# Is Kerr effect in glass observable?

If I apply high electric field to a glass piece can I observe Kerr effect to some extent or will it be too small for glass to observe with eyes? I want to perform simple experiment without any costly material e.g non-linear crystals .

The Kerr effect is a phenomenon in which the refractive index of a material changes because of an applied electrical field. The change in the refractive index is proportional to the square of the applied electric field. ie. $$\Delta n\propto E^2$$

So your eyes will be able to see this effect only if the change in refractive index causes a change in the refractive angle large enough to be detected by your eyes. As the light is entering a denser medium from a less denser one (air to glass), it will move towards the normal. Hence, the refractive angle will have to decrease, if the refractive index increases (Snell's Law: $n_1Sin\theta_1=n_2Sin\theta_2$). You can calculate how much change you need in the refractive index and find the appropiate amount of Electric field that would cause an observable change (it would be pretty high I am guessing, very difficult to produce in a lab).

• What if I use a Tesla Coil? Can I use it to generate the electric field to demonstrate the Kerr effect in glass? – user43794 Apr 3 '14 at 18:43
• If it is large enough to produce the necessary electric field, I don't see why not. I think you should do the calculations first because before that you can't be sure what magnitude of the field is required. – user42733 Apr 3 '14 at 19:00

According to the Encyclopaedia of Laser Physics the variation of the refractive index of silica glass with light intensity is given by:

$$\Delta n = n_2 I$$

where $n_2$ has the value $3 \times 10^{-16}$ cm$^2$/W. If we assume that a $1$% change in the refractive index would be needed to be observable you would need a power of about $5 \times 10^{13}$ W/cm$^2$.

This is a staggeringly high power density, but it's not as inaccessible as it sounds. In practice we would use a reasonably powered light source and focus it down to a tiny spot to achieve the high power densities required. However unless you have a very high power laser and extraordinarily good optics I suspect this would be out of your reach.

• Agreed these are high densities: 1W into a good 0.3NA objective will get you about $1\mathrm{TW m^{-2}}$ at the focus. So you need about 500kW - not too inaccessible with a pulsed Q-switched laser. – WetSavannaAnimal Apr 3 '14 at 9:09
• Thanks for the answer but it looks like source of high electric field is light itself. I am looking at simple experiment where applied electric field cause change in refractive index which can be shown to people. Is there any other material which is cheap <$100? – user43794 Apr 3 '14 at 18:36 If a wet cell is OK, the largest common Kerr constant is held by anethole. You will still need about 30 kV. http://www.nist.gov/data/PDFfiles/jpcrd435.pdf and http://www.ict.kth.se/courses/IO2651/docs/ElectroOptics_paper.pdf DOI:10.1063/1.3559614 DOI:10.1080/02678292.2013.836253 • thanks it is helpful to know materials which can be used for demonstration. – user43794 Apr 6 '14 at 18:21 Parth Vader is correct, for experiments with DC Kerr effect in glass the applied field usually needs to be a few kV/cm. Perhaps the effects of birefringence could be observable by eye at much lower field strengths. The Kerr effect induces birefringence a change in the refractive index for light polarised parallel to the applied field$n_\|$, relative to the index for the orthogonal polarisation$n_\perp$. Observing the change in angle of refraction brought about by changes in the refractive index would be difficult when the change is small, a more sensitive measurement uses the difference between$n_\|$and$n_\perp$to change the polarisation of the light. If the Kerr cell is placed between crossed linear polarisers the intensity of the transmitted light will be proportional$E^2$where$E$is the applied field, a dark adapted eye is a surprisingly sensitive detector and good at discriminating some light from no light in a dark room. Experimental Setup: Light, a laser pointer would be the best source, passes though a linear polariser oriented at$45^\circ$to the direction of the applied field then through the glass then though another linear polariser oriented at$90^\circ$to the first. At$0\$ field strength the light is absorbed at the last polariser, otherwise the Kerr effect rotates the polarisation and some light is transmitted.

The amount the polarisation rotates is determined by the field strength and also the length of glass the beam propagates through where the Kerr effect is active, so at a given field strength the longer the Kerr cell the more the effect is "visible".

Its not clear what motivates your simple experiment, but this approach should make the DC Kerr effect observable by eye cheaply and possibly avoid dangerously high voltages.

• @John Rennie I think you are referring to the optical Kerr effect the question is about the DC Kerr effect. – baldrik Apr 3 '14 at 9:07
• This does not seem to answer my question. The answer is about change in polarization angle when an electric field is applied (my guess). – user43794 Apr 3 '14 at 18:28
• Ok the change in polarisation angle is due to the Kerr effect though. Give more details of what you want to do with your experiment. Its also not clear what aspect of the Kerr effect you want to observe you want to apply an external field right? What do you want to see that do? – baldrik Apr 3 '14 at 18:35
• @user43794 The change in polarisation is due to the change of refractive index due the Kerr effect. A polarised measurement is the most sensitive way to see the Kerr effect. – baldrik Apr 3 '14 at 18:52
• If I have a polarized light beam in vacuum passing through two charged plates will it not alter polarization? I am just trying to see if they are two separate phenomenon or just one due to Kerr effect. – user43794 Apr 4 '14 at 5:21

The quadratic EO was observed by Kerr with candles :) so it is possible.

Filamentation would require laser though.