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phys: fix other intensity mentions

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eggy 2021-01-11 16:13:33 -05:00
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@ -701,7 +701,7 @@ $$E=E_0\cos\theta$$
And so: And so:
$$I=I_0\cos^2\theta$$ $$I=I_0\cos^2\theta$$
When **unpolarised light** passes through a polariser, the average result of $I\cos\theta$ is $\frac{1}{2}$, so the intensity of polarised light is **half** of the intensity of unpolarised light. When **unpolarised light** passes through a polariser, the average result of $I\cos^2\theta$ is $\frac{1}{2}$, so the intensity of polarised light is **half** of the intensity of unpolarised light.
When unpolarised light reflects off of a **smooth non-metallic** surface it will be at least partially polarised. When unpolarised light reflects off of a **smooth non-metallic** surface it will be at least partially polarised.