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(Soft question) does light intensity oscillate really fast since it is a wave?
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(Soft question) does light intensity oscillate really fast since it is a wave?
Question about the wave nature of lightIntensity of light wave penetrating soft tissueHow does light oscillate?Doesn't light have to be a wave since it has a wave length? (In contrast to the wave–particle duality)How does distance affect light intensity?What does “intensity of light” mean?Why does the intensity of light decrease as you move away from a particular point (described in question)?Relationship between intensity and amplitude of light waveFeynman’s Treatment of an Opaque Wallwhy does the intensity of light does not vary with time in youngs double slit experiment
$begingroup$
If you shine light on a wall, what will be seen is a "patch" with constant intensity. However, if light is viewed as a wave, then it is oscillations of the electromagnetic field changing from 0 to the amplitude and back really fast. So my question is, if I were able to look at the world at extreme slow motion, a quadrillion times slower or so, and I shined a beam of light at a wall, will I see a "patch" with oscillating intensity, with maximum brightness at the peaks of the wave and minimum when the field is 0? If so, is the constant brightness seen normally just our puny mortal eyes capturing only the average of this oscillating brightness?
visible-light waves electromagnetic-radiation intensity
asked 31 mins ago
AdgornAdgorn
113
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$endgroup$
add a comment |
$begingroup$
If you shine light on a wall, what will be seen is a "patch" with constant intensity. However, if light is viewed as a wave, then it is oscillations of the electromagnetic field changing from 0 to the amplitude and back really fast. So my question is, if I were able to look at the world at extreme slow motion, a quadrillion times slower or so, and I shined a beam of light at a wall, will I see a "patch" with oscillating intensity, with maximum brightness at the peaks of the wave and minimum when the field is 0? If so, is the constant brightness seen normally just our puny mortal eyes capturing only the average of this oscillating brightness?
visible-light waves electromagnetic-radiation intensity
asked 31 mins ago
AdgornAdgorn
113
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If you shine light on a wall, what will be seen is a "patch" with constant intensity. However, if light is viewed as a wave, then it is oscillations of the electromagnetic field changing from 0 to the amplitude and back really fast. So my question is, if I were able to look at the world at extreme slow motion, a quadrillion times slower or so, and I shined a beam of light at a wall, will I see a "patch" with oscillating intensity, with maximum brightness at the peaks of the wave and minimum when the field is 0? If so, is the constant brightness seen normally just our puny mortal eyes capturing only the average of this oscillating brightness?
visible-light waves electromagnetic-radiation intensity
asked 31 mins ago
AdgornAdgorn
113
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If you shine light on a wall, what will be seen is a "patch" with constant intensity. However, if light is viewed as a wave, then it is oscillations of the electromagnetic field changing from 0 to the amplitude and back really fast. So my question is, if I were able to look at the world at extreme slow motion, a quadrillion times slower or so, and I shined a beam of light at a wall, will I see a "patch" with oscillating intensity, with maximum brightness at the peaks of the wave and minimum when the field is 0? If so, is the constant brightness seen normally just our puny mortal eyes capturing only the average of this oscillating brightness?
visible-light waves electromagnetic-radiation intensity
visible-light waves electromagnetic-radiation intensity
asked 31 mins ago
AdgornAdgorn
113
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 31 mins ago
AdgornAdgorn
113
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 31 mins ago
AdgornAdgorn
113
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 31 mins ago
AdgornAdgorn
113
asked 31 mins ago
AdgornAdgorn
113
113
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In a plane polarized EMW the electric and the magnetic fields are phase-shifted like $sin(omega t)$ and $cos(omega t)$, and the light intensity is a sum of squares, so you get a constant intensity$ Ipropto E^2+B^2= textconst$.
However, some devices are much more sensitive to the electric field than to the magnetic one (photo effect, for example), so they "feel" oscillations.
There are also some devices (magnetic antennas, for example) that are more sensitive to the magnetic field (some radio-receivers).
answered 20 mins ago
Vladimir KalitvianskiVladimir Kalitvianski
11.2k11334
$endgroup$
$begingroup$
I've heard this before, but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
1 min ago
add a comment |
$begingroup$
You would need a coherent beam, because in waves it is not only intensity but also phase that makes a difference. In an incoherent beam, as sunlight, you would not get any changes in this thought experiment, because the average intensity would hold even at wavelength distances.
