Reading a CD that has no reflective layer - test-drb@ccmp.vtda.org

18 Nov 2014

[John Foust, quoting me]
...
   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?  I'm not
quite sure what you mean here.   
The goal is changes in reflectivity on a fairly small scale.

There's no reason in principle this couldn't be done with a smooth
surface by making the surface not all the same material.  (A
larger-scale example of this is ink on paper: inked paper reflects less
light than un-inked paper does.)  I was speculating that, on the scales
of interest here, it's easier/cheaper to produce a non-smooth surface
of a uniform material (how pressed CDs are actually made) than to
produce a smooth surface of a nonuniform material (the alternative).

...
  Wikipedia has an explanation of the process: 
...

http://en.wikipedia.org/wiki/Compact_Disc_manufacturing 
Yes, but it does not discuss why that process was chosen over various
alternatives.

...
 > Is it possible that it's clear only in visible
light, with some sort
> of reflective layer present in the (infrared, IIRC) wavelengths
> used? 
...
  But as you mentioned, detection normally uses the
phase shift, and
 that's dependent on the frequency of the light. 
Yes, but I think the light normally used for CD reading is outside the
visible range.  This raises the possibility that the disc might be
clear in the visible range but not in the range used for reading.

Your experience trying to read it argues against that, though.  Oh
well, it was a nice theory while it lasted.

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