Correct physics behind the colors on CD (compact disc)?
$begingroup$
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
asked 4 hours ago
solidbastardsolidbastard
544
$endgroup$
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
asked 4 hours ago
solidbastardsolidbastard
544
$endgroup$
$begingroup$
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
4 hours 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
3 hours 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
asked 4 hours ago
solidbastardsolidbastard
544
$endgroup$
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
optics diffraction
asked 4 hours ago
solidbastardsolidbastard
544
asked 4 hours ago
solidbastardsolidbastard
544
asked 4 hours ago
solidbastardsolidbastard
544
asked 4 hours ago
solidbastardsolidbastard
544
asked 4 hours ago
solidbastardsolidbastard
544
544
$begingroup$
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
4 hours 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
3 hours ago
add a comment |
$begingroup$
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
4 hours 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
3 hours ago
$begingroup$
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
4 hours ago
$begingroup$
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
4 hours 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
3 hours 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
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 4 hours ago
PieterPieter
8,70531436
$endgroup$
$begingroup$
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
4 hours 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
4 hours 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
4 hours 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
4 hours ago
add a comment |
$begingroup$
We often think that light goes in straight lines, and reflection angle equals incidence angle, but that is only a statistical property of bulk light. An individual photon can go in any direction after a disturbance (at least we can't predict it), but paths whose lengths are much longer than the straight line path will cancel because the phase at the detector (eye) is affected by the path length. When the apparent distances between obstacles match the wavelength (color) of light, that color is reinforced (not cancelled, because the cancelling photons went through the gaps between the obstacles). For any colored spot that you see, there are several paths whose lengths differ only by a multiple of the wavelength. Other colors don't appear there because their wavelengths are not exact divisors of the multiple path lengths, so they can cancel out. Reflection and diffraction merely refer to the two sides of the obstacles: reflected light bounces off the obstacles and diffracted light goes through the gaps.
On a CD, the important distance is that between the 'data tracks', not the thickness of the disk or the spacing between 'bits' (a blank CD-R already diffracts). A CD is pressed with tracks (either with bit pits, or just a groove for photosensitive dye), and the aluminized layer is placed over that. It is the gap between tracks that reflects the most light (or passes it if you remove the backing). Depending on viewing angle, the apparent distance between tracks
will be a multiple of certain wavelengths, which will be the color you see there.
answered 2 hours ago
amIamI
1114
$endgroup$
$begingroup$
Thank you @aml So it is due to the reflection on a surface? I get it the that I se the color becuse difference of those paths is multiple of wavelengths. But this difference in paths on compact disc (CD) is because CD has a small pits/bumps, so light on top of the layer and bottom of the layer has a path difference. And when that path difference is multiple of wavelength I see a color? Isn't this the same a oil mark, like a thin film? Or I got it wrong? btw thanks for the reply one more time!
$endgroup$
– solidbastard
2 hours ago
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
1 hour ago
$begingroup$
Not fully. Because I still can not put this fully why are there colors on CD, although I am reading many articles on this and literature. I know why is diffraction and interference happening. But still with small holes (pits) on cd + reflective layer. Can not connect all the dots. Because for the diffraction I need to have difference in paths that light travels. And I guess you said it is the gap in a space between two tracks on cd.(that usually goes in spiral) that is making the effect of colors. $dsintheta=mlambda$ and depending on theta angle i see different wavelength
$endgroup$
– solidbastard
1 hour ago
$begingroup$
Sorry I can't draw -- in cross section the shiny gaps between tracks would look like '_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _'. Now put a lamp and eyeball anywhere and draw lines connecting each to the center of each track gap. Now measure each light-to-eye path and see how nearby paths differ by a similar amount.
$endgroup$
– amI
1 hour ago
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
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votes
$begingroup$
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.
answered 4 hours ago
PieterPieter
8,70531436
$endgroup$
$begingroup$
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
4 hours 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
4 hours 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
4 hours 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
4 hours 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.
answered 4 hours ago
PieterPieter
8,70531436
$endgroup$
$begingroup$
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
4 hours 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
4 hours 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
4 hours 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
4 hours 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.
answered 4 hours ago
PieterPieter
8,70531436
$endgroup$
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.
