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How do polarized sunglasses reduce intensity? When unpolarized sunlight enters our eyes through linearly polarized sunglasses the intensity is reduced. How can I understand it from the expression of the intensity? I would prefer some mathematics.

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The boring bit is that one polarisation gets through and the other doesn't, so the intensity is cut by half.

But of course suitably opaque glasses could do the same much more cheaply.

The interesting bit comes from reflection of (unpolarised) sunlight off puddles, or swimming pools, or the sea. The reflected light is predominately horizontally polarised. It depends on the angle. At a particular angle, the Brewster angle, when the reflected ray and the refracted ray are at 90 degrees to each other, there is no vertical polarisation in the reflection. So Polaroid sunglasses which accept the vertical and reject the horizontal component cut out half of general light, but much more than half of light reflected off horizontal sheets of water. So you can see what's happening in the pool without being blanked by the glare reflected from the surface.

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    $\begingroup$ There's a little more to the boring part because sunlight is not split up equally into two perpendicular polarizations aligned with your glasses ;) $\endgroup$ – Jasper May 29 '18 at 19:14
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When one talks about polarization, one conventionally thinks about the direction of the electric field oscillation of an electromagnetic wave. A light wave train consists of mutually perpendicular electric and magnetic fields whose planes of oscillation are perpendicular to the direction of propagation. A non-polarized light source is emitting gazillions (exactly!) of wave trains for which the planes of the electric fields (and consequently the magnetic fields) are uniformly distributed from $0$ to $2\pi$. The intensity of light is proportional to the square of electric field amplitude, so $$I \propto E^2.$$ For any particular plane of oscillation (and hence any particular linear polarization), the intensity in that plane is proportional to the square of component the electric field amplitude in that plane. Note that the electric field is a vector, so we can break that vector into components parallel and perpendicular to any polarization plane we wish. Armed with this concept we can write $$I\left(\theta\right)\propto \left(E\cos\theta\right)^2$$ Now, because, the unpolarized light is randomly (uniformly?) distributed over all angles, the total intensity for any particular plane is simply the average intensity in that plane coming from all contributing components: $$I_{polarized}\propto\dfrac{1}{2\pi}\int\limits_0^{2\pi}E^2\cos^2\theta\ d\theta=\dfrac{E^2}{2}$$ So, polarized glasses, in their first action, cut the intensity by half (50%) by plane-polarizing the non-polarized light.

In their second action, their plane of polarization is to allow the vertical (perpendicular to the side-by-side alignment of the lenses) electric field component to pass through. The reason for this is that most optical glare which affects the average human viewing experience is due to reflection of light from horizontal surfaces. Consider, for example, sunlight reflecting from the hood of a care or from the dashboard, or light reflecting from a lake surface or swimming pool.

When light reflects from a surface, the light is partially linearly polarized with the electric field parallel, or transverse, to the surface. If the reflection is at a specific angle, called the Brewster angle, it is totally linearly polarized transverse to the reflection surface, or TE polarized. Therefore, light reflected from horizontal surfaces is thus partially or totally linearly polarized horizontally. The polarized sunglasses will then filter out this glare preferentially more than 50%.

The specific value of the Brewster angle depends on the material of the reflecting surface, as does the amount of partial polarization at other angles.

Additionally, sunlight is partially linearly polarized by scattering from molecules in the atmosphere, but this is only slightly noticeable if you look at different parts of the "blue" sky on a generally clear day. This polarization doesn't affect the glare factor much (at least in my experience).

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    $\begingroup$ You can observe the scattering polarization of the sky with a polarizing sheet (or a pair of polarizing sunglasses). Just hold it out so that you look at any patch of sky 90 degrees from the sun and then twist it back and forth. $\endgroup$ – dmckee Jun 1 '18 at 17:52
  • $\begingroup$ @dmckee That comes in handy when viewing the moon at first or last quarter, a polarizing sheet really improves the contrast. $\endgroup$ – PM 2Ring Jun 1 '18 at 19:11
  • $\begingroup$ BTW, bees use sky polarization in their navigation. Allegedly, the sky polarization can be detected in a hazy sky, but I bet it's a skill that requires a lot of practice. It's claimed that Viking navigators used the birefringent crystal Iceland spar, aka calcite, to determine the sun direction on hazy days. See en.wikipedia.org/wiki/Sunstone_(medieval) $\endgroup$ – PM 2Ring Jun 1 '18 at 19:22

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