CD Players: How Do They Interfere?

[Cheman]

CD players...

I've learned about CD players, and that the bumps cause the reflected light to be out of phase with the other infra red rays which form the laaser. But surely for destructive interference to occur you need the light to be in the same plce - yet all these light rays are parallel. ie - whether they have been put out of phase or not. So how do they cross over and interfer?

Thanks.

 

[NoTime]

The "bumps" are holes in a reflective layer.
They are simply less reflective.
CD and DVD disks modulate the reflection intensity.
They do not work by interference.

 

[cepheid]

Yeah...interference only explains why the CDs have rainbow colours when a whole bunch of normal light shines on them. I don't think that phenomenon has much to do with the reading of the CD via laser. My understanding was that lasers will either reflect off the surface or not, depending on what part they are hitting...(bump, etc). That's what gives you the binary encoding. Oh ok...NoTime explained it. It's not that zero laser light is reflected at certain points, but there's a definite high and low intensity. Cool.

 

[Chi Meson]

What Cheman has said is mostly right. BUT:

It is not infra red light, it is ultra violet light that is used to read the information on a CD/DVD.

The rainbow of reflected colors created is due to the phenomenon of diffraction, but this is a secondary effect of the closeness of the CD "tracks."

As I understand it, the "pits" are indeed as reflective as the "bumps," but due to the difference in depth (pit to bump) the reflected beam will indeed be in or out of phase with a "reference" beam. THis is why it must be ultraviolet light, because 1/2 wavelength of the light has to be the difference between pit and bump height(in order to cause constructive or desructive interference). Infra red and visible light is too long in wavelength to be useful. This process also allows "layering." More than one wavelength of light can be used, and pits of differing depth can be created, so that on one track you can get two or more sets of information recorded.

ZapperZ will hopefully correct me if I have erred

 

[Adrian Baker]

As I understand it, the "pits" are indeed as reflective as the "bumps," but due to the difference in depth (pit to bump) the reflected beam will indeed be in or out of phase with a "reference" beam. THis is why it must be ultraviolet light, because 1/2 wavelength of the light has to be the difference between pit and bump height(in order to cause constructive or desructive interference). Infra red and visible light is too long in wavelength to be useful.

ZapperZ will hopefully correct me if I have erred

I hope you are right - as that's what I teach my students!

 

[NoTime]

It is not infra red light, it is ultra violet light that is used to read the information on a CD/DVD.

Errrr.
Where did you get these ideas
They would love to use UV lasers on DVDs or CDs for that matter.
Reasonable priced lasers like that do not exist.
Currently it is mostly red lasers and was infrared.
The new Blue Ray stuff coming out uses blue lasers with a big step up in storage density.
It's about how small a spot the beam can be focused to.
Multi layer DVDs are done with objective focusing.

 

[NoTime]

I hope you are right - as that's what I teach my students!

None of the technical references I have ever read said this.

 

[Chi Meson]

Errrr.
Where did you get these ideas
It's about how small a spot the beam can be focused to.
Multi layer DVDs are done with objective focusing.

Well it's a darn good thing I speak up when I do! Otherwise I'd never have a chance of getting corrected! Indeed the wavelength is 790 nm, quite red, like my face!

 

[cepheid]

Wow...this thread is indeed confusing...I freely admit now having no idea how a CD works.

So we've established that the wavelength of the laser beam is 790mm...it is red.

What else? Does it work on the principles of comparison of phase with a reference beam?

Or do they just modulate reflection intensity?

And I thought the colours were due to "thin-film interference". How does diffraction enter into it? What is the distinction between diffration and interference anyway? Isn't the latter one of the effects of the former?

 

[Adrian Baker]

None of the technical references I have ever read said this.

I meant I teach how CD's work by interference. It is certainly how they work in all the references I have read.

I fully agree with the comments on the red wavelength used.

 

[Chi Meson]

Wow...this thread is indeed confusing...I freely admit now having no idea how a CD works.

So we've established that the wavelength of the laser beam is 790mm...it is red.

