What's the deal with coloring CD's and/or tray mechanism

kijanki
"As I explained earlier - spilled light would appear as mostly red but it shows as white (no red at all), because all camera sensors can see infrared and filters are not able to filter it completely. If you cannot understand it I cannot help it. Also LED transmitters have narrow band, like this one: http://www.vishay.com/docs/81009/tsal6100.pdf

As you can see there is no light emitted below 840nm (Fig. 8) while visible light ends at 750nm. If you compare two cameras one with the film and another with digital sensor set to the same sensitivity you’ll find that infrared won’t be visible on the picture made with film camera but will show as white light on the one with digital sensor. It is not because of "spilled" light but because digital sensor is sensitive to infrared and film isn’t. I’m not sure if your understanding of this is as poor as one of the class D amps, but it is getting close. No sense to discuss it any further."

We are not on the same page. Not even close. I define infrared light as above 700 nm. Therefore if there is red in the CD laser light it MUST be because the 760 nm nominal wavelength is NOT monochromatic. It has SOME RED light in it. And that RED LIGHT has wavelenths BELOW 700 nm since that is where visible light ENDS. This conversation can serve no purpose anymore.

Exactly - don't waste your time - or money.

People who do not understand electronics or physics can be convinced to believe it works, just as snake oil was sold to people decades ago.

randy-11
Exactly - don’t waste your time - or money.

People who do not understand electronics or physics can be convinced to believe it works, just as snake oil was sold to people decades ago.

Hey, mods! Randy’s stalking me again. Deep six this Druid.

No matter what the wavelength of the laser used in cd players, wouldn’t any scattered light picked up by the photodiodes be miniscule compared to the direct reflection that the photodiodes read and wouldn’t any halfway well designed cd player be able to ignore the miniscule amounts of scattered light and only send on to the DAC 1s and 0s read from directly reflected light?

tomcy6
"No matter what the wavelength of the laser used in cd players, wouldn’t any scattered light picked up by the photodiodes be miniscule compared to the direct reflection that the photodiodes read and wouldn’t any halfway well designed cd player be able to ignore the miniscule amounts of scattered light and only send on to the DAC 1s and 0s read from directly reflected light?"

The problem is that any scattered light that gets into the detector - scattered light more than 75% of full reflected value is accepted by the detector - is accepted as REAL signal. If it’s less than 75%, the scattered light isn’t accepted and can be ignored. Recall the photodetector is essentially seeing OFF and ON signals, a series of "no signal" and "signal." These ON and OFF signals are converted to meaningful sequences of 1s and 0s later. Thus the data stream is corrupted, even if slightly, by scattered light during the optical reading phase - every so often an OFF is interpreted as ON. The whole issue of course is, how often does the error occur and is it audible. 

There are some players that address this issue by painting the surfaces around the laser assembly turquoise and, as I mentioned earlier, at least one CD manufacturer that paints the CDs labels turquoise. By the way painting the CD label black or painting the outer edge of the CD black ruins the sound. Since most CD player manufacturers are either unaware of this problem with scattered light and sound quality or else don’t address it if they are aware, virtually all players, even high end players, suffer this problem. Furthermore, as I mentioned earlier, even if turquoise or green or cyan is used to counter the RED portion of the laser light the INVISIBLE scattered laser light can still get through, the invisible light that extends from 700 nm to around 850 nm or higher. Since the wavelength of the CD laser is centered at 760 nm most of the scattered light is INVISIBLE.