(Soft question) does light intensity oscillate really fast since it is a wave?












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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?










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    $begingroup$


    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?










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      $begingroup$


      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?










      share|cite|improve this question







      New contributor




      Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      $endgroup$




      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






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      asked 32 mins ago









      AdgornAdgorn

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          3 Answers
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          $begingroup$

          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= text{const}$.



          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).






          share|cite|improve this answer









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          • $begingroup$
            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





















          1












          $begingroup$

          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.






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            0












            $begingroup$

            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.






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              3 Answers
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              active

              oldest

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              3 Answers
              3






              active

              oldest

              votes









              active

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              active

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              2












              $begingroup$

              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= text{const}$.



              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).






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                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


















              2












              $begingroup$

              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= text{const}$.



              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).






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                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
















              2












              2








              2





              $begingroup$

              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= text{const}$.



              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).






              share|cite|improve this answer









              $endgroup$



              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= text{const}$.



              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).







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              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




















              • $begingroup$
                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













              1












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 21 mins ago









                  anna vanna v

                  161k8153453




                  161k8153453























                      0












                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        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.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          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.






                          share|cite|improve this answer









                          $endgroup$



                          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.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 12 mins ago









                          EL_DONEL_DON

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