17 Nov
2014
17 Nov
'14
8:51 p.m.
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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17 Nov
17 Nov
11:57 p.m.
I think you're actually more likely to be able to read it in modern drives,
which are designed to handle lower reflectivity as found on burnt CD-R and
CD-R/W discs.
On Tue, Nov 18, 2014 at 9:51 AM, Mouse <mouse at rodents-montreal.org> wrote:
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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18 Nov
18 Nov
12:07 a.m.
Slap a layer of Mylar on the back and give it a shot..? Might even work
with something of lower reflectivity, like white paper..
On Mon, Nov 17, 2014 at 5:57 PM, Don Hills <dmhills at gmail.com> wrote:
I think you're actually more likely to be able to
read it in modern drives,
which are designed to handle lower reflectivity as found on burnt CD-R and
CD-R/W discs.
On Tue, Nov 18, 2014 at 9:51 AM, Mouse <mouse at rodents-montreal.org> wrote:
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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