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The intensity of light (as calculated from time average of the poynting vector) is given by $I = (1/2) \epsilon v E_0^2$. Here the intensity is dependent on the velocity of light in the medium. The refractive index also depends on the velocity of light. So is it safe to say that the intensity of light depends on the refractive index of the medium?

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  • $\begingroup$ It should be noted that the refractive index does not depend on the velocity of light. It is a multiplicative constant upon which the velocity of light depends. (Similar to the fact that density is not dependent upon volume, despite the fact that it links volume and mass, but rather they are dependent on it). $\endgroup$ – DoublyNegative Jun 19 at 16:56
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Since $n=\sqrt{\varepsilon_r\mu_r}$ (the relative permeability $\mu_r$ being almost always $1$), and $v=\frac{c}{n}$, you can also write $I = \frac{nc\varepsilon_0}{2} E_0^2$. (We used the decomposition $\varepsilon=\varepsilon_r\varepsilon_0$.)

So, the intensity depends linearly on the refractive index.

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  • $\begingroup$ But what would this mean? That a higher refractive index will have higher intensity of light? So a detector under water(n~1.33) will see higher intensity than that in air(n=1)? $\endgroup$ – Rick_2047 Aug 6 '14 at 11:45
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    $\begingroup$ Yes. What happens, when a beam of light enters water from air? Photons will slow down, and by doing so move closer together. So you have a higher photon density in water, and thus a higher intensity. $\endgroup$ – M.Herzkamp Aug 6 '14 at 11:52
  • $\begingroup$ Dude!!! That was such an intuitive explanation. $\endgroup$ – Rick_2047 Aug 6 '14 at 11:58
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    $\begingroup$ This is somewhere between highly misleading and utterly incorrect. For a constant electric-field amplitude, yes, the intensity "depends linearly on the refractive index", but if you shine a laser through a piece of glass, the beam does not magically get more intense in the region with higher $n$. Instead, the intensity remains constant (it is an energy flux, and energy is conserved), and the electric-field amplitude $E_0$ decreases. $\endgroup$ – Emilio Pisanty Nov 27 '18 at 23:43
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    $\begingroup$ @M.Herzkamp, the "photons as bullets with definite positions" concept is very questionable, especially here in what obviously a classical physics question. Would you use $E=hf$ or $E=hc/\lambda$ for photon's energy? The first is same as in vacuum, the other is lower. $\endgroup$ – Ján Lalinský Nov 27 '18 at 23:56
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As mentioned in the other answers, if the medium is linear then the refractive index is independent of the intensity of light, and the intensity can be related to the electric field amplitude through $I = \frac{nc\varepsilon_0}{2} E_0^2$.

However, that does not mean, as the (incorrect) accepted answer implies, that the intensity "depends linearly on $n$". If you shine a laser through a piece of glass, the beam does not magically get more intense in the region with higher refractive index; instead, the intensity remains constant (it is an energy flux, and energy is conserved), and the electric-field amplitude $E_0$ decreases.

As such, in the linear-optical regime, and absent reflection losses at the boundary between media, the intensity does not depend on the refractive index.


Having gotten over the boring part, however, and addressing the broader issue raised in the question's title,

Relation between intensity of light and refractive index

there are indeed regimes when the intensity of light has an interesting relationship with the refractive index, though it goes the other way around $-$ the refractive index depends on the intensity.

To be more specific, this happens when the light is intense enough that nonlinear effects can kick in, because of something called the Kerr effect: if the intensity is high enough, then the refractive index will increase by a small amount $\Delta n$ which is normally proportional to the intensity: $$ n(I) = n_0 + n_2 I. $$ This is important, because when lasers reach that kind of intensity, this typically only happens in the middle of the beam, and there the added $\Delta n$ makes the medium seem optically thicker, much like a convex lens would (an effect known as a Kerr lens), so it will tend to focus the beam into a tighter spot.

