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2

You are getting confused by the fact that we use an imaginary number (a mathematical construct) to describe a physical phenomenon. But it is just that - a mathematical trick. You can perfectly well describe a attenuating wave with $$e^{ikx}e^{-k'x}$$ Which can be made more compact by combining the two k's into $$e^{i(k+ik')x}$$ If you are OK with the ...


1

Recall that Maxwell's equations (in the absence of losses) require only that $n^2 = \epsilon\mu$. So when you take the square root, you are mathematically allowed to take either the positive or the negative square root. Of course, then the question becomes, why would you want to take the negative square root? Clearly, this is only an issue if $\epsilon\mu ...


1

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


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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 ...


1

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 ...


2

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 $5 \times 10^{13}$ ...


1

Irrespective of where the light comes from, your eye judges the colour of light by the way it stimulates the three types of cone cell. The cone cells have broad and overlapping spectral responses: (picture from the Wikipedia article I've linked) but to a reasonable approximation you can think of the three types of cone cell as responding to red, green and ...


20

You are right. Rainbows can occur all over the sky. However the traditional one and two internal reflections of the primary and secondary bows send light back towards the sun and hence their bows appear opposite the sun and centered on the antisolar point. The reflection of the main light makes these bows stand out. And only the light that enters a droplet ...


2

Check this: You can see that the angle difference is much bigger when the light gets reflected, and can be differentiated much easier. Also another factor which probably plays a very important role is that the sun is in front of you, and its simply too strong in comparison with the light from the droplet, thus making it even more difficult to inspect this ...


0

This is quite a subtle issue. The more charges medium has in unit volume, the more it produces secondary EM waves. Common belief backed by the success of dispersion theory is that the relation $\mathbf j = c\mathbf E$ is valid, where $c$ is a constant dependent on the frequency of the wave, $\mathbf j$ is current density and $\mathbf E$ is total macroscopic ...


4

The simplest picture is that light always travels at the speed of light. But in a material it travels at the speed of light until it hits an atom. It is then absorbed and re-emitted in the same direction, which takes a small amount of time. The more this happens, the slower the effective average speed. The denser the material, the more atoms there are in the ...


1

Light gets refracted at ANY interface between two transparent mediums, following the Snell-Descartes law: $n_1\sin(i_1)=n_2\sin(i_2)$ That's it. The equation also accounts for the fact that there is no change of direction at normal incidence ($i_1=0 \implies i_2=0$).



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