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Correct physics behind the colors on CD (compact disc)?

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1

$begingroup$

So, we all know if you shine some light on CD (compact disc) or DVD, you can see color from all the specter red to violet.

What is bothering me is what is the reason of why do we see the colors on compact disc?

I know that CD has tiny pits/bumps (order of size is micrometers).
So when I shine a light on CD, I see different colors and they are changing I move the disc, which is because I change the angle $theta$ between my eyes and disc; which is followed by diffraction condition:

$dsintheta = mlambda$

So my, my main concern is are colors over there because light travels different distances when in comes to the surface of CD because of bumps/pits?

Can I interpret diffraction colors on CD same as color on thin films?
Because although one part of CD is quite reflective surface the other is almost fully transparent.
Is this like combination of reflection and diffraction?
I mean reflection is there because of reflective surface, and even because of some thickness of aluminum I see colors (which is like thin films)?

And now comes the bumps/pits on cd? Because of them I also see colors or?

Thanks for help, really trying to break this into smaller pieces but constantly going around...

optics diffraction

share|cite|improve this question

asked 1 hour ago

solidbastardsolidbastard

494

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So my, my main concern is are colors over there because light travels different distances when in comes to the surface of CD because of bumps/pits? Are you considering how different colors have different wavelengths $lambda$
  $endgroup$
  – Aaron Stevens
  58 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, I get that part. My question was are we seeing colors because of bumps/pits on CD, or aluminium reflective coating. Because bumps/pits would be explained as diffraction and aluminum coating woudl be explained as thin film, I guess?
  $endgroup$
  – solidbastard
  53 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

So, we all know if you shine some light on CD (compact disc) or DVD, you can see color from all the specter red to violet.

What is bothering me is what is the reason of why do we see the colors on compact disc?

I know that CD has tiny pits/bumps (order of size is micrometers).
So when I shine a light on CD, I see different colors and they are changing I move the disc, which is because I change the angle $theta$ between my eyes and disc; which is followed by diffraction condition:

$dsintheta = mlambda$

So my, my main concern is are colors over there because light travels different distances when in comes to the surface of CD because of bumps/pits?

Can I interpret diffraction colors on CD same as color on thin films?
Because although one part of CD is quite reflective surface the other is almost fully transparent.
Is this like combination of reflection and diffraction?
I mean reflection is there because of reflective surface, and even because of some thickness of aluminum I see colors (which is like thin films)?

And now comes the bumps/pits on cd? Because of them I also see colors or?

Thanks for help, really trying to break this into smaller pieces but constantly going around...

optics diffraction

share|cite|improve this question

asked 1 hour ago

solidbastardsolidbastard

494

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So my, my main concern is are colors over there because light travels different distances when in comes to the surface of CD because of bumps/pits? Are you considering how different colors have different wavelengths $lambda$
  $endgroup$
  – Aaron Stevens
  58 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, I get that part. My question was are we seeing colors because of bumps/pits on CD, or aluminium reflective coating. Because bumps/pits would be explained as diffraction and aluminum coating woudl be explained as thin film, I guess?
  $endgroup$
  – solidbastard
  53 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

So, we all know if you shine some light on CD (compact disc) or DVD, you can see color from all the specter red to violet.

What is bothering me is what is the reason of why do we see the colors on compact disc?

I know that CD has tiny pits/bumps (order of size is micrometers).
So when I shine a light on CD, I see different colors and they are changing I move the disc, which is because I change the angle $theta$ between my eyes and disc; which is followed by diffraction condition:

$dsintheta = mlambda$

So my, my main concern is are colors over there because light travels different distances when in comes to the surface of CD because of bumps/pits?

Can I interpret diffraction colors on CD same as color on thin films?
Because although one part of CD is quite reflective surface the other is almost fully transparent.
Is this like combination of reflection and diffraction?
I mean reflection is there because of reflective surface, and even because of some thickness of aluminum I see colors (which is like thin films)?

And now comes the bumps/pits on cd? Because of them I also see colors or?

Thanks for help, really trying to break this into smaller pieces but constantly going around...

optics diffraction

share|cite|improve this question

asked 1 hour ago

solidbastardsolidbastard

494

$endgroup$

share|cite|improve this question

asked 1 hour ago

solidbastardsolidbastard

494

share|cite|improve this question

asked 1 hour ago

solidbastardsolidbastard

494

asked 1 hour ago

solidbastardsolidbastard

494

asked 1 hour ago

solidbastardsolidbastard

494

1 Answer
1

active

oldest

votes

6

$begingroup$

The pits are in parallel circular tracks with a distance of 1.6 micrometers. These act as a diffraction grating. Normally this is a reflection grating, but one can make a transmission grating by removing the metal layer (easiest in recordable disks).

