# Hologram questions

1. Aug 1, 2006

### fargoth

after many experiments with digital holography i finally got the best interference image i could...

but after analyzing it, i got my object in two locations!
it was supposed to be on the lower right side of the image, but i got it on the upper right side too....

any idea why it happened?

if you need more info, i can explain the setup i used to get the image:
(without getting into detail about how i control the intensity of each beam and make sure there are no distubances):
first i split the laser beam, and light my object with one beam (the object beam), the second beam is reflected by glass from (almost) behind the object, and a digital camera without it's lenses pictures the interference of the two rays.

after i got my image, i use matlab to construct the image back - i simulate a beam with the same wavelength of the original beam that light through the image - it's pretty easy actually, i only need to use inverse fourier transform on my image multiplied by something which holds the info about the distance of the camera and the focal length i want to use for my simulated eye.

the script is very strait forward, so if anyone want it, i can post it here.

anyway, i just can't figure out why i got the second object...

2. Aug 1, 2006

### Staff: Mentor

Just a wild guess, but maybe you need to use a single-sided transform. Is the reflection backwards? Could it be a reflection about the zero spatial frequency axis in the transform?

3. Aug 1, 2006

### fargoth

i don't think it's backwards... here, i'll post it:

the default ifft function in matlab is nonsymetric, so i don't think that's the reason...

any other ideas?

#### Attached Files:

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• ###### hologram2.jpg
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Last edited: Aug 1, 2006
4. Aug 1, 2006

### fargoth

just to make sure it's not the fault of my script, i'll post it here, so you can take a look:

Code (Text):

I=I(750:1773, 1:1024);

C=complex(I);
C=double(C);
lam = 632.817e-9;
%lam is the wavelength
d=1.128;

%d is the distance of the camera from my object
dx=6.8e-6;
%dx is the pixel seperation of my camera

j=1:1024;
K= -i*pi/lam/d*(j.^2*dx.^2)';
L= -i*pi/lam/d*(j.^2*dx.^2);
KL=exp(K)*exp(L);
C=KL.*C;

D=ifft2(C);
%inverse FFT
F=abs(D);

F1=F(1:512,1:512);
F2=F(1:512,513:1024);
F3=F(513:1024,1:512);
F4=F(513:1024,513:1024);
F=[F4,F3;F2,F1];
clear F4 F3 F2 F1 D KL K L j

F=F.^0.5;
%making the faded image of my hologram brighter.

imshow(F)

5. Aug 2, 2006

### fargoth

I found the cause!

when i picture the interference i get $$I=O^2+R^2+2ORcos(\Delta \phi)$$
where R is the reference beam's amplitude, and O is the object beam's.
so i don't really get the phase difference of the two - i get the cosine of it, and the cosine have the same value for positive phase differences as for negative ones.

6. Aug 2, 2006

### Danger

Good to hear that you sorted it out, fargoth. My thought was that you might not be using front-surface mirrors, and thus getting muliple reflections.