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In the 1953 science fiction novel Childhood’s End by Arthur C. Clarke, characters use two crossed fields in outer space to block some of the solar radiation traveling towards earth: “Somehow, out in space, the light of the Sun had been polarized by two crossed fields so that no radiation could pass” (page 12, here's the link: https://books.google.com/books?id=duCDX5kelHsC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false) I was wondering what is meant when Clarke writes that the radiation was “polarized by two crossed fields.” What are the ways that this effect could be achieved in theory (not necessarily in practice)? Also, I am studying first year college physics, and I am asking this question out of my own curiosity. Thank you very much.

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I can see something here.

So, from the sun let say you have unpolarised light. This means it could be going in every direction. I have drawn a (crap) diagram to give you an idea that the light is polarised in every direction.

Light polarised in every direction.

Normally, I think (don't quote me), in polarisers you have long thin iodine molecules which block light in every other polarisation apart from the one in which they are aligned.

enter image description here

Now, if we add another polariser, at 90%s to the original, we have now no light left.

enter image description here

Fundamentally, I suspect this will be an electric interaction with the iodine molecules. Therefore, maybe. (I like to believe anyway).

And in the real life, we have an image like this, see the polarisers are dark (letting no light through), when their axis cross.

enter image description here

Anyway, this is probably what he was getting at.

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  • $\begingroup$ You are using molecules as polarizers not fields. Therefore your answer doesn't answer the question regarding crossed fields. $\endgroup$
    – freecharly
    Commented Oct 22, 2016 at 23:55
  • $\begingroup$ I did use that as an analogy, see where I mention it is a probably an electromagnetic effect from the molecules themselves - be that their very fundamental charges. In fact, it must be because the light itself is a travelling EM wave, the only way we can kill of a E&B field like that is by it interacting with another E/M field. $\endgroup$
    – Tomi
    Commented Oct 23, 2016 at 0:47

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