Here is an image that I made in transmission with a cover disk without a recording layer. The mercury street light can be seen in the middle.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

PieterPieter

8,65531435

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. So no think films? I think I was more confused with the reflection; but now realized the only difference is that for transmission grating I need a screen to see a pattern, and over here (on CD) I see it on reflective surface..
  $endgroup$
  – solidbastard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @solidbastard You do not need a screen, just hold the transmission grating about 10 cm in front of your eye. One can also see rings in reflection, with the source behind one's head.
  $endgroup$
  – Pieter
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks I do this. But I do not get nice circles, only lines. Do you have CD or DVD there? So, bottom line. When light falls on CD, it is as grating, diffraction comes because of difference in distance that light travels or?
  $endgroup$
  – solidbastard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lines are fine! One can see circles when the hole is centered in front of the eye. Yes, the effect is due to differences in distance to the parallel tracks. See Fig 2 on page 20 of this: fusioned.gat.com/images/pdf/EMcurriculum.pdf
  $endgroup$
  – Pieter
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

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6

$begingroup$

The pits are in parallel circular tracks with a distance of 1.6 micrometers. These act as a diffraction grating. Normally this is a reflection grating, but one can make a transmission grating by removing the metal layer (easiest in recordable disks).

Here is an image that I made in transmission with a cover disk without a recording layer. The mercury street light can be seen in the middle.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

PieterPieter

8,65531435

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. So no think films? I think I was more confused with the reflection; but now realized the only difference is that for transmission grating I need a screen to see a pattern, and over here (on CD) I see it on reflective surface..
  $endgroup$
  – solidbastard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @solidbastard You do not need a screen, just hold the transmission grating about 10 cm in front of your eye. One can also see rings in reflection, with the source behind one's head.
  $endgroup$
  – Pieter
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks I do this. But I do not get nice circles, only lines. Do you have CD or DVD there? So, bottom line. When light falls on CD, it is as grating, diffraction comes because of difference in distance that light travels or?
  $endgroup$
  – solidbastard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lines are fine! One can see circles when the hole is centered in front of the eye. Yes, the effect is due to differences in distance to the parallel tracks. See Fig 2 on page 20 of this: fusioned.gat.com/images/pdf/EMcurriculum.pdf
  $endgroup$
  – Pieter
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

6

$begingroup$

The pits are in parallel circular tracks with a distance of 1.6 micrometers. These act as a diffraction grating. Normally this is a reflection grating, but one can make a transmission grating by removing the metal layer (easiest in recordable disks).

Here is an image that I made in transmission with a cover disk without a recording layer. The mercury street light can be seen in the middle.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

PieterPieter

8,65531435

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. So no think films? I think I was more confused with the reflection; but now realized the only difference is that for transmission grating I need a screen to see a pattern, and over here (on CD) I see it on reflective surface..
  $endgroup$
  – solidbastard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @solidbastard You do not need a screen, just hold the transmission grating about 10 cm in front of your eye. One can also see rings in reflection, with the source behind one's head.
  $endgroup$
  – Pieter
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks I do this. But I do not get nice circles, only lines. Do you have CD or DVD there? So, bottom line. When light falls on CD, it is as grating, diffraction comes because of difference in distance that light travels or?
  $endgroup$
  – solidbastard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lines are fine! One can see circles when the hole is centered in front of the eye. Yes, the effect is due to differences in distance to the parallel tracks. See Fig 2 on page 20 of this: fusioned.gat.com/images/pdf/EMcurriculum.pdf
  $endgroup$
  – Pieter
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

The pits are in parallel circular tracks with a distance of 1.6 micrometers. These act as a diffraction grating. Normally this is a reflection grating, but one can make a transmission grating by removing the metal layer (easiest in recordable disks).

Here is an image that I made in transmission with a cover disk without a recording layer. The mercury street light can be seen in the middle.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

PieterPieter

8,65531435

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

PieterPieter

8,65531435

answered 1 hour ago

PieterPieter

8,65531435

answered 1 hour ago

PieterPieter

8,65531435