Understanding the principle

Light and its behaviour and properties
Ytre
Posts: 7
Joined: Fri Oct 16, 2020 5:49 am

Understanding the principle

Post by Ytre »

Do I think correctly when I assume that the property of the hologram is associated with the difference in the intervals between the interference maximums registrated on the film?
Din
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Joined: Thu Mar 12, 2015 4:47 pm

Re: Understanding the principle

Post by Din »

Ytre wrote: Fri Oct 16, 2020 5:56 am Do I think correctly when I assume that the property of the hologram is associated with the difference in the intervals between the interference maximums registrated on the film?
Correct. The difference between the intervals is a measure of the spatial frequency of the recording. This is also a function of interbeam beam angle. Consider having a point source, at some angle, and a planar beam at some other angle, both hitting the film. The interaction of the planar beam and the point source creates a set of lines, whose interval between maxima is a function of the angle between the beams and the wavelength. If you have another point source, each point source creates it's own set of lines. For an actual object, the object may be considered as a continuous set of point sources, each creating it's own set of lines. The resolution of the film is a measure of how the film can record the lines. If the lines are too fine, some film may not be able to record them.
Ytre
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Joined: Fri Oct 16, 2020 5:49 am

Re: Understanding the principle

Post by Ytre »

If the red line was a film, the maximums would lie on it as I showed in picture? Is it correct? Just trying to understand how angle affects the distance.
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Screenshots_2020-10-16-20-58-18.png
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lobaz
Posts: 280
Joined: Mon Jan 12, 2015 6:08 am
Location: Pilsen, Czech Republic

Re: Understanding the principle

Post by lobaz »

Yes. But keep in mind that analysis of interference from a drawing of a few wavefronts (here, circles) is quite limited.
A better idea is to analyse the pattern in this way:
We have two point sources (pinholes in the second screen). Light starting at these sources interferes. What happens at a particular point P? It depends on distances between P from both point sources; let us denote the distances r1 and r2.
When r1 = r2, the waves add constructively. When r1 = r2 + wavelength/2, the waves destroy each other. When r1 = r2 + wavelength, the waves add constructively again; and so on. Thus, if you draw lines for which r1 - r2 = 0, r1 - r2 = wavelength, r1 - r2 = 2*wavelength, and so on, you get points in space where the waves add constructively. Such lines are hyperbolas with foci at point light sources.
If you can use Octave or Matlab, a small script does it for you:

Code: Select all

% Area we watch will be 10 x 10 units
x = linspace(0, 10, 500);
y = linspace(-5, 5, 500)';

% The point light sources will be located at [0, 2] and [0, -2]
r1 = sqrt(x.^2 + (y-2).^2);
r2 = sqrt(x.^2 + (y+2).^2);

% Calculate difference between r1 and r2 at all points.
% Use modulo function as differences 0, wavelength, 2*wavelength, etc. behave the same way
wavelength = 0.5;
imagesc(x, y, mod(r1 - r2, wavelength)); 
axis('image');

% Colormap will be white for difference 0 and wavelength,
% black for difference wavelength / 2
c = abs(linspace(-1, 1, 100))';
colormap([c, c, c]);
colorbar();
interference.png
interference.png (60.12 KiB) Viewed 5616 times
Please note that the image does NOT show the actual interference pattern as we did not count with quadratic attenuation of light intensity. It just shows points where lights are in and out of phase.
Ytre
Posts: 7
Joined: Fri Oct 16, 2020 5:49 am

Re: Understanding the principle

Post by Ytre »

Unfortunatelly I dont have mathlab. My mathlab exists only in my head. I purposely simplified process by using two-dimentional model so far as I showed in pic. I dont understand what did you mean by hyperbola? Is it projection on film? Could you explain it in detail? I couldnt understand this on howstuffworks article. Not good explanation for newbie
Ytre
Posts: 7
Joined: Fri Oct 16, 2020 5:49 am

Re: Understanding the principle

Post by Ytre »

It seems I have imagined it correctly. Its because of spherical fronts in space they intersect each other. Ok. I have it now understood at last.
lobaz
Posts: 280
Joined: Mon Jan 12, 2015 6:08 am
Location: Pilsen, Czech Republic

Re: Understanding the principle

Post by lobaz »

Ytre wrote: Fri Oct 16, 2020 3:03 pm I dont understand what did you mean by hyperbola? Is it projection on film? Could you explain it in detail?
https://en.wikipedia.org/wiki/Hyperbola
Din
Posts: 402
Joined: Thu Mar 12, 2015 4:47 pm

Re: Understanding the principle

Post by Din »

If you have two point sources, as in your diagram, then you will get a sinusoidal variation of light bands and dark bands. The defining equations is:

m*λ = d*sin(θ)

where m is called the 'order' of the pattern (the pattern repeats - dark, light, dark, light etc) ), d is the distance between two slits, λ is the wavelength and θ is the angle between any point on the screen and a normal (straight line) from half way between the slits. So, let us say that you set up your two slits (if you're using a laser, you don't need the first slit) separated by 0.02 mm and you measure the third bright point on your screen to be at an angle of 5.45 degrees from the centre of the slit system, what is the wavelength?

λ = d*sin(θ)/m = {(0.02)*sin(5.45)}/3 = 0.000633

The intensity variation is sinusoidal, and is given by

I(t) = 4*I* cos²(θ/2)
where I(t) is the total intensity at the screen, and I is the intensity at each slit. So, if you imagine a cos² curve, the fringe system follows the curve.

This is for a two slit system, as you've drawn. However, in holography, we usually have a planar (flat) wave and a diverging wave from a point source. In this case, the fringe system is hyperbolic, the lines of equal intensity are shaped like hyperbolas. Depending on where you place the film, the film will intersect the hyperbolas and form the fringe system. Below is a diagram from "Optical Holography" by Collier, Lin and Burckhardt
fringes.jpg
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Ytre
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Re: Understanding the principle

Post by Ytre »

It is not entirely clear from this figure, but I understood principal of how hyperboles are formed. If both waves were spherical, then the hyperbola would lie in the plane perpendicular to the film, and in the case of one plane wave it is in the plane of film. I got it. Is it possible to demonstrate a hologram in real time without using film? After all, the effect will be exactly the same as with the film, i.e. in the plane of space, the same patterns will form and the maximums will be the place of formation of spherical waves. Or not?
Din
Posts: 402
Joined: Thu Mar 12, 2015 4:47 pm

Re: Understanding the principle

Post by Din »

The fringe system is actually 3 dimensional, they're hyperboloids, not hyperbolae. If the system is just 2-dimensional - across one plane - then only one line across the film would expose.
Ytre wrote: Sun Oct 18, 2020 5:42 am Is it possible to demonstrate a hologram in real time without using film?
It's not possible to demonstrate an actual object, because the spatial frequency is high, ie the spacing between fringe lines is small, in the order of microns. If you look at the above equation, λ = d*sin(θ), you'll see that even with a 10 degree beam separation, the lines are spaced at 7 microns, too close to see with the unaided eye (in comparison, a human hair is about 25 microns wide). It's always possible to observe interference lines with any interferometer. In fact, most holographers test the stability of their system using a Michelson interferometer. In a Michelson interferometer, the two beams are inline, ie the interbeam angle is close to zero. So you can see the lines on a wall. If you want to see the effect of changing the angle between the two beams, you can tilt one of the mirrors in the interferometer.
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