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Why do we see a rainbow of colors reflected off a CD or DVD?
Can thick-film reflection holograms be used to create true mirrors?Rainbow reflected on TV screen? (pic attached)How do you see a rainbow?Physics of detection of coherent light in an incoherent background using the more compact Fabry Perot interfeometer?Musical notes and colors of a rainbowWhy can we only “see” reflected light?Why is UV light visible when reflected off paper?Why does a CDDVD work as a transmission diffraction grating?Radiance for light reflected off a wallHow does a recordable CD that uses organic dyes to store data produce iridescent rainbow colors?
$begingroup$
So, we all know if you shine some light on CD (compact disc) or DVD, you can see all the colors from 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?
optics diffraction
edited Mar 7 at 12:14
knzhou
44.8k11122217
asked Mar 6 at 21:15
solidbastardsolidbastard
11917
$endgroup$
add a comment |
$begingroup$
So, we all know if you shine some light on CD (compact disc) or DVD, you can see all the colors from 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?
optics diffraction
edited Mar 7 at 12:14
knzhou
44.8k11122217
asked Mar 6 at 21:15
solidbastardsolidbastard
11917
$endgroup$
$begingroup$
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
Mar 6 at 22:06
$begingroup$
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
Mar 6 at 22:11
$begingroup$
Note that you can use this to identify whether your light source is incandescent, or whether there is something else going on to give you illumination. An incandescent lightbulb will give you a smooth rainbow, while fluorescent gas tubes and LEDs will give you distinct, separate reflections in only a few different colours (predominantly red, green and blue).
$endgroup$
– Arthur
Mar 7 at 12:19
add a comment |
$begingroup$
So, we all know if you shine some light on CD (compact disc) or DVD, you can see all the colors from 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?
optics diffraction
edited Mar 7 at 12:14
knzhou
44.8k11122217
asked Mar 6 at 21:15
solidbastardsolidbastard
11917
$endgroup$
So, we all know if you shine some light on CD (compact disc) or DVD, you can see all the colors from 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?
optics diffraction
optics diffraction
edited Mar 7 at 12:14
knzhou
44.8k11122217
asked Mar 6 at 21:15
solidbastardsolidbastard
11917
edited Mar 7 at 12:14
knzhou
44.8k11122217
asked Mar 6 at 21:15
solidbastardsolidbastard
11917
edited Mar 7 at 12:14
knzhou
44.8k11122217
edited Mar 7 at 12:14
knzhou
44.8k11122217
edited Mar 7 at 12:14
knzhou
44.8k11122217
44.8k11122217
asked Mar 6 at 21:15
solidbastardsolidbastard
11917
asked Mar 6 at 21:15
solidbastardsolidbastard
11917
asked Mar 6 at 21:15
solidbastardsolidbastard
11917
11917
$begingroup$
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
Mar 6 at 22:06
$begingroup$
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
Mar 6 at 22:11
$begingroup$
Note that you can use this to identify whether your light source is incandescent, or whether there is something else going on to give you illumination. An incandescent lightbulb will give you a smooth rainbow, while fluorescent gas tubes and LEDs will give you distinct, separate reflections in only a few different colours (predominantly red, green and blue).
$endgroup$
– Arthur
Mar 7 at 12:19
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
Mar 6 at 22:06
$begingroup$
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
Mar 6 at 22:11
$begingroup$
Note that you can use this to identify whether your light source is incandescent, or whether there is something else going on to give you illumination. An incandescent lightbulb will give you a smooth rainbow, while fluorescent gas tubes and LEDs will give you distinct, separate reflections in only a few different colours (predominantly red, green and blue).
$endgroup$
– Arthur
Mar 7 at 12:19
$begingroup$
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
Mar 6 at 22:06
$begingroup$
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
Mar 6 at 22:06
$begingroup$
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
Mar 6 at 22:11
$begingroup$
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
Mar 6 at 22:11
$begingroup$
Note that you can use this to identify whether your light source is incandescent, or whether there is something else going on to give you illumination. An incandescent lightbulb will give you a smooth rainbow, while fluorescent gas tubes and LEDs will give you distinct, separate reflections in only a few different colours (predominantly red, green and blue).
$endgroup$
– Arthur
Mar 7 at 12:19
$begingroup$
Note that you can use this to identify whether your light source is incandescent, or whether there is something else going on to give you illumination. An incandescent lightbulb will give you a smooth rainbow, while fluorescent gas tubes and LEDs will give you distinct, separate reflections in only a few different colours (predominantly red, green and blue).
$endgroup$
– Arthur
Mar 7 at 12:19
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 Mar 6 at 21:26
PieterPieter
8,97331536
$endgroup$
$begingroup$
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
Mar 6 at 21:36
$begingroup$
@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
Mar 6 at 21:39
$begingroup$
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
Mar 6 at 21:48
$begingroup$
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
Mar 6 at 21:56
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 Mar 6 at 23:47
amIamI
1314
$endgroup$
$begingroup$
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
Mar 6 at 23:53
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
Mar 7 at 0:29
$begingroup$
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
Mar 7 at 0:43
$begingroup$
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
Mar 7 at 0:53
1
$begingroup$
@solidbastard The pits do not have to do with the colors from the diffraction grating, which are also there from empty tracks. (The pits are not really holes, they have a depth of a quarter of a wavelength at the working wavelength in plastic, to optimize data contrast, but that is a different subject.)