In a coherent laser beam you should see in your thought experiment what is shown towards the end of this youtube video, the sinusoidal pattern of impinging intensity . After all, mathematics allows us to materialize thought experiments as this one.
answered 21 mins ago
anna vanna v
161k8153453
$endgroup$
add a comment |
$begingroup$
The EM field strength in a light wave does indeed cross through zero in between + and - peaks, just like the surface of a pond goes through its natural rest height in between going up and down as ripples go by. If you slowed the wave down, you'd change its frequency, which is the same as changing its color. You could change from blue, to red, to infrared, and all the way through radio waves and other invisible colors. So no, you could not see the EM field changing as a flashing light for a slow wave (the slow changes couldn't simulate your vision receptors) , but you could set up an electric field meter and measure the change in field as the (no longer visible) wave went by. What does it really mean for the field to cross through zero? Not much; just like having the surface of a pond cross through its equilibrium height doesn't mean the ripples are gone, neither does a moment of 0 electric field mean the light wave is gone.
answered 11 mins ago
EL_DONEL_DON
2,3142726
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In a plane polarized EMW the electric and the magnetic fields are phase-shifted like $sin(omega t)$ and $cos(omega t)$, and the light intensity is a sum of squares, so you get a constant intensity$ Ipropto E^2+B^2= textconst$.
However, some devices are much more sensitive to the electric field than to the magnetic one (photo effect, for example), so they "feel" oscillations.
There are also some devices (magnetic antennas, for example) that are more sensitive to the magnetic field (some radio-receivers).
answered 20 mins ago
Vladimir KalitvianskiVladimir Kalitvianski
11.2k11334
$endgroup$
$begingroup$
I've heard this before, but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
1 min ago
add a comment |
$begingroup$
In a plane polarized EMW the electric and the magnetic fields are phase-shifted like $sin(omega t)$ and $cos(omega t)$, and the light intensity is a sum of squares, so you get a constant intensity$ Ipropto E^2+B^2= textconst$.
However, some devices are much more sensitive to the electric field than to the magnetic one (photo effect, for example), so they "feel" oscillations.
There are also some devices (magnetic antennas, for example) that are more sensitive to the magnetic field (some radio-receivers).
answered 20 mins ago
Vladimir KalitvianskiVladimir Kalitvianski
11.2k11334
$endgroup$
$begingroup$
I've heard this before, but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
1 min ago
add a comment |
$begingroup$
In a plane polarized EMW the electric and the magnetic fields are phase-shifted like $sin(omega t)$ and $cos(omega t)$, and the light intensity is a sum of squares, so you get a constant intensity$ Ipropto E^2+B^2= textconst$.
However, some devices are much more sensitive to the electric field than to the magnetic one (photo effect, for example), so they "feel" oscillations.
There are also some devices (magnetic antennas, for example) that are more sensitive to the magnetic field (some radio-receivers).
answered 20 mins ago
Vladimir KalitvianskiVladimir Kalitvianski
11.2k11334
$endgroup$
In a plane polarized EMW the electric and the magnetic fields are phase-shifted like $sin(omega t)$ and $cos(omega t)$, and the light intensity is a sum of squares, so you get a constant intensity$ Ipropto E^2+B^2= textconst$.
However, some devices are much more sensitive to the electric field than to the magnetic one (photo effect, for example), so they "feel" oscillations.
There are also some devices (magnetic antennas, for example) that are more sensitive to the magnetic field (some radio-receivers).
answered 20 mins ago
Vladimir KalitvianskiVladimir Kalitvianski
11.2k11334
answered 20 mins ago
Vladimir KalitvianskiVladimir Kalitvianski
11.2k11334
answered 20 mins ago
Vladimir KalitvianskiVladimir Kalitvianski
11.2k11334
answered 20 mins ago
Vladimir KalitvianskiVladimir Kalitvianski
11.2k11334
11.2k11334
$begingroup$
I've heard this before, but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
1 min ago
add a comment |
$begingroup$
I've heard this before, but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
1 min ago
$begingroup$
I've heard this before, but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
1 min ago
$begingroup$
I've heard this before, but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
1 min ago
add a comment |
$begingroup$
You would need a coherent beam, because in waves it is not only intensity but also phase that makes a difference. In an incoherent beam, as sunlight, you would not get any changes in this thought experiment, because the average intensity would hold even at wavelength distances.
In a coherent laser beam you should see in your thought experiment what is shown towards the end of this youtube video, the sinusoidal pattern of impinging intensity . After all, mathematics allows us to materialize thought experiments as this one.
answered 21 mins ago
anna vanna v
161k8153453
$endgroup$
add a comment |
$begingroup$
You would need a coherent beam, because in waves it is not only intensity but also phase that makes a difference. In an incoherent beam, as sunlight, you would not get any changes in this thought experiment, because the average intensity would hold even at wavelength distances.
In a coherent laser beam you should see in your thought experiment what is shown towards the end of this youtube video, the sinusoidal pattern of impinging intensity . After all, mathematics allows us to materialize thought experiments as this one.
answered 21 mins ago
anna vanna v
161k8153453
$endgroup$
add a comment |
$begingroup$
You would need a coherent beam, because in waves it is not only intensity but also phase that makes a difference. In an incoherent beam, as sunlight, you would not get any changes in this thought experiment, because the average intensity would hold even at wavelength distances.