What else? Does it work on the principles of comparison of phase with a reference beam?

Or do they just modulate reflection intensity?

And I thought the colours were due to "thin-film interference". How does diffraction enter into it? What is the distinction between diffration and interference anyway? Isn't the latter one of the effects of the former?

OK, I am now speaking with more certainty, having just brushed up on the details. 790 nm is infra-red, just below the deepest visible red.

The "reference beam" in the case of CDs and DVDs is the part of the laser beam that is reflected from the surface of the disk (aka the "land"). THe beam that goes into the pits will travel a total extra distance of 1/2 a wavelength through the plastic (so the depth of the pit must be 1/4 wavelength in the plastic). The beam reflected from the "land" will superimpose with the beam reflected from a pit (or a flat "non-pit") to cause destructive interference (or not).

In this regard, it is very much like thin film interference, but apparently, the reflection from the top surface of the plastic layer is NOT used as the "reference" since it would be of low intensity.


In regard to my previous comment about multiple depth layering: it seems that this is where my confusion started. This form of layering is what would require UV in order to detect the nanometer differences in the depths of pits. My apologies.

 

[NoTime]

There is a reference beam. Two actually.
The original laser beam is split into three parts. Two side beams and the main reader beam. The two side beams are used to control the lens positioning servo to focus the main reader beam and have separate photodiodes to read them(triangulation).
AFAIK no part of this system uses interference.

Mass produced CD disks have holes physically punched by machinery, not lasers.
The CDR & CDRW are heated enough to introduce a phase change in the media. A change in color (reflectivity) there is no actual hole.

If you have a reference that says otherwise, I would be interested in it.

 

[Adrian Baker]

AFAIK no part of this system uses interference.

If you have a reference that says otherwise, I would be interested in it.

I'm no expert on CD players, but the students textbook that we use, "Advancing Physics AS" - IoP Publishing, includes a diagramatic explanation, with text, that clearly shows Interference as the way in which the signal is read. The light waves are shown to either cancel or reinforce.

I'll check this with the Institute of Physics who publish the book as I'm not happy using an 'incorrect' text ...

 

[Chi Meson]

Ditto here, except I use the Cutnell & Johnson text. An entire section is devoted to "Compact Discs and the use of interference" (chapter 27.8, pp. 855 ff). In which the pit depth of 130 nm is specifically mentioned in order to cause destructive interference with the 790 nm beam (index of refracction in the plastic n= 1.5).

The calc-based textbook by Serway and Jewett also refer to this desructive interference ocurring as the basis for one of their chapter problems; same numbers same results.

HOWEVER:
THe Serway and Faughn text specifically says that the beam from the pits reflects "away in a randon direction" causing merely a fluctuation in the intensity of the reflected beam. Evidently Faughn knows something Jewett doesn't know (or vice versa).

I have a feeling that this is one of those this-used-to-be-the-case" scenarios where the depth of pits were originally designed with destructive interference in mind (I distincty remember learning this back in my first year of college, 1985, when CDs were still new). In the years since, DI has no longer been an important consideration and so has been "dropped"; some textbooks are slow to change. Can anyone corroborate?

 

[NoTime]

HOWEVER:
THe Serway and Faughn text specifically says that the beam from the pits reflects "away in a randon direction" causing merely a fluctuation in the intensity of the reflected beam. Evidently Faughn knows something Jewett doesn't know (or vice versa).

Interesting. This one seems too be more in line with what I know.

Perhaps there was a change in design?
Or perhaps there was the initial development theory which got implemented by an engineer with different ideas?
I suspect the tolerances required for the interference idea to work were simply cost prohibitive.
Nm tolerance over a 13 cm surface = expensive.
I vaguely remember something about a design change to reduce the discard rate, prior to the release of CD players.
Maybe that is the point of discontinuity.
I don't know.

I will see if I can find some of the technical doc I have.
Won't be able to even try until next week though.
CD readers have been too cheap to bother messing with for a good 10 years now.

