Reading a CD that has no reflective layer

Mouse mouse at Rodents-Montreal.ORG
Mon Nov 17 14:51:55 CST 2014

> As I understand it, the depth of the pits are about a fifth of the
> wavelength of the light used to read them, so the detector sees a
> phase shift.

I thought the pits were, in theory, 1/4 wavelength deep, so that the
reflection from the pit is 180 degrees out of phase with the reflection
from the surrounding area, producing destructive interference (ie,
manifesting as a drop in reflectivity).  Of course, .25 is about .2....

I don't understand why this technique was used.  Perhaps it's
easier/cheaper to produce a nonsmooth surface made of a uniform
material than to produce a smooth surface of a nonuniform material?
(That's the other way I'd expect to produce reflectivity variations.)

Is it possible that it's clear only in visible light, with some sort of
reflective layer present in the (infrared, IIRC) wavelengths used?

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