{"id":5766,"date":"2011-07-28T10:29:05","date_gmt":"2011-07-28T10:29:05","guid":{"rendered":"http:\/\/scienceblogs.com\/principles\/2011\/07\/28\/how-good-are-polarized-sunglas\/"},"modified":"2011-07-28T10:29:05","modified_gmt":"2011-07-28T10:29:05","slug":"how-good-are-polarized-sunglas","status":"publish","type":"post","link":"http:\/\/chadorzel.com\/principles\/2011\/07\/28\/how-good-are-polarized-sunglas\/","title":{"rendered":"How Good Are Polarized Sunglasses?"},"content":{"rendered":"

\"i-657b3aad5096f591d534bed7db3dd91c-broken_glasses.jpg\"A while back, I explained how polarized sunglasses work<\/a>, the short version of which is that light reflected off the ground in front of you tends to be polarized, and by blocking that light, they reduce the effects of glare. This is why fishermen wear polarized sunglasses (they make it easier to see through the surface of water) and why they’re good for driving (they cut down on glare off the road ahead). I almost exclusively buy polarized sunglasses, because I like this feature.<\/p>\n

But let’s say you have a pair of polarized sunglasses that broke, because they were cheap to begin with (such as the ones at right). At that point, they’re useless, right? Not if you’re a physicist– you can use them to answer the burning question, “How good are the polarizers in those sunglasses, anyway?”<\/p>\n

The answer turns out to be “surprisingly good.” Here’s the apparatus you need to do this test:<\/p>\n

\"i-e92872494f8e4cd0edb930cb26057d60-polarizer_test_apparatus.jpg\"<\/p>\n

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On the left, the black tube is a helium-neon laser I borrowed from the teaching labs. The light from the laser passes through a research-grade polarizer (a Glan-Thompson<\/a>, also borrowed from the teaching lab) on a rotating mount, then one lens from the broken sunglasses taped to another rotating mount, then onto a photodiode. The white-and-blue oscilloscope at the left served to make the necessary voltage measurements, which is overkill, but we have a bunch of them kicking around, and all the voltmeters in the basement labs had dead batteries (as did the nice power meter, because my summer student left it turned on, and we’re out of the special batteries it takes).<\/p>\n

The procedure is simple: the laser passes through the G-T polarizer, which establishes a linear polarization for the beam, then it goes through the sunglasses lens, which transmits a fraction of the light that depends on the angle between its polarization axis and the G-T polarizer. The figure you use to measure the quality of the polarizer is the “extinction ratio,” which is the ratio of the minimum and maximum transmitted intensities. For an ideal polarizer, this would be zero– perfectly polarized light would be completely blocked by an ideal polarizer at 90 degrees from the light polarization– but nothing is perfect, so there’s always a little bit of light leaking through.<\/p>\n

To establish a baseline for this, the first thing I did was to test two G-T polarizers together (part of the reason for using that type in the first place was that we have two identical ones in the optics lab). With the two G-T polarizers in the beam, the diode output was 0.134V with a neutral-density filter with OD 30 (that is, a filter that reduced the intensity of the light by a factor of 1000) in front of the diode. Why knock the power down this way? Because if I didn’t, the diode would saturate at its maximum value, and I wouldn’t get an accurate reading. This way, the diode is safely in a regime where the output voltage increases linearly with the input intensity.<\/p>\n

Rotating the second G-T polarizer to minimize the transmitted intensity, and removing the ND filter, the diode output was 0.00152 V. The background level was 0.00056 V (due to light from the oscilloscope display and other such factors. The ratio of these intensities is then:<\/p>\n

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Extinction Ratio = (0.00096 V)\/(1000*0.13344V) = 0.0000072<\/p>\n<\/blockquote>\n

(Notice that there are no units on the final number, because it’s a ratio. This is why it doesn’t matter that my diode output is in the somewhat arbitrary “volts” rather than “Watts” or “Watts\/cm2<\/sup>” as you would expect for a laser power or intensity. the units divide out in the end, so as long as it’s linear, we’re free to use volts.)<\/p>\n

That’s an extinction ratio of a hair under 10-5<\/sup>, which is what they advertise for commercial G-T polarizers<\/a>. Which, you will note, will set you back $300.<\/p>\n

So, how do the polarized shades stack up? The maximum transmitted intensity in this case was 0.0352 with the ND filter in place– about a quarter of the intensity for the G-T polarizers. Which is what you expect for sunglasses– they block a lot of the light of either polarization, because that’s the whole point.<\/p>\n

The minimum transmitted intensity was 0.00512V, with no filter, and the same background level. This gives an extinction ratio of 0.00013. Which is worse than the G-T polarizers, to be sure, but comparable to the “economy” polarizers ThorLabs sells<\/a> for $100-ish. That’s pretty good for a $20 pair of shades I got at Target.<\/p>\n

(A lot of the price from ThorLabs is because you’re getting something that transmits more than 25% of the light of the correct polarization, of course. It’s easy to make stuff that absorbs lots of light, but requires more care to make something that is mostly clear but still polarizes effectively.)<\/p>\n

So, if you’ve ever wondered how good those polarizing shade you bought at your local discount mega-mart are, the answer is, they’re pretty good. And if you were setting up an optics experiment on the cheap– making your own quantum eraser<\/a>, say– you could do worse than cutting up a pair of cheap polarized sunglasses.<\/p>\n

And because this is the Internet, here’s a thematically appropriate video:<\/p>\n