answered 4 hours ago
PieterPieter
8,70531436
edited 4 hours ago
answered 4 hours ago
PieterPieter
8,70531436
answered 4 hours ago
PieterPieter
8,70531436
answered 4 hours ago
PieterPieter
8,70531436
8,70531436
$begingroup$
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
4 hours 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
4 hours 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
4 hours 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
4 hours ago
add a comment |
$begingroup$
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
4 hours 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
4 hours 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
4 hours 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
4 hours ago
$begingroup$
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
4 hours ago
$begingroup$
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
4 hours 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
4 hours 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
4 hours 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
4 hours 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
4 hours 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
4 hours 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
4 hours ago
add a comment |
$begingroup$
We often think that light goes in straight lines, and reflection angle equals incidence angle, but that is only a statistical property of bulk light. An individual photon can go in any direction after a disturbance (at least we can't predict it), but paths whose lengths are much longer than the straight line path will cancel because the phase at the detector (eye) is affected by the path length. When the apparent distances between obstacles match the wavelength (color) of light, that color is reinforced (not cancelled, because the cancelling photons went through the gaps between the obstacles). For any colored spot that you see, there are several paths whose lengths differ only by a multiple of the wavelength. Other colors don't appear there because their wavelengths are not exact divisors of the multiple path lengths, so they can cancel out. Reflection and diffraction merely refer to the two sides of the obstacles: reflected light bounces off the obstacles and diffracted light goes through the gaps.
On a CD, the important distance is that between the 'data tracks', not the thickness of the disk or the spacing between 'bits' (a blank CD-R already diffracts). A CD is pressed with tracks (either with bit pits, or just a groove for photosensitive dye), and the aluminized layer is placed over that. It is the gap between tracks that reflects the most light (or passes it if you remove the backing). Depending on viewing angle, the apparent distance between tracks
will be a multiple of certain wavelengths, which will be the color you see there.
answered 2 hours ago
amIamI
1114
$endgroup$
$begingroup$
Thank you @aml So it is due to the reflection on a surface? I get it the that I se the color becuse difference of those paths is multiple of wavelengths. But this difference in paths on compact disc (CD) is because CD has a small pits/bumps, so light on top of the layer and bottom of the layer has a path difference. And when that path difference is multiple of wavelength I see a color? Isn't this the same a oil mark, like a thin film? Or I got it wrong? btw thanks for the reply one more time!
$endgroup$
– solidbastard
2 hours ago
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
1 hour ago
$begingroup$
Not fully. Because I still can not put this fully why are there colors on CD, although I am reading many articles on this and literature. I know why is diffraction and interference happening. But still with small holes (pits) on cd + reflective layer. Can not connect all the dots. Because for the diffraction I need to have difference in paths that light travels. And I guess you said it is the gap in a space between two tracks on cd.(that usually goes in spiral) that is making the effect of colors. $dsintheta=mlambda$ and depending on theta angle i see different wavelength
$endgroup$
– solidbastard
1 hour ago
$begingroup$
Sorry I can't draw -- in cross section the shiny gaps between tracks would look like '_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _'. Now put a lamp and eyeball anywhere and draw lines connecting each to the center of each track gap. Now measure each light-to-eye path and see how nearby paths differ by a similar amount.
$endgroup$
– amI
1 hour ago
add a comment |
$begingroup$
We often think that light goes in straight lines, and reflection angle equals incidence angle, but that is only a statistical property of bulk light. An individual photon can go in any direction after a disturbance (at least we can't predict it), but paths whose lengths are much longer than the straight line path will cancel because the phase at the detector (eye) is affected by the path length. When the apparent distances between obstacles match the wavelength (color) of light, that color is reinforced (not cancelled, because the cancelling photons went through the gaps between the obstacles). For any colored spot that you see, there are several paths whose lengths differ only by a multiple of the wavelength. Other colors don't appear there because their wavelengths are not exact divisors of the multiple path lengths, so they can cancel out. Reflection and diffraction merely refer to the two sides of the obstacles: reflected light bounces off the obstacles and diffracted light goes through the gaps.
On a CD, the important distance is that between the 'data tracks', not the thickness of the disk or the spacing between 'bits' (a blank CD-R already diffracts). A CD is pressed with tracks (either with bit pits, or just a groove for photosensitive dye), and the aluminized layer is placed over that. It is the gap between tracks that reflects the most light (or passes it if you remove the backing). Depending on viewing angle, the apparent distance between tracks
will be a multiple of certain wavelengths, which will be the color you see there.
answered 2 hours ago
amIamI
1114
$endgroup$
$begingroup$
Thank you @aml So it is due to the reflection on a surface? I get it the that I se the color becuse difference of those paths is multiple of wavelengths. But this difference in paths on compact disc (CD) is because CD has a small pits/bumps, so light on top of the layer and bottom of the layer has a path difference. And when that path difference is multiple of wavelength I see a color? Isn't this the same a oil mark, like a thin film? Or I got it wrong? btw thanks for the reply one more time!
