^18*(Subscript[I, Diffracted])[a,b,d],{θ,α-W/(2R),α+W/(2R)},PlotPoints->500,PlotRange->All,AxesLabel -> {θ, Intensity }]

Detector on Right

(Subscript[I, Right])=NIntegrate[(Subscript[I, Diffracted])[a,b,d],{θ,α-W/(2R) ,α+W/(2R) }]

Detector on Left

(Subscript[I, Left])=NIntegrate[(Subscript[I, Diffracted])[a,b,d],{θ,-α-W/(2R) ,-α+W/(2R) }]

Integral Over the Two Detectors

(Subscript[I, TwoDetectors])=(Subscript[I, Right])+(Subscript[I, Left])

(Subscript[I, EntireRegion])=(Subscript[L, DiffractedSum])

In order to find the fraction of light over the two detectors compared to the light over the entire region, we must:

(Subscript[I, FractionTwoDetectors])=(Subscript[I, TwoDetectors])/(Subscript[I, EntireRegion])

Fractional Area Under Each Detector

(Subscript[A, UnderEachDetector])=(Subscript[I, FractionTwoDetectors])/2

Percentage of Area Both Detectors Take Up
(0.15% from calculation of:)

(Subscript[I, FractionTwoDetectors])*100

Fraction of Diffracted Photons at Each Detector

(Subscript[F, DiffractedAtEachDetector])=(Subscript[F, StoppedByWires])*(Subscript[I, Right])/(Subscript[I, EntireRegion])

Fraction of Diffracted Photons Away From Detectors

(Subscript[F, DiffractedAwayDetectors])=(Subscript[F, StoppedByWires])-2*(Subscript[F, DiffractedAtEachDetector])

Fraction of Photon Losses At Either Detector

(Subscript[F, PhotonLossesEitherDetector])=(Subscript[F, DiffractedAwayDetectors])+(Subscript[F, StoppedByWires])

Photon Count At Each Detector
Using conservation of energy in terms of the number of photons:
2N photons is equal to the losses at the wires plus all the diffracted photons away from the detectors plus the two detector counts.

2N=2N((Subscript[F, StoppedByWires]))+2N((Subscript[F, DiffractedAwayDetectors]))+2N((Subscript[F, AtEachDetector]))

Solving for the fraction at each detector, we get:

2N((Subscript[F, AtEachDetector]))=2N-2N((Subscript[F, StoppedByWires]))-2N((Subscript[F, DiffractedAwayDetectors]))

Dividing both sides by 2N yields:

(Subscript[F, AtEachDetector])=1-((Subscript[F, StoppedByWires]))-((Subscript[F, DiffractedAwayDetectors]))

Decrease In Photon Count

(which should agree with our previously calculated value of (Subscript[F, PhotonLossesEitherDetector]) a few lines above)

(Subscript[F, PhotonLossesEitherDetector])

(Subscript[P, CountDecrease])=1-(Subscript[F, AtEachDetector])

Photons with Full Which-Way Information

(Fraction of Photons at Each Detector Not Due to Diffraction)

(Subscript[F, AtEachDetectorNotFromDiffraction])=(Subscript[F, AtEachDetector])-(2*(Subscript[F, DiffractedAtEachDetector]))

Total Number of Photons minus 2 times the wire absorption

(Subscript[P, TotalMinusWireAbsorption])=1-(2*(Subscript[F, StoppedByWires]))

The number of photons at each detector subtracting all the diffraction and wire absorption

(Subscript[P, TotalNotFromDiffraction])=(Subscript[P, TotalMinusWireAbsorption])

WichWayInformation=(Subscript[F, AtEachDetectorNotFromDiffraction])/(Subscript[F, AtEachDetector])

0.9999999946661668`
1-0.9999999946661668

5.333833152931788`*^-9
five photons in a billion go to the wrong detector!