How Good Are Polarized Sunglasses?

i-657b3aad5096f591d534bed7db3dd91c-broken_glasses.jpgA while back, I explained how polarized sunglasses work, 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.

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?”

The answer turns out to be “surprisingly good.” Here’s the apparatus you need to do this test:

i-e92872494f8e4cd0edb930cb26057d60-polarizer_test_apparatus.jpg

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, 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).

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.

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.

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:

Extinction Ratio = (0.00096 V)/(1000*0.13344V) = 0.0000072

(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” 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.)

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

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.

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 for $100-ish. That’s pretty good for a $20 pair of shades I got at Target.

(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.)

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, say– you could do worse than cutting up a pair of cheap polarized sunglasses.

And because this is the Internet, here’s a thematically appropriate video:

Comments

  1. #1 Rob Monkey
    July 28, 2011

    Love the posts on polarized sunglasses, I’ve always enjoyed explaining how they work to people, especially since you can show them by rotating a cell phone in front of them while they wear the glasses (I assume LCDs only put out a certain orientation of light, and that’s why they darken significantly at different angles). Incidentally, as someone with eyes that are somewhat sensitive to bright light, I love my polarized glasses. I can wear them on sunny days and not have my eyes get tired, but also on bright but cloudy days and they don’t take away too much light to see effectively.

  2. #2 Blaise Pascal
    July 28, 2011

    Are there any similar experiments which can be done with the cheap circularly polarized glasses you get at the Real3D movie theaters?

  3. #3 AnonymousCoward
    July 28, 2011

    I would be interested to see how polarizing filters for photography stack up since they do need to allow light in.

    Oh, and while driving in the winter, keep in mind that reflections from ice are polarized.

  4. #4 Skwid
    July 28, 2011

    Ha! Great post, Chad.

  5. #5 Kemanorel
    July 28, 2011

    @1

    That’s exactly how they work, but they polarize based on color. They have 3 per pixil to allow different amounts of the 3 colors through. That’s why when you poke the screen the colors distort around your finger; you shift the polarizers.

    I don’t recommend it as you can damage the screen, but if you poke different colors it’s kind of interesting to see the result.

  6. #6 Kevin
    July 28, 2011

    Couldn’t you do this experiment using the other lens from your glasses instead of the G-T? (I suppose you still need the laser and other stuff…) How do the extinction ratios (of the sunglasses and of the G-T) vary with the wavelength? That might be relevant for visual perception when wearing the sunglasses.

  7. #7 MRW
    July 28, 2011

    “OD 30 (that is, a filter that reduced the intensity of the light by a factor of 1000)”

    That should be 3, not 30 (10^3=1000).

  8. #8 D. C. Sessions
    July 28, 2011

    I’m assuming your photodiode has a shunt resistor, since the primary effect is photocurrent. Back when I was designing optoelectronics, we got the best linearity by applying a fixed reverse voltage to the diode and measuring the current (optical electron-hole pair generation.)

    That was also the fastest configuration for other reasons, but not really an issue for this application :-)

  9. #9 Eric Lund
    July 28, 2011

    MRW @7: I don’t know about optics, but in other signal propagation contexts it is usual to use dB. In that case 30 would be correct: 10^(30/10) = 1000.

  10. #10 Chad Orzel
    July 28, 2011

    “OD 30 (that is, a filter that reduced the intensity of the light by a factor of 1000)”

    That should be 3, not 30 (10^3=1000).

    Yes, it should. There was supposed to be a decimal point there, but it didn’t take.

    I’m assuming your photodiode has a shunt resistor, since the primary effect is photocurrent.

    Yeah, it’s a photodiode module from ThorLabs, with a battery and little amplifier circuit in it. It outputs a voltage that saturates around 12-15V. We’ve got a bunch of them around, though I think they have a capacitance problem that plagues one of my experiments. I really need to get some bare diodes and try them at some point.

  11. #11 Lord
    July 28, 2011

    The quick test is to look at the reflection of light from pavement at Brewster’s angle and watch it brighten as turned away from horizontal.

  12. #12 hcubic
    July 29, 2011

    Here is another simple experiment you can try. Take your polarized sunglasses (or a research-grade polarizer, for that matter), and look the shadow in the light of an LCD projector that is the staple of every classroom these days.

  13. #13 CCPhysicist
    July 29, 2011

    Wish I had that stuff laying around for in-class demos!

    I’d love to use it to quantify the difference between three sets of polarizing filters we have for demos and hands-on activities. One set is quite old and came from a name manufacturer. The difference between it and the others is the fraction of the “correct” polarization it passes.

    You can have a lot of fun with the circular polarizers used in the RealD glasses, starting with a comparison of what happens when you put a linear polarizer in front or behind them.

  14. #14 Jasper Janssen
    July 31, 2011

    Coud you try this with the throwaway polarized glasses from a 3D cinema? Those *are* made to transmit as much light as possible but still have good extinction, while still being under a dollar each.

    I believe there’s one system with left/right circular, and one with horiz/vert for either eye.