$endgroup$
– solidbastard
2 hours ago
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
1 hour ago
$begingroup$
Not fully. Because I still can not put this fully why are there colors on CD, although I am reading many articles on this and literature. I know why is diffraction and interference happening. But still with small holes (pits) on cd + reflective layer. Can not connect all the dots. Because for the diffraction I need to have difference in paths that light travels. And I guess you said it is the gap in a space between two tracks on cd.(that usually goes in spiral) that is making the effect of colors. $dsintheta=mlambda$ and depending on theta angle i see different wavelength
$endgroup$
– solidbastard
1 hour ago
$begingroup$
Sorry I can't draw -- in cross section the shiny gaps between tracks would look like '_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _'. Now put a lamp and eyeball anywhere and draw lines connecting each to the center of each track gap. Now measure each light-to-eye path and see how nearby paths differ by a similar amount.
$endgroup$
– amI
1 hour ago
add a comment |
$begingroup$
We often think that light goes in straight lines, and reflection angle equals incidence angle, but that is only a statistical property of bulk light. An individual photon can go in any direction after a disturbance (at least we can't predict it), but paths whose lengths are much longer than the straight line path will cancel because the phase at the detector (eye) is affected by the path length. When the apparent distances between obstacles match the wavelength (color) of light, that color is reinforced (not cancelled, because the cancelling photons went through the gaps between the obstacles). For any colored spot that you see, there are several paths whose lengths differ only by a multiple of the wavelength. Other colors don't appear there because their wavelengths are not exact divisors of the multiple path lengths, so they can cancel out. Reflection and diffraction merely refer to the two sides of the obstacles: reflected light bounces off the obstacles and diffracted light goes through the gaps.
On a CD, the important distance is that between the 'data tracks', not the thickness of the disk or the spacing between 'bits' (a blank CD-R already diffracts). A CD is pressed with tracks (either with bit pits, or just a groove for photosensitive dye), and the aluminized layer is placed over that. It is the gap between tracks that reflects the most light (or passes it if you remove the backing). Depending on viewing angle, the apparent distance between tracks
will be a multiple of certain wavelengths, which will be the color you see there.
answered 2 hours ago
amIamI
1114
$endgroup$
We often think that light goes in straight lines, and reflection angle equals incidence angle, but that is only a statistical property of bulk light. An individual photon can go in any direction after a disturbance (at least we can't predict it), but paths whose lengths are much longer than the straight line path will cancel because the phase at the detector (eye) is affected by the path length. When the apparent distances between obstacles match the wavelength (color) of light, that color is reinforced (not cancelled, because the cancelling photons went through the gaps between the obstacles). For any colored spot that you see, there are several paths whose lengths differ only by a multiple of the wavelength. Other colors don't appear there because their wavelengths are not exact divisors of the multiple path lengths, so they can cancel out. Reflection and diffraction merely refer to the two sides of the obstacles: reflected light bounces off the obstacles and diffracted light goes through the gaps.
On a CD, the important distance is that between the 'data tracks', not the thickness of the disk or the spacing between 'bits' (a blank CD-R already diffracts). A CD is pressed with tracks (either with bit pits, or just a groove for photosensitive dye), and the aluminized layer is placed over that. It is the gap between tracks that reflects the most light (or passes it if you remove the backing). Depending on viewing angle, the apparent distance between tracks
will be a multiple of certain wavelengths, which will be the color you see there.
answered 2 hours ago
amIamI
1114
edited 1 hour ago
answered 2 hours ago
amIamI
1114
answered 2 hours ago
amIamI
1114
answered 2 hours ago
amIamI
1114
1114
$begingroup$
Thank you @aml So it is due to the reflection on a surface? I get it the that I se the color becuse difference of those paths is multiple of wavelengths. But this difference in paths on compact disc (CD) is because CD has a small pits/bumps, so light on top of the layer and bottom of the layer has a path difference. And when that path difference is multiple of wavelength I see a color? Isn't this the same a oil mark, like a thin film? Or I got it wrong? btw thanks for the reply one more time!
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– solidbastard
2 hours ago
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I hope the paragraph I added about CDs helps.
