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I'm trying to figure out what happens to the maximum intensity of a fringe pattern in the Fraunhoffer regime when one of the slits in the Young's double slit experiment is widened. The other slit is kept at a constant (finite) slit width, and the separation between the center of the slits is also kept constant.

So far, I've attempted numerical solutions by plotting the slit function in MATLAB and taking the FFT of the slit function to get the corresponding fringe pattern. It seems that the maximum intensity just climbs linearly with the ratio of the two slit widths.

My attempted numerical solution

I haven't attempted an analytical solution and I don't quite trust my programming enough to accept the linear answer I got above. What is the theoretical outcome of this experiment?

P.S: Here's my MATLAB code if you want to check my numerical solution:

clear 

for hw = 1:100 %hw is half width of slit with variable area
    hw0 = 20; %other slit kept at 20 units half width

    X = [-300:300]; 
    Xamp = 0; %amplitude of slit function
    Xamp(201-hw0:200+hw0) = 1; %left slit (unchanged)
    Xamp(401-hw:400+hw) = 1; %right slit (dependent on hw)
    Xamp(401+hw:601) = 0;

    ff = fft(Xamp); %fft to give interference pattern
    rightf = ff(1:300);
    leftf = ff(301:601);
    trunf = [leftf,rightf]; %truncation to center about 0

    trunfval = abs(trunf);

    relC(hw) = max(trunfval);
    ratio(hw) = hw/hw0; %finds ratio of hole areas
end

plot(ratio, relC)
title('Maximum intensity of pattern VS ratio of area of holes')
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2 Answers 2

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In this limit, all the light interferes constructively at $\theta = 0$, the direction perpendicular to the slits. The amplitude is therefore linear in the combined width of the two slits (IE the amplitude is proportional to the sum of the widths). The peak intensity is the square of this.

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  • $\begingroup$ So Amp VS (w1 + w2) is linear. Is there a way to figure out the same graph for Amp vs (w1/w2)? $\endgroup$
    – Sheepnut
    Commented Oct 30, 2015 at 0:10
  • $\begingroup$ just use algebra $\endgroup$
    – Mark Eichenlaub
    Commented Oct 30, 2015 at 0:19
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You can find the maximum and minimum intensities of the patterns quite easily in terms of the relative slit widths. The width of the slit is directly proportional to the amount of light passing through it so if you x and y are your respective slit widths, then you can express the amplitude of light from the two slits as the square roots of x and y respectively. Since the maximum intensity is simply the square of the sum of the amplitudes, it can be found in relation to the relative slit widths quite easily. The minimum intensity is simply the square of the difference of the amplitudes so it too can be computed quite easily. Hope this helps

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