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

Geoffkait, pure matte black has 100% absorption of all colors. Other colors are used only to reflect only one color. Targets for infrared, for counting number of pulses in rotational speed measurements, are usually gold/matte black (including one I designed: http://himmelstein.com/images/product-datasheets/0807545dc201B8701.pdf )

As I mentioned before you can see infrared with your phone camera and test it for yourself.

kijanki
"Geoffkait, pure matte black has 100% absorption of all colors."

that cannot be true since adding gree to the CD tray improves the sound. Almost all CD trays are black or black matte. As I said you can see for yourself with a red laser poster how much is reflected off the black tray.nalso, of the black matte tray was actually 100% effective the green pen wouldn’t make any difference. Nor would painting the tray cyan. And we know that’s not true.

"Other colors are used only to reflect only one color."

I don’t think we’re on the same page. You want to absorb color not reflect it. For a given color there is a complementary absorbing color.

Furthermore, dare I even say it? Black around the outer edge of the CD actually hurts the sound. 😀

"Targets for infrared, for counting number of pulses in rotational speed measurements, are usually gold/matte black (including one I designed: http://himmelstein.com/images/product-datasheets/0807545dc201B8701.pdf )

As I mentioned before you can see infrared with your phone camera and test it for yourself."

the reason you see the infrared is because it’s near infrared next to the visible part of the spectrum thus part of the signal is in the red zone. Same for a CD laser. Hel-loo!

As I said you can see for yourself with a red laser poster how much is reflected off the black tray.nalso, of the black matte tray was actually 100% effective the green pen wouldn’t make any difference.

It doesn't, according to many.  You are trying to make up theory based on some silly tweak, that according to many doesn't work at all.

the reason you see the infrared is because it’s near infrared next to the visible part of the spectrum thus part of the signal is in the red zone.

 No, spectrum of the infrared emitting diode is very narrow and it does not project white light.  You still cannot see it with naked eye.  The portion that you can see if you look directly into transmitter will be dark red.  The reason camera can see it is because camera sensors are sensitive to infrared up to around 1100nm.  You see white light because red green and blue sensor filters are not very strong and all of them pass some infrared.

"Other colors are used only to reflect only one color."

I don’t think we’re on the same page. You want to absorb color not reflect it. For a given color there is a complementary absorbing color. 

You still don't get it.  Perhaps this drawing will help you:
http://archive.cnx.org/resources/b3c8a8818e60b70ff638c267249826b1e5c0747f/Figure_27_03_02.jpg

The bandwidth of the laser is not narrow. It’s actually fairly wide, certainly not monochromatic. As is the photodetector bandwidth. That’s why red scattered laser light gets into the detector. Neither one is monochromatic. And that’s precisely why the CD laser appears RED even though it’s wavelength is 760 nm. Which is invisible. Hel-loo! I must be smarter than those other people you’re listening to. You’re following the wrong sheep.