$endgroup$
– Pieter
Mar 7 at 6:59
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 Mar 6 at 21:26
PieterPieter
8,97331536
$endgroup$
$begingroup$
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
Mar 6 at 21:36
$begingroup$
@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
Mar 6 at 21:39
$begingroup$
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
Mar 6 at 21:48
$begingroup$
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
Mar 6 at 21:56
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 Mar 6 at 21:26
PieterPieter
8,97331536
$endgroup$
$begingroup$
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
Mar 6 at 21:36
$begingroup$
@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
Mar 6 at 21:39
$begingroup$
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
Mar 6 at 21:48
$begingroup$
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
Mar 6 at 21:56
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 Mar 6 at 21:26
PieterPieter
8,97331536
$endgroup$
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 Mar 6 at 21:26
PieterPieter
8,97331536
edited Mar 6 at 21:34
answered Mar 6 at 21:26
PieterPieter
8,97331536
answered Mar 6 at 21:26
PieterPieter
8,97331536
answered Mar 6 at 21:26
PieterPieter
8,97331536
8,97331536
$begingroup$
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
Mar 6 at 21:36
$begingroup$
@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
Mar 6 at 21:39
$begingroup$
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
Mar 6 at 21:48
$begingroup$
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
Mar 6 at 21:56
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
Mar 6 at 21:36
$begingroup$
@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
Mar 6 at 21:39
$begingroup$
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
Mar 6 at 21:48
$begingroup$
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
Mar 6 at 21:56
$begingroup$
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
Mar 6 at 21:36
$begingroup$
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
Mar 6 at 21:36
$begingroup$
@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
Mar 6 at 21:39
$begingroup$
@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
Mar 6 at 21:39
$begingroup$
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
Mar 6 at 21:48
$begingroup$
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
Mar 6 at 21:48
$begingroup$
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
Mar 6 at 21:56
$begingroup$
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
Mar 6 at 21:56
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 Mar 6 at 23:47
amIamI
1314
$endgroup$
$begingroup$
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
Mar 6 at 23:53
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
Mar 7 at 0:29
$begingroup$
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
Mar 7 at 0:43
$begingroup$
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
Mar 7 at 0:53
1
$begingroup$
@solidbastard The pits do not have to do with the colors from the diffraction grating, which are also there from empty tracks. (The pits are not really holes, they have a depth of a quarter of a wavelength at the working wavelength in plastic, to optimize data contrast, but that is a different subject.)
$endgroup$
– Pieter
Mar 7 at 6:59
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 Mar 6 at 23:47
amIamI
1314
$endgroup$
$begingroup$
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
Mar 6 at 23:53
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
Mar 7 at 0:29
$begingroup$
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
Mar 7 at 0:43
$begingroup$
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
Mar 7 at 0:53
1
$begingroup$
@solidbastard The pits do not have to do with the colors from the diffraction grating, which are also there from empty tracks. (The pits are not really holes, they have a depth of a quarter of a wavelength at the working wavelength in plastic, to optimize data contrast, but that is a different subject.)
$endgroup$
– Pieter
Mar 7 at 6:59
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 Mar 6 at 23:47
amIamI
1314
$endgroup$
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 Mar 6 at 23:47
amIamI
1314
edited Mar 7 at 0:29
answered Mar 6 at 23:47
amIamI
1314
answered Mar 6 at 23:47
amIamI
1314
answered Mar 6 at 23:47
amIamI
1314
1314
$begingroup$
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
Mar 6 at 23:53
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
Mar 7 at 0:29
$begingroup$
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
Mar 7 at 0:43
$begingroup$
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
Mar 7 at 0:53
1
$begingroup$
@solidbastard The pits do not have to do with the colors from the diffraction grating, which are also there from empty tracks. (The pits are not really holes, they have a depth of a quarter of a wavelength at the working wavelength in plastic, to optimize data contrast, but that is a different subject.)
$endgroup$
– Pieter
Mar 7 at 6:59
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
Mar 6 at 23:53
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
Mar 7 at 0:29
$begingroup$
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
Mar 7 at 0:43
$begingroup$
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
Mar 7 at 0:53
1
$begingroup$
@solidbastard The pits do not have to do with the colors from the diffraction grating, which are also there from empty tracks. (The pits are not really holes, they have a depth of a quarter of a wavelength at the working wavelength in plastic, to optimize data contrast, but that is a different subject.)
$endgroup$
– Pieter
Mar 7 at 6:59
$begingroup$
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
Mar 6 at 23:53
$begingroup$
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
Mar 6 at 23:53
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
Mar 7 at 0:29
$begingroup$
I hope the paragraph I added about CDs helps.
$endgroup$
– amI
Mar 7 at 0:29
$begingroup$
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
Mar 7 at 0:43
$begingroup$
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
Mar 7 at 0:43
$begingroup$
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
Mar 7 at 0:53
$begingroup$
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
Mar 7 at 0:53
1
1
$begingroup$
@solidbastard The pits do not have to do with the colors from the diffraction grating, which are also there from empty tracks. (The pits are not really holes, they have a depth of a quarter of a wavelength at the working wavelength in plastic, to optimize data contrast, but that is a different subject.)
$endgroup$
– Pieter
Mar 7 at 6:59
$begingroup$
@solidbastard The pits do not have to do with the colors from the diffraction grating, which are also there from empty tracks. (The pits are not really holes, they have a depth of a quarter of a wavelength at the working wavelength in plastic, to optimize data contrast, but that is a different subject.)
$endgroup$
– Pieter
Mar 7 at 6:59
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
Mar 6 at 22:06
$begingroup$
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
Mar 6 at 22:11
$begingroup$
Note that you can use this to identify whether your light source is incandescent, or whether there is something else going on to give you illumination. An incandescent lightbulb will give you a smooth rainbow, while fluorescent gas tubes and LEDs will give you distinct, separate reflections in only a few different colours (predominantly red, green and blue).
$endgroup$
– Arthur
Mar 7 at 12:19