So, what happens if you focus the beam more tightly? Well, it will get more intense, so the self-focusing will increase and the Kerr lensing will get more severe - and if you're not careful, you can get into a regime with runaway self-focusing where the beam gets tighter and tighter until the intensity exceeds the damage threshold of the material and you burn a hole in your medium. And, if it's not your lucky day, the light will then diffract off of that hole only to re-self-focus a bit further down the line, and eventually it will destroy your entire beamline.

enter image description here

To emphasize the importance of this, if you look at the largest available peak laser intensity over the past few decades, there's a very, very flat line lasting some fifteen years between the late sixties and 1985: this is the threshold where self-focusing makes it impossible to amplify the light further without the laser destroying itself, a problem which was only solved with the advent of chirped pulse amplification.

For a more recent take on that topic, see What is Chirped Pulse Amplification, and why is it important enough to warrant a Nobel Prize?

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  • $\begingroup$ This is interesting, but does not answer the question at all. $\endgroup$ – Ján Lalinský Nov 27 '18 at 23:30
  • $\begingroup$ @JánLalinský If you mean that this doesn't touch on the relationship between intensity and the refractive index, then you're welcome to your opinion. If you (or whoever else it was) are going to downvote correct content, though, I would suggest starting with the incorrect content in the accepted answer. $\endgroup$ – Emilio Pisanty Nov 27 '18 at 23:39
  • $\begingroup$ It is better now that you've addressed the actual question, so I've removed my downvote. However, "the intensity remains constant" part is not very clear. Remains constant compared to what? Intensity outside the glass? Or compared to inside the glass, but in place with lower index of refraction? There will be either reflection or wavefront deformation and both can result in change of intensity. $\endgroup$ – Ján Lalinský Nov 28 '18 at 0:16
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The other answers here seem incomplete, because they ignore how you might actually do an experiment. The formula in question is correct, as is the response saying equivalently $I=n\times c\times \epsilon_0\times E_0^2/2$, but if you only ask what happens when you raise $n$ without considering what happens to $E_0$ you will get the wrong idea. The equation seems to say that if n increases then the intensity will increase. But actually, if you go from air to water the Fresnel equations show that $E_0$ changes by a factor $2/(1+n)$ where $n$ is the water refractive index, so Intensity changes by a factor $4n/(1+n)^2$ which is always less than one (for positive $n$).

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    $\begingroup$ and by the way, the photons slowing down argument is also only partially correct. They slow down and bunch together, but their momentum has dropped so they pack less punch. $\endgroup$ – Benedict Murdin May 30 '17 at 13:39
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So is it safe to say that the intensity of light depends on the refractive index of the medium?

All other things in the formula being equal (electric field amplitude, dielectric constant), yes. However, this may be difficult to achieve with a single light ray in a single experiment. A propagating light ray experiencing change of $n$ will also experience change of $\epsilon_0$ and will also change its electric field amplitude.

When light ray enters a dielectric with higher $n$, only part of the light energy will "get in" and propagate inside the dielectric. So intensity inside may be less than intensity outside, despite higher $n$. It depends on what percentage will get through the boundary, which in turn depends on the details of the angle of incidence and quality of boundary.

The answer can also depend on whether one wants to include the energy of the excited dielectric matter into the definition of light intensity, or consider it separately (this may make sense, since part of it is essentially kinetic energy of charged particles, not EM energy).

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  • $\begingroup$ This is misleading. All other things being equal (total energy flux, dielectric constant), no. Or which experiment do you have in mind in which the electric field amplitude (as opposed to the energy flux) is kept constant? It should be detailed explicitly if that's the scheme you're thinking of. $\endgroup$ – Emilio Pisanty Nov 27 '18 at 23:46
  • $\begingroup$ Thank you for the criticism, I've added a clarification. I've responded in the context of the question, that is, the standard formula for light intensity. There is no single experiment specified, that's why I've included the paragraphs saying "it depends". $\endgroup$ – Ján Lalinský Nov 28 '18 at 0:04

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