Polarisation

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take two polaroid sheets, the dark filters used in some sunglasses, and turn one until no light gets through. Now slide a third sheet in between them, at an angle. One more filter, so it should stay dark, right? But no, the light comes back! How can adding a filter let more light through? Polarisation questions do come up in JEE papers, so let's find out.

0:28Unpolarised and plane-polarised light

So, first, light is a transverse wave. Its electric field, written E with a small arrow on top because it is a vector, is perpendicular to the direction of travel. The upside down T on screen is the perpendicular sign. Okay? In unpolarised light, the field takes all directions in that perpendicular plane, changing at random. In the figure, that is the star of lines on the left. Right? But if the field stays along one line only, the light is plane polarised: the double headed arrow in the figure. Now tell me, can sound waves in air be polarised? No, they cannot, very good! Sound in air is longitudinal, it vibrates along the direction of travel, and only transverse waves can be polarised. Clear?

1:17Polaroids and Malus's law

So, a polaroid passes the part of the field along its pass axis, the dark bar on each disc, and absorbs the rest. Now, unpolarised light of intensity I nought, the capital I with a small zero below it, falls on the first polaroid, the polariser. How much comes out? Half of it, I nought over two, yes! And it is half for any orientation, because unpolarised light has all directions equally. With me? Now this polarised light, of intensity I, meets a second polaroid, the