Does the inverse-square law apply to linearly polarized light?
$begingroup$
It's a stupid question but: We did and experiment using linearly polarized microwave radiation generators and receivers. Our teacher asked to check experimentally if the receiver measurements are proportional to the intensity of the radiation I or to the intensity of the electric field $E$. To check that we took the measurements M of the receiver to different distances R between the receiver and the generator. She said that if the receiver goes with I the diagram of $M=f(frac{1}{R^2})$ will fit better to linear fitting than the $M=f(frac{1}{R})$ diagram. If the reverse happens the receiver goes with $E$. So that's why I am confused. Shouldn't the $frac{1}{R^2}$ relationship indicate that the receiver's measurements are proportional to $E$?
polarization
edited Dec 30 '18 at 9:41
Wasserwaage
3591416
asked Dec 30 '18 at 2:14
JimmyNikJimmyNik
183
$endgroup$
add a comment |
$begingroup$
It's a stupid question but: We did and experiment using linearly polarized microwave radiation generators and receivers. Our teacher asked to check experimentally if the receiver measurements are proportional to the intensity of the radiation I or to the intensity of the electric field $E$. To check that we took the measurements M of the receiver to different distances R between the receiver and the generator. She said that if the receiver goes with I the diagram of $M=f(frac{1}{R^2})$ will fit better to linear fitting than the $M=f(frac{1}{R})$ diagram. If the reverse happens the receiver goes with $E$. So that's why I am confused. Shouldn't the $frac{1}{R^2}$ relationship indicate that the receiver's measurements are proportional to $E$?
polarization
edited Dec 30 '18 at 9:41
Wasserwaage
3591416
asked Dec 30 '18 at 2:14
JimmyNikJimmyNik
183
$endgroup$
$begingroup$
What exactly is M in your question ?
$endgroup$
– Lelouch
Dec 30 '18 at 7:34
add a comment |
$begingroup$
It's a stupid question but: We did and experiment using linearly polarized microwave radiation generators and receivers. Our teacher asked to check experimentally if the receiver measurements are proportional to the intensity of the radiation I or to the intensity of the electric field $E$. To check that we took the measurements M of the receiver to different distances R between the receiver and the generator. She said that if the receiver goes with I the diagram of $M=f(frac{1}{R^2})$ will fit better to linear fitting than the $M=f(frac{1}{R})$ diagram. If the reverse happens the receiver goes with $E$. So that's why I am confused. Shouldn't the $frac{1}{R^2}$ relationship indicate that the receiver's measurements are proportional to $E$?
polarization
edited Dec 30 '18 at 9:41
Wasserwaage
3591416
asked Dec 30 '18 at 2:14
JimmyNikJimmyNik
183
$endgroup$
It's a stupid question but: We did and experiment using linearly polarized microwave radiation generators and receivers. Our teacher asked to check experimentally if the receiver measurements are proportional to the intensity of the radiation I or to the intensity of the electric field $E$. To check that we took the measurements M of the receiver to different distances R between the receiver and the generator. She said that if the receiver goes with I the diagram of $M=f(frac{1}{R^2})$ will fit better to linear fitting than the $M=f(frac{1}{R})$ diagram. If the reverse happens the receiver goes with $E$. So that's why I am confused. Shouldn't the $frac{1}{R^2}$ relationship indicate that the receiver's measurements are proportional to $E$?
polarization
polarization
edited Dec 30 '18 at 9:41
Wasserwaage
3591416
asked Dec 30 '18 at 2:14
JimmyNikJimmyNik
183
edited Dec 30 '18 at 9:41
Wasserwaage
3591416
asked Dec 30 '18 at 2:14
JimmyNikJimmyNik
183
edited Dec 30 '18 at 9:41
Wasserwaage
3591416
edited Dec 30 '18 at 9:41
Wasserwaage
3591416
edited Dec 30 '18 at 9:41
Wasserwaage
3591416
3591416
asked Dec 30 '18 at 2:14
JimmyNikJimmyNik
183
asked Dec 30 '18 at 2:14
JimmyNikJimmyNik
183
asked Dec 30 '18 at 2:14
JimmyNikJimmyNik
183
183
$begingroup$
What exactly is M in your question ?
$endgroup$
– Lelouch
Dec 30 '18 at 7:34
add a comment |
$begingroup$
What exactly is M in your question ?
