How to get circular fringes?

Holography related topics.
lobaz
Posts: 280
Joined: Mon Jan 12, 2015 6:08 am
Location: Pilsen, Czech Republic

Re: How to get circular fringes?

Post by lobaz »

I updated the script so you can play with the interference fringes as you wish :)
I set the on-axis reference light R (100 m away from the hologram, amplitude 1), and three point light sources A, B, C according to Figure 10 in Kock486.pdf
A - located at [0, 3 mm, -1 m], amplitude 0.1
B - located at [0, 0, -2 m], amplitude 0.1
C - located at [0, -3 mm, -1.5 m], amplitude 0.1

When only R and A are shining (B and C have amplitude 0), the hologram located at z=0, size 8x8 mm, looks like this:
R+A
R+A
RA.jpg (27.03 KiB) Viewed 1569 times
When only R and B are shining (A and C have amplitude 0), the hologram looks like this. Note that the fringes are not as close to each other because B is further than A.
R+B
R+B
RB.jpg (18.4 KiB) Viewed 1569 times
When only R and C are shining (A and B have amplitude 0), the hologram looks like this. Note that until now, all the holograms have low contrast because the reference is 100x stronger that the object light (intensity is the square of the amplitude).
R+C
R+C
RC.jpg (23.51 KiB) Viewed 1569 times
lobaz
Posts: 280
Joined: Mon Jan 12, 2015 6:08 am
Location: Pilsen, Czech Republic

Re: How to get circular fringes?

Post by lobaz »

Let us turn on A and C. Note that fringes sometimes sum up and create more dense black, sometimes they cancel each other. Anyway, the sum of A and C (and R, of course) is not as simple as adding previous images together.
R+A+C
R+A+C
RAC.jpg (32.25 KiB) Viewed 1569 times
Finally, we can turn on all lights (A, B, C and R). The result starts to be quite hard to read. Note how the contrast of the hologram is improving.
R+A+B+C
R+A+B+C
RABC.jpg (31.64 KiB) Viewed 1569 times
And the script, of course. Enjoy!
Petr

# the wavelength
lambda = 532e-9;

# the extent of the holographic plate
x=(-200:200)*0.02e-3;
y=(-200:200)*0.02e-3;

# point light sources: x, y, z coordinates and the amplitude (squate root of the intensity)
# in succession:
# R - on-axis reference light, almost collimated (pinhole 100 m away)
# A - top light, the closest to the hologram
# B - middle light, the furthest from the hologram
# C - bottom light, somewhere between A and B
lights = [0, 0, -100, 1;
0, 3e-3, -1, 0.1;
0, 0, -2, 0.1;
0, -3e-3, -1.5, 0.1];

# the calculation
[xx,yy] = meshgrid(x, y);
wave = zeros(length(y), length(x));
for lightNo = 1:rows(lights)
x0 = lights(lightNo(1), 1);
y0 = lights(lightNo(1), 2);
z0 = lights(lightNo(1), 3);
wave = wave + lights(lightNo, 4) * exp(j*2*pi/lambda*sqrt((xx-x0).**2 + (yy-y0).**2 + z0**2));
end
interference = wave .* conj(wave);

# just for display purposes
interference(1,1) = 0;

#display
imagesc(x, y, flipud(interference));
colormap(gray)
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