Patterns

[Alan Sailer]

One of my recent dennisyuk holograms has a very pronounced pattern like contour lines. It looks
like the fringes from an interference pattern. Unlike what I would expect from an interference pattern
the lines move over the object when I tilt the plate.

One website refers to the dreaded wood grain pattern and how antihalation coatings will prevent it.
I am not clear at all what an antihalation coating is.

The dennisyuk plate was angle was perpendicular to the beam. The edges of the plate have no
masking/blackout paint.

Cheers.

[BobH]

"Woodgrain" fringes do not move as one tilts the plate. Uniform contour lines suggest the laser did not have a long enough coherence length. Those lines may move as the plate is tilted, but I don't have enough experience with that problem to be certain. Lines that move as the plate is tilted can also suggest object motion.

Antihalation coating is an absorptive coating on some silver halide films to prevent the Fresnel reflection off the final material-to-air interface (despite its not being a Fresnel lens as has been stated here as necessary for such reflections). Using a "P" polarized reference beam near Brewster's angle achieves the same result but possibly with slightly reduced diffraction efficiency.

[Din]

One of my recent dennisyuk holograms has a very pronounced pattern like contour lines. It looks
like the fringes from an interference pattern. Unlike what I would expect from an interference pattern
the lines move over the object when I tilt the plate.

One website refers to the dreaded wood grain pattern and how antihalation coatings will prevent it.
I am not clear at all what an antihalation coating is.

The dennisyuk plate was angle was perpendicular to the beam. The edges of the plate have no
masking/blackout paint.

Cheers.

It's not woodgrain. It's probably object motion.

Antihalation coatings are designed to prevent "ringing" of the laser beam, as it hits the back of the material, reflects, and hits the front, and then reflects again to hit the back. These continuous reflections will expose silver halide, creating a set of lines that look like woodgrain - the grain in a piece of wood. If you're using polymer, it's unlikely to be woodgrain; woodgrain manifests itself by a darkening of the material ever time it's hit with light during the ''ringing" process. The actinic reaction of polymers is not absorption-dependent.

Light entering an interface between two media at the Brewster angle with TM polarisation will inhibit reflections, as per the Fresnel equations. But, it's necessary to know the index between the two materials to calculate the Brewster angle. However, bear in mind that the polarisation state depends on the plane of incidence. If the plane of incidence is twisted, you cannot get pure TM polarisation.

However, there is no such thing as "Fresnel reflection", as opposed to any other kind of reflection; the phenomena of reflection is defined by the equation:i = r ( angle of incidence = angle of reflection ). The Fresnel equations determine the intensity - well, actually, the amplitude - of reflected light, based on the polarisation state of the light, assuming the polarisation state is either TM or TE. Any other plarisation state is not covered by the Fresnel equations, though the equations can be used for linear polarised light that is not TM or TE, by taking a weighted average. Elliptically polarised light - most forms of light - will not obey the Fresnel equations. But, again, the TM/TE state is based on the plane of incidence; if the plane of incidence is twisted, the Fresnel equations will not apply without an averaging process. The Brewster effect is due to the fact that if the E vector of the reflected light is perpendicular to the refracted light, the reflectivity drops to zero because the resulting intensity is the dot product of the the two vectors.

The Fresnel equations for the reflectivity of TM and TE polarisation are:

R(TE) = |{n(1) cos(θ(i)) - n(2) cos(θ(t)}/n(1) cos(θ(i)) + n(2) cos(θ(t))}|²

R(TM) = |{n(1) cos(θ(t)) - n(2) cos(θ(i)}/{n(1) cos(θ(t)) - n(2) cos(θ(i))}|²

where n(1) and n(2) are the indices of the two media, θ(i) is the angle of incidence and θ(t) is the angle of transmission.
Of course T = 1 - R.
Also, notice that if n(1) cos(θ(t)) = n(2) cos(θ(i) - the Brewster condition - the reflectivity, R(TM) = 0.

By the by, there are interesting effects of the Fresnel equations. On a rainbow, If you calculate the light hitting the back of a raindrop, the Fresnel equations predict that the TM component will vanish, leaving only the TE component ( http://hyperphysics.phy-astr.gsu.edu/hb ... owpol.html ). However, nobody calls a rainbow a "Fresnel reflection". Well, almost nobody.

If the sun is about an hour or two before sunset, light reflected off a body of water - say an ocean - will be polarised as per the Fresnel equations. But, nobody refers to sunlight reflecting off the ocean as a "Fresnel reflection".

[Alan Sailer]

Thanks Bob. The lines are not straight. They roughly follow the contours of the object like a topographic map.