$endgroup$
– Lelouch
Dec 30 '18 at 7:34
$begingroup$
What exactly is M in your question ?
$endgroup$
– Lelouch
Dec 30 '18 at 7:34
$begingroup$
What exactly is M in your question ?
$endgroup$
– Lelouch
Dec 30 '18 at 7:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Nope. Point sources of waves have fields falling off as $1/r$ while intensities fall off as $1/r^2$. To see why, think about conservation of energy.
A point source is spherically symmetric, and so the intensity should only be a function of the distance $I(r)$. Intensity is defined as the area flux density of energy due to the wave. Consider a sphere of radius $r$ around the point source. The total power going through this sphere should be the product of the area and intensity:
$$P=4pi r^2 I(r).$$
And since the power leaving the sphere should be the same at any distance (all the energy has to leave the sphere in order to be conserved) then the intensity is given by
$$I(r)=frac{P}{4pi}frac{1}{r^2}.$$
Which clearly falls off as $1/r^2$. To see why the field (electric field, in your case) goes as $1/r$, remember that intensity and field $phi$ are related as
$$I=kappa|phi|^2,$$
With $kappa$ being a constant. As such, we can then say that
$$|phi|=sqrtfrac{P}{4pikappa}frac{1}{r},$$
Which falls off as $1/r$.
answered Dec 30 '18 at 3:53
Gabriel GolfettiGabriel Golfetti
1,3221713
$endgroup$
add a comment |
$begingroup$
Static electric fields fall off as $1/R^{2}$, but the ${bf E}$ field associated with a spherical wave does not. The electric field of a radiating spherical wave falls off as $1/R$. In finding the solutions of problems with accelerating charges, the difference in behavior can often be used to separate space into a "quasi-static zone" (close to the sources where the ${bf E}$ field falls off as $1/R^{2}$) and a "radiation zone" (farther away, where ${bf E}$ falls as $1/R$).
If you are measuring a wave's intensity as a function of distance from the source, your should observe the $1/R^{2}$ behavior. The reasons for this is that the amount of energy carried by an outgoing wave is proportional to ${bf E}^{2}$, and thus falls off as $1/R^{2}$ even though ${bf E}$ itself goes down as $1/R$.
answered Dec 30 '18 at 3:48
BuzzBuzz
3,54431726
$endgroup$
$begingroup$
Ok I think I got it but does it have to do with linear polarization?
$endgroup$
– JimmyNik
Dec 30 '18 at 14:08
$begingroup$
@JimmyNik No, these properties are the same whatever the polarization state.
$endgroup$
– Buzz
Dec 30 '18 at 14:21
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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votes
$begingroup$
Nope. Point sources of waves have fields falling off as $1/r$ while intensities fall off as $1/r^2$. To see why, think about conservation of energy.
A point source is spherically symmetric, and so the intensity should only be a function of the distance $I(r)$. Intensity is defined as the area flux density of energy due to the wave. Consider a sphere of radius $r$ around the point source. The total power going through this sphere should be the product of the area and intensity:
$$P=4pi r^2 I(r).$$
And since the power leaving the sphere should be the same at any distance (all the energy has to leave the sphere in order to be conserved) then the intensity is given by
$$I(r)=frac{P}{4pi}frac{1}{r^2}.$$
Which clearly falls off as $1/r^2$. To see why the field (electric field, in your case) goes as $1/r$, remember that intensity and field $phi$ are related as
$$I=kappa|phi|^2,$$
With $kappa$ being a constant. As such, we can then say that
$$|phi|=sqrtfrac{P}{4pikappa}frac{1}{r},$$
Which falls off as $1/r$.
answered Dec 30 '18 at 3:53
Gabriel GolfettiGabriel Golfetti
1,3221713
$endgroup$
add a comment |
$begingroup$
Nope. Point sources of waves have fields falling off as $1/r$ while intensities fall off as $1/r^2$. To see why, think about conservation of energy.