$endgroup$
– amI
1 hour ago
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Not fully. Because I still can not put this fully why are there colors on CD, although I am reading many articles on this and literature. I know why is diffraction and interference happening. But still with small holes (pits) on cd + reflective layer. Can not connect all the dots. Because for the diffraction I need to have difference in paths that light travels. And I guess you said it is the gap in a space between two tracks on cd.(that usually goes in spiral) that is making the effect of colors. $dsintheta=mlambda$ and depending on theta angle i see different wavelength
$endgroup$
– solidbastard
1 hour ago
$begingroup$
Sorry I can't draw -- in cross section the shiny gaps between tracks would look like '_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _'. Now put a lamp and eyeball anywhere and draw lines connecting each to the center of each track gap. Now measure each light-to-eye path and see how nearby paths differ by a similar amount.
$endgroup$
– amI
1 hour ago
add a comment |
$begingroup$
Thank you @aml So it is due to the reflection on a surface? I get it the that I se the color becuse difference of those paths is multiple of wavelengths. But this difference in paths on compact disc (CD) is because CD has a small pits/bumps, so light on top of the layer and bottom of the layer has a path difference. And when that path difference is multiple of wavelength I see a color? Isn't this the same a oil mark, like a thin film? Or I got it wrong? btw thanks for the reply one more time!
$endgroup$
– solidbastard
2 hours ago
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
1 hour ago
$begingroup$
Not fully. Because I still can not put this fully why are there colors on CD, although I am reading many articles on this and literature. I know why is diffraction and interference happening. But still with small holes (pits) on cd + reflective layer. Can not connect all the dots. Because for the diffraction I need to have difference in paths that light travels. And I guess you said it is the gap in a space between two tracks on cd.(that usually goes in spiral) that is making the effect of colors. $dsintheta=mlambda$ and depending on theta angle i see different wavelength
$endgroup$
– solidbastard
1 hour ago
$begingroup$
Sorry I can't draw -- in cross section the shiny gaps between tracks would look like '_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _'. Now put a lamp and eyeball anywhere and draw lines connecting each to the center of each track gap. Now measure each light-to-eye path and see how nearby paths differ by a similar amount.
$endgroup$
– amI
1 hour ago
$begingroup$
Thank you @aml So it is due to the reflection on a surface? I get it the that I se the color becuse difference of those paths is multiple of wavelengths. But this difference in paths on compact disc (CD) is because CD has a small pits/bumps, so light on top of the layer and bottom of the layer has a path difference. And when that path difference is multiple of wavelength I see a color? Isn't this the same a oil mark, like a thin film? Or I got it wrong? btw thanks for the reply one more time!
$endgroup$
– solidbastard
2 hours ago
$begingroup$
Thank you @aml So it is due to the reflection on a surface? I get it the that I se the color becuse difference of those paths is multiple of wavelengths. But this difference in paths on compact disc (CD) is because CD has a small pits/bumps, so light on top of the layer and bottom of the layer has a path difference. And when that path difference is multiple of wavelength I see a color? Isn't this the same a oil mark, like a thin film? Or I got it wrong? btw thanks for the reply one more time!
$endgroup$
– solidbastard
2 hours ago
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
1 hour ago
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
1 hour ago
$begingroup$
Not fully. Because I still can not put this fully why are there colors on CD, although I am reading many articles on this and literature. I know why is diffraction and interference happening. But still with small holes (pits) on cd + reflective layer. Can not connect all the dots. Because for the diffraction I need to have difference in paths that light travels. And I guess you said it is the gap in a space between two tracks on cd.(that usually goes in spiral) that is making the effect of colors. $dsintheta=mlambda$ and depending on theta angle i see different wavelength
$endgroup$
– solidbastard
1 hour ago
$begingroup$
Not fully. Because I still can not put this fully why are there colors on CD, although I am reading many articles on this and literature. I know why is diffraction and interference happening. But still with small holes (pits) on cd + reflective layer. Can not connect all the dots. Because for the diffraction I need to have difference in paths that light travels. And I guess you said it is the gap in a space between two tracks on cd.(that usually goes in spiral) that is making the effect of colors. $dsintheta=mlambda$ and depending on theta angle i see different wavelength
$endgroup$
– solidbastard
1 hour ago
$begingroup$
Sorry I can't draw -- in cross section the shiny gaps between tracks would look like '_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _'. Now put a lamp and eyeball anywhere and draw lines connecting each to the center of each track gap. Now measure each light-to-eye path and see how nearby paths differ by a similar amount.
$endgroup$
– amI
1 hour ago
$begingroup$
Sorry I can't draw -- in cross section the shiny gaps between tracks would look like '_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _'. Now put a lamp and eyeball anywhere and draw lines connecting each to the center of each track gap. Now measure each light-to-eye path and see how nearby paths differ by a similar amount.
$endgroup$
– amI
1 hour ago
add a comment |
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$begingroup$
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
4 hours 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
3 hours ago