by Din » Sun Jul 24, 2016 8:26 am
hjalmar1234 wrote:Why does crosstalk only occur with transmission holograms?
Because in transmission holograms the diffraction mechanism is different from reflection holograms. There are two diffraction mechanisms. Technically, one is called Raman Nath, and one is called Bragg. In the Raman Nath type of diffraction, everything diffracts, but at different angles. Transmission holograms diffract by Raman nath diffraction and reflection holograms diffract with Bragg diffraction.
A transmission hologram has what's called a "spatial frequency", which is the number of lines/unit distance - generally, in holography, this is given by lines/millimetre - and is determined by the angles between the reference and the object. When you reconstruct the hologram, this frequency determines the angle through which any wavelength diffracts. I don't know how much maths you have, but the angle through which any wavelength diffracts is given by the "grating equation" (lambda)*f = sin(theta), where lambda is the wavelength you're hitting the transmission hologram with, f is the spatial frequency already recorded in the hologram, and theta is the diffraction angle. So, for example, if the spatial frequency of your hologram is 1000 lines/millimetre, and you hit the hologram with green light (lambda = 0.000532, the extra number of zeroes to keep the units consistent), you have sin(theta) = (1000)*(0.000532) = 0.532. So, the angle whose sin is 0.532 is 32.14 degrees. So, hitting this hologram with a wavelength of 532nm will cause a diffraction through 32 degrees. It doesn't matter what wavelength you hit the transmission hologram with, everything diffracts. So, if you record a transmission hologram with any single wavelength,ie recording only one spatial frequency, then all wavelengths will diffract. If you then reconstruct the hologram with three specific wavelengths, say 633nm, 532nm and 457nm, they'll all diffract and give you three images. If you make the transmission hologram with three wavelengths, then you'll record three spatial frequencies. If you now hit the transmission hologram with three wavelengths, any one frequency will give you three images, all three frequencies will give you nine images.
In Bragg diffraction, the planes, or sheets, within the hologram will function as filters, filtering out all wavelengths except one, at a specific angle.These planes are tilted by some angle and separated by some distance. This tilt angle and this separation determines which angle and which wavelength will diffract. This is called "Bragg selection". Bragg selection allows diffraction for only one wavelength at only one angle. No other wavelength will diffract. So, if you have the right wavelength at the wrong angle, there will be no diffraction. So, if you record a reflection hologram with 633nm, 532nm and 457nm, at some angle, then Bragg selection ensures that the hologram will only diffract if it's hit with those wavelengths and at the angle with which you shot it.
[quote="hjalmar1234"]Why does crosstalk only occur with transmission holograms?[/quote]
Because in transmission holograms the diffraction mechanism is different from reflection holograms. There are two diffraction mechanisms. Technically, one is called Raman Nath, and one is called Bragg. In the Raman Nath type of diffraction, everything diffracts, but at different angles. Transmission holograms diffract by Raman nath diffraction and reflection holograms diffract with Bragg diffraction.
A transmission hologram has what's called a "spatial frequency", which is the number of lines/unit distance - generally, in holography, this is given by lines/millimetre - and is determined by the angles between the reference and the object. When you reconstruct the hologram, this frequency determines the angle through which any wavelength diffracts. I don't know how much maths you have, but the angle through which any wavelength diffracts is given by the "grating equation" (lambda)*f = sin(theta), where lambda is the wavelength you're hitting the transmission hologram with, f is the spatial frequency already recorded in the hologram, and theta is the diffraction angle. So, for example, if the spatial frequency of your hologram is 1000 lines/millimetre, and you hit the hologram with green light (lambda = 0.000532, the extra number of zeroes to keep the units consistent), you have sin(theta) = (1000)*(0.000532) = 0.532. So, the angle whose sin is 0.532 is 32.14 degrees. So, hitting this hologram with a wavelength of 532nm will cause a diffraction through 32 degrees. It doesn't matter what wavelength you hit the transmission hologram with, everything diffracts. So, if you record a transmission hologram with any single wavelength,ie recording only one spatial frequency, then all wavelengths will diffract. If you then reconstruct the hologram with three specific wavelengths, say 633nm, 532nm and 457nm, they'll all diffract and give you three images. If you make the transmission hologram with three wavelengths, then you'll record three spatial frequencies. If you now hit the transmission hologram with three wavelengths, any one frequency will give you three images, all three frequencies will give you nine images.
In Bragg diffraction, the planes, or sheets, within the hologram will function as filters, filtering out all wavelengths except one, at a specific angle.These planes are tilted by some angle and separated by some distance. This tilt angle and this separation determines which angle and which wavelength will diffract. This is called "Bragg selection". Bragg selection allows diffraction for only one wavelength at only one angle. No other wavelength will diffract. So, if you have the right wavelength at the wrong angle, there will be no diffraction. So, if you record a reflection hologram with 633nm, 532nm and 457nm, at some angle, then Bragg selection ensures that the hologram will only diffract if it's hit with those wavelengths and at the angle with which you shot it.