A point source is spherically symmetric, and so the intensity should only be a function of the distance $I(r)$. Intensity is defined as the area flux density of energy due to the wave. Consider a sphere of radius $r$ around the point source. The total power going through this sphere should be the product of the area and intensity:
$$P=4pi r^2 I(r).$$
And since the power leaving the sphere should be the same at any distance (all the energy has to leave the sphere in order to be conserved) then the intensity is given by
$$I(r)=frac{P}{4pi}frac{1}{r^2}.$$
Which clearly falls off as $1/r^2$. To see why the field (electric field, in your case) goes as $1/r$, remember that intensity and field $phi$ are related as
$$I=kappa|phi|^2,$$
With $kappa$ being a constant. As such, we can then say that
$$|phi|=sqrtfrac{P}{4pikappa}frac{1}{r},$$
Which falls off as $1/r$.
answered Dec 30 '18 at 3:53
Gabriel GolfettiGabriel Golfetti
1,3221713
$endgroup$
add a comment |
$begingroup$
Nope. Point sources of waves have fields falling off as $1/r$ while intensities fall off as $1/r^2$. To see why, think about conservation of energy.
A point source is spherically symmetric, and so the intensity should only be a function of the distance $I(r)$. Intensity is defined as the area flux density of energy due to the wave. Consider a sphere of radius $r$ around the point source. The total power going through this sphere should be the product of the area and intensity:
$$P=4pi r^2 I(r).$$
And since the power leaving the sphere should be the same at any distance (all the energy has to leave the sphere in order to be conserved) then the intensity is given by
$$I(r)=frac{P}{4pi}frac{1}{r^2}.$$
Which clearly falls off as $1/r^2$. To see why the field (electric field, in your case) goes as $1/r$, remember that intensity and field $phi$ are related as
$$I=kappa|phi|^2,$$
With $kappa$ being a constant. As such, we can then say that
$$|phi|=sqrtfrac{P}{4pikappa}frac{1}{r},$$
Which falls off as $1/r$.
answered Dec 30 '18 at 3:53
Gabriel GolfettiGabriel Golfetti
1,3221713
$endgroup$
Nope. Point sources of waves have fields falling off as $1/r$ while intensities fall off as $1/r^2$. To see why, think about conservation of energy.
A point source is spherically symmetric, and so the intensity should only be a function of the distance $I(r)$. Intensity is defined as the area flux density of energy due to the wave. Consider a sphere of radius $r$ around the point source. The total power going through this sphere should be the product of the area and intensity:
$$P=4pi r^2 I(r).$$
And since the power leaving the sphere should be the same at any distance (all the energy has to leave the sphere in order to be conserved) then the intensity is given by
$$I(r)=frac{P}{4pi}frac{1}{r^2}.$$
Which clearly falls off as $1/r^2$. To see why the field (electric field, in your case) goes as $1/r$, remember that intensity and field $phi$ are related as
$$I=kappa|phi|^2,$$
With $kappa$ being a constant. As such, we can then say that
$$|phi|=sqrtfrac{P}{4pikappa}frac{1}{r},$$
Which falls off as $1/r$.
answered Dec 30 '18 at 3:53
Gabriel GolfettiGabriel Golfetti
1,3221713
edited Dec 30 '18 at 4:15
answered Dec 30 '18 at 3:53
Gabriel GolfettiGabriel Golfetti
1,3221713
answered Dec 30 '18 at 3:53
Gabriel GolfettiGabriel Golfetti
1,3221713
answered Dec 30 '18 at 3:53
Gabriel GolfettiGabriel Golfetti
1,3221713
1,3221713
add a comment |
add a comment |
$begingroup$
Static electric fields fall off as $1/R^{2}$, but the ${bf E}$ field associated with a spherical wave does not. The electric field of a radiating spherical wave falls off as $1/R$. In finding the solutions of problems with accelerating charges, the difference in behavior can often be used to separate space into a "quasi-static zone" (close to the sources where the ${bf E}$ field falls off as $1/R^{2}$) and a "radiation zone" (farther away, where ${bf E}$ falls as $1/R$).
If you are measuring a wave's intensity as a function of distance from the source, your should observe the $1/R^{2}$ behavior. The reasons for this is that the amount of energy carried by an outgoing wave is proportional to ${bf E}^{2}$, and thus falls off as $1/R^{2}$ even though ${bf E}$ itself goes down as $1/R$.
answered Dec 30 '18 at 3:48
BuzzBuzz
3,54431726
$endgroup$
$begingroup$
Ok I think I got it but does it have to do with linear polarization?
$endgroup$
– JimmyNik
Dec 30 '18 at 14:08
$begingroup$
@JimmyNik No, these properties are the same whatever the polarization state.