Cheers.

[Alan Sailer]

Din,

Object motion is possible. But wouldn't the fringes from an interference hologram stay put?

I wish the interference holograms I tried to make would have worked. It would have answered
this question. I gave it three tries and then got tired of wasting film.

Cheers.

[BobH]

However, there is no such thing as "Fresnel reflection", as opposed to any other kind of reflection; the phenomena of reflection is defined by the equation:i = r ( angle of incidence = angle of reflection ).

https://www.rp-photonics.com/fresnel_reflections.html

[Din]

Object motion is possible. But wouldn't the fringes from an interference hologram stay put?

No, they wouldn't. You get these interference lines because the light from the object at one position (the object light) will interfere with the light from the object from another position, causing the object light to interfere with itself if the motion is small (about half a wavelength) creating interference lines on the image. However, a complex object moves in complex ways. So, if you look at the object at some angle, you'll see a specific set of interference lines, dependent on how the object moved in the viewing direction. If you move your head and see the image from another direction, you'll see a different set of lines, depending on how the object moved with respect to the new viewing direction. If the motion is continuous, the lines seem to move.

I wish the interference holograms I tried to make would have worked. It would have answered
this question. I gave it three tries and then got tired of wasting film.

Not sure what you mean by "interference holograms" There is a technique of non-destructive testing known as 'holographic interferometry". The idea is that you make a hologram of some object (no motion lines!), then you replace the plate as accurately as you can while leaving the original object in place. Now, any small motion of the object will create interference lines between the real object and the image - provided the motion is within about a wavelength or two. It's used for architectural modelling, musical instruments and failure modes of some model, like a bridge for example. If you want to try this, remember that the movement must be very small. These days, this technique uses digital holography, so the interference pattern can be captured. (https://www.wileyindustrynews.com/en/ne ... rferometry )

[Joe Farina]

I would say, almost certainly, laser instability. I would wager that you're not seeing this effect with a helium-neon laser. Anyone who has made holograms with laser diodes probably knows this situation well. I have seen it frequently with laser diodes, and also occasionally with OPSL's like Sapphire. They are called "contour fringes" because they seem to follow the contours of the object. The laser is losing its coherence during the length of the exposure, and the coherence length corresponds to the light/dark contour bands across the object. With some diode lasers (which are prone to mode-hopping) the width of the bands can be something like 1 to 3mm wide. On other lasers like Sapphire, the bands can be more like 4 or 5mm. This loss of coherence can be unpredictable for some lasers, such as some types of Sapphire. With my 488-200 (not the "single frequency" version), I would say this happens about 10% of the time. Not good of course, but I can live with it. I have seen this reported on this forum even with such lasers as the Melles-Griot 457nm laser (58-BLD-605) which isn't known to mode-hop (mine didn't). Someone also reported that even a 532nm ring-cavity laser would occasionally mode-hop. I have a paper in which the authors use this effect to generate contour-fringes, deliberately. They varied the size of the fringes by making small changes to the wavelength of a dye laser.

[Alan Sailer]

Din,

Yes holographic interferometry. I put a weight on an object, exposed for a while and gently removed the weight.
Either my weight wasn't heavy enough or the removing caused changes in the objects position, but I never got any fringes.

BobH,

If your higher quality lasers are unstable, my twenty year old CrystaLaser could certainly be unstable. My bands are about
1-2mm wide. If I get some gumption I will try setting up my He-Ne.

Another reason to get out of this hobby. The requirements (since it is all relying on fractions of a wavelengths
of light) are just too crazy. I've got parts for a spatial filter coming in the mail and my final plan is to try and
make a few single wavelength holograms that I could be proud of.

As a side note I am finding that holography is an art that leans too much on the technical aspects for my taste. It's
too easy to loose my intentions in a fog of technical confusion. Years ago I did some nice work with high speed photography.
It obviously required some technical learning but once that was done I could concentrate on ideas for fun pictures.

It also helped that the exposure time for the pictures was on millionth of a second...no sipping tea waiting for a
hologram to expose.

Cheers.

[BobH]

I also like to keep it simple and would encourage you to disregard unnecessary explanation and calculation.

Holography can be very simple. I made a hologram of a big crack in the floor of my San Jose garage a few years ago. Many people on the Facebook pages are getting great results with a variety of lasers. It's such a shame that's the place on the internet for it now as I hate it too. Check out the "Archive" tab on my site listed below. Under "Home Holography" you'll find some projects I've done over the years.

You can also visit Ed Wesly's pages for lots of info on how to get good results at home and cheap.

https://laserhistorymuseum.com/