Number of layers in DCG

JohnFP · Post by **JohnFP** » Fri Jun 18, 2004 3:37 pm

How many layers of "mirrors" are in a reflection hologram in DCG with reference beam and object beam 180 apart, reference beam perpendicular to plane of film, film 25 microns thick and 488nm is the wavelength. I know there has to be a formula somewhere but I cannot find it. Would is simply be the a layer every 488nm?

Kaveh · Post by **Kaveh** » Fri Jun 18, 2004 7:27 pm

Separation of 'mirrors' will be half a wavelength. But the wavelength is reduced in the emulsion because of the refractive index. So assuming a refractive index of 1.5, say, the mirror or fringe separation will be

.5*(488/1.5) = 162nm (roughly)

JohnFP · Post by **JohnFP** » Fri Jun 18, 2004 9:54 pm

Thanks.

Makes sense as the fringe from where it starts to become a fringe to where it stops would be about half. And I never thought of the refractive index.

Now, is it possible to make a grating that is...oh, lets say 10 fringes deep (emulsion thickness of 1.5 microns) so that you get 10 distinct replay colors. In theory, I know getting an emulsion that thin and being able to control the shrinking that precisely would be very difficult.

Has there been any work with emulsion thicknesses in the sub micron in DCG?

Kaveh · Post by **Kaveh** » Sat Jun 19, 2004 6:36 am

JohnFP wrote:Now, is it possible to make a grating that is...oh, lets say 10 fringes deep (emulsion thickness of 1.5 microns) so that you get 10 distinct replay colors. In theory, I know getting an emulsion that thin and being able to control the shrinking that precisely would be very difficult.

There is no one-to-one relationship between the number of fringes or planes recorded, and the number of colors in the replay. What you see in the replayed image is a continuous range of colors, with a peak decided by the distance between the planes. If there was no shrinkage during processing, and if the average refractive index of the emulsion did not change, then this peak wavelength will be the same as that of the recording wavelength, say 488nm.

Now the more planes you have, i.e. the thicker the emulsion, the narrower will be your band of wavelengths, so you will get an image with a purer, more saturated color. In general, with a thin emulsion of say 1.5 microns, you only have a few planes taking part in the Bragg diffraction, and the image will be broadband.

You should think of all of these planes working together to diffract the light, and the more of them there are working in tandem, the purer the reconstruction wavelength.

Regarding the refractive index, it only affects the fringe separation in reflection holograms. In transmission, although the wavelength is still reduced in the recording material, the inter-beam angle also reduces, and the two cancel out. This is why you need so much more resolution for reflection holograms. Not only are the fringes half a wavelength apart because the beams are opposing, the refractive index reduces the fringe separation further.

Tell me if any of this is not clear.