$endgroup$
– Buzz
Dec 30 '18 at 14:21
add a comment |
$begingroup$
Static electric fields fall off as $1/R^{2}$, but the ${bf E}$ field associated with a spherical wave does not. The electric field of a radiating spherical wave falls off as $1/R$. In finding the solutions of problems with accelerating charges, the difference in behavior can often be used to separate space into a "quasi-static zone" (close to the sources where the ${bf E}$ field falls off as $1/R^{2}$) and a "radiation zone" (farther away, where ${bf E}$ falls as $1/R$).
If you are measuring a wave's intensity as a function of distance from the source, your should observe the $1/R^{2}$ behavior. The reasons for this is that the amount of energy carried by an outgoing wave is proportional to ${bf E}^{2}$, and thus falls off as $1/R^{2}$ even though ${bf E}$ itself goes down as $1/R$.
answered Dec 30 '18 at 3:48
BuzzBuzz
3,54431726
$endgroup$
$begingroup$
Ok I think I got it but does it have to do with linear polarization?
$endgroup$
– JimmyNik
Dec 30 '18 at 14:08
$begingroup$
@JimmyNik No, these properties are the same whatever the polarization state.
$endgroup$
– Buzz
Dec 30 '18 at 14:21
add a comment |
$begingroup$
Static electric fields fall off as $1/R^{2}$, but the ${bf E}$ field associated with a spherical wave does not. The electric field of a radiating spherical wave falls off as $1/R$. In finding the solutions of problems with accelerating charges, the difference in behavior can often be used to separate space into a "quasi-static zone" (close to the sources where the ${bf E}$ field falls off as $1/R^{2}$) and a "radiation zone" (farther away, where ${bf E}$ falls as $1/R$).
If you are measuring a wave's intensity as a function of distance from the source, your should observe the $1/R^{2}$ behavior. The reasons for this is that the amount of energy carried by an outgoing wave is proportional to ${bf E}^{2}$, and thus falls off as $1/R^{2}$ even though ${bf E}$ itself goes down as $1/R$.
answered Dec 30 '18 at 3:48
BuzzBuzz
3,54431726
$endgroup$
Static electric fields fall off as $1/R^{2}$, but the ${bf E}$ field associated with a spherical wave does not. The electric field of a radiating spherical wave falls off as $1/R$. In finding the solutions of problems with accelerating charges, the difference in behavior can often be used to separate space into a "quasi-static zone" (close to the sources where the ${bf E}$ field falls off as $1/R^{2}$) and a "radiation zone" (farther away, where ${bf E}$ falls as $1/R$).
If you are measuring a wave's intensity as a function of distance from the source, your should observe the $1/R^{2}$ behavior. The reasons for this is that the amount of energy carried by an outgoing wave is proportional to ${bf E}^{2}$, and thus falls off as $1/R^{2}$ even though ${bf E}$ itself goes down as $1/R$.
answered Dec 30 '18 at 3:48
BuzzBuzz
3,54431726
answered Dec 30 '18 at 3:48
BuzzBuzz
3,54431726
answered Dec 30 '18 at 3:48
BuzzBuzz
3,54431726
answered Dec 30 '18 at 3:48
BuzzBuzz
3,54431726
3,54431726
$begingroup$
Ok I think I got it but does it have to do with linear polarization?
$endgroup$
– JimmyNik
Dec 30 '18 at 14:08
$begingroup$
@JimmyNik No, these properties are the same whatever the polarization state.
$endgroup$
– Buzz
Dec 30 '18 at 14:21
add a comment |
$begingroup$
Ok I think I got it but does it have to do with linear polarization?
$endgroup$
– JimmyNik
Dec 30 '18 at 14:08
$begingroup$
@JimmyNik No, these properties are the same whatever the polarization state.
$endgroup$
– Buzz
Dec 30 '18 at 14:21
$begingroup$
Ok I think I got it but does it have to do with linear polarization?
$endgroup$
– JimmyNik
Dec 30 '18 at 14:08
$begingroup$
Ok I think I got it but does it have to do with linear polarization?
$endgroup$
– JimmyNik
Dec 30 '18 at 14:08
$begingroup$
@JimmyNik No, these properties are the same whatever the polarization state.
$endgroup$
– Buzz
Dec 30 '18 at 14:21
$begingroup$
@JimmyNik No, these properties are the same whatever the polarization state.
$endgroup$
– Buzz
Dec 30 '18 at 14:21
add a comment |
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$begingroup$
What exactly is M in your question ?
$endgroup$
– Lelouch
Dec 30 '18 at 7:34