 

[Wol377]

Without certainty I thought I would bring up a point. Now that Cds are widly distributed in the commercial market, maybe useing a second laser to reflect off the surface (or "land") for interferance was flawed. You see, I remember some time back a product called.. "skip doctor" This product was meant to shave a layer of the cd to eliminate stratches. I'm almost certain it would shave a slice off larger then a wavelength of IR light. Here is a site showing to the product I am talking about:
http://www.digitalinnovations.com/skipdr/SkipDR_manual.html

 

[Adrian Baker]

I questioned the Institute of Physics who wrote the textbook I use. They said:

"An excellent and accessible source is J C G Lesurf Information and
Measurement (IoP Publishing 1995). Chapter 10 is called "The CD Player as a
measurement system".

I quote from pages 84-85:

"The optical system employs a highly coherent light source and the pits are
made approximately a quarter wavelength deep. The readout beam axis is
nominally aligned to be perpendicular to the disc plane. When there are no
pit-land edges in trhe spot, all of the reflected beam will mshare the same
phase. The phase of the reflected beam will however change by 180 degrees
when the spot moves from pit to land, or vice versa.

When the optical spot traverses a pit-land edge the magnitude of the beam
reflected back into the sensor will momentarily drop almost to zero...The
reflected beam then consists of two portions, equal in magnitude but
opposite in phase.

The chapter goes on to describe some further possible variations and
subtleties, but the principle stays the same."


Well that may be the case, but it doesn't answer the point I put to them that this MAY have been the case originally but do they still work this way know?

Is anybody out there a CD player designer?

 

[Adrian Baker]

More info. I got the following reply in an email from a fellow Physics teacher:

Adrian
I think that you will find that CD- CD-R and CD-RW use a different
method. In the latter two there is a dye material which can be
changed in absorbency by the heat from a laser. In the CD-RW a higher
temperature can be used to reset the colour back to the original and so
it can be used again. It takes a slightly more powerful laser to read
these disks and so earlier players will not play them. If mass
produced commercial disks have changed to the CD-R method I suspect
that they would no longer work on some players, It would also seem
quite difficult to mass produce CD-R disks and so they are probably
only suitable for shortish runs.

The plot thickens... Obviously the question "How do Cds work?" is not such a simple one as I thought!

 

[NoTime]

I have been making CDRs for a long time.
They played fine in my 1985 era 1X CD player, which finally died about a year ago.
It lasted that long because I refurbished the optics a few times.
The optics are not consistent with the concept using interference.
For $600 it was kind of painful to finally heave it.

The CDRW disks will not play.
As I understand it the data structures for CDRW are different.
The diagnostics I got from trying indicate that the data can be read by an older players, just not understood.
However, I did not do extensive testing on this aspect.
Even 5 years ago a CD player that could read CDRW was rare and generally available only for computers.

 

[chroot]

I'm almost certain that CD players to not rely on interference, but only on intensity modulation. I work for a major semiconductor manufacturer which produces the laser diode drivers used in CD-R and CD-RW drives, so someone here certainly knows. I will ask around.

- Warren

 

[theriddler876]

hmn I think it's basically a scantron reader except with lasers and such, for example the machine goes to problem one, say the answer is supposed to be a, so it fires light at it, if it's bubbled in, the graphite will absob most of it, and the sensor will pick up less light, thus the answer being correct, but if it's not bubbled in it will reflect most light, and the sensor will detect it, and mark it wrong, now cd's works in 1's and 0's, now if you shoot a laser at it, a sensor will see if it gets the laser back, if it does it is a one, if it doesn't it is a zero. so the laser is fired at all times, but as the cd spins, the "bumps" interrupt the reception to the sensor, thus it would read 101010101010110101, and such, then the decoder does it's decoding into sound and there you have it. now with blue ray and such, the wavelength is smaller, which means you would fit more "bumps" into the same amount of space, thus more ones and zeros thus more information

 

[Newonda]

Does anyone have a new answer to this 6 year old question? The physics book I am now using still talks about CD players using interference. Is this how today's CD and DVD players work or not?