In a coherent laser beam you should see in your thought experiment what is shown towards the end of this youtube video, the sinusoidal pattern of impinging intensity . After all, mathematics allows us to materialize thought experiments as this one.
answered 21 mins ago
anna vanna v
161k8153453
$endgroup$
You would need a coherent beam, because in waves it is not only intensity but also phase that makes a difference. In an incoherent beam, as sunlight, you would not get any changes in this thought experiment, because the average intensity would hold even at wavelength distances.
In a coherent laser beam you should see in your thought experiment what is shown towards the end of this youtube video, the sinusoidal pattern of impinging intensity . After all, mathematics allows us to materialize thought experiments as this one.
answered 21 mins ago
anna vanna v
161k8153453
answered 21 mins ago
anna vanna v
161k8153453
answered 21 mins ago
anna vanna v
161k8153453
answered 21 mins ago
anna vanna v
161k8153453
161k8153453
add a comment |
add a comment |
$begingroup$
The EM field strength in a light wave does indeed cross through zero in between + and - peaks, just like the surface of a pond goes through its natural rest height in between going up and down as ripples go by. If you slowed the wave down, you'd change its frequency, which is the same as changing its color. You could change from blue, to red, to infrared, and all the way through radio waves and other invisible colors. So no, you could not see the EM field changing as a flashing light for a slow wave (the slow changes couldn't simulate your vision receptors) , but you could set up an electric field meter and measure the change in field as the (no longer visible) wave went by. What does it really mean for the field to cross through zero? Not much; just like having the surface of a pond cross through its equilibrium height doesn't mean the ripples are gone, neither does a moment of 0 electric field mean the light wave is gone.
answered 11 mins ago
EL_DONEL_DON
2,3142726
$endgroup$
add a comment |
$begingroup$
The EM field strength in a light wave does indeed cross through zero in between + and - peaks, just like the surface of a pond goes through its natural rest height in between going up and down as ripples go by. If you slowed the wave down, you'd change its frequency, which is the same as changing its color. You could change from blue, to red, to infrared, and all the way through radio waves and other invisible colors. So no, you could not see the EM field changing as a flashing light for a slow wave (the slow changes couldn't simulate your vision receptors) , but you could set up an electric field meter and measure the change in field as the (no longer visible) wave went by. What does it really mean for the field to cross through zero? Not much; just like having the surface of a pond cross through its equilibrium height doesn't mean the ripples are gone, neither does a moment of 0 electric field mean the light wave is gone.
answered 11 mins ago
EL_DONEL_DON
2,3142726
$endgroup$
add a comment |
$begingroup$
The EM field strength in a light wave does indeed cross through zero in between + and - peaks, just like the surface of a pond goes through its natural rest height in between going up and down as ripples go by. If you slowed the wave down, you'd change its frequency, which is the same as changing its color. You could change from blue, to red, to infrared, and all the way through radio waves and other invisible colors. So no, you could not see the EM field changing as a flashing light for a slow wave (the slow changes couldn't simulate your vision receptors) , but you could set up an electric field meter and measure the change in field as the (no longer visible) wave went by. What does it really mean for the field to cross through zero? Not much; just like having the surface of a pond cross through its equilibrium height doesn't mean the ripples are gone, neither does a moment of 0 electric field mean the light wave is gone.
answered 11 mins ago
EL_DONEL_DON
2,3142726
$endgroup$
The EM field strength in a light wave does indeed cross through zero in between + and - peaks, just like the surface of a pond goes through its natural rest height in between going up and down as ripples go by. If you slowed the wave down, you'd change its frequency, which is the same as changing its color. You could change from blue, to red, to infrared, and all the way through radio waves and other invisible colors. So no, you could not see the EM field changing as a flashing light for a slow wave (the slow changes couldn't simulate your vision receptors) , but you could set up an electric field meter and measure the change in field as the (no longer visible) wave went by. What does it really mean for the field to cross through zero? Not much; just like having the surface of a pond cross through its equilibrium height doesn't mean the ripples are gone, neither does a moment of 0 electric field mean the light wave is gone.
answered 11 mins ago
EL_DONEL_DON
2,3142726
answered 11 mins ago
EL_DONEL_DON
2,3142726
answered 11 mins ago
EL_DONEL_DON
2,3142726
answered 11 mins ago
EL_DONEL_DON
2,3142726
2,3142726
add a comment |
add a comment |
Adgorn is a new contributor. Be nice, and check out our Code of Conduct.
Adgorn is a new contributor. Be nice, and check out our Code of Conduct.
Adgorn is a new contributor. Be nice, and check out our Code of Conduct.
Adgorn is a new contributor. Be nice, and check out our Code of Conduct.
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