Alan Sailer wrote: ↑Thu Dec 19, 2024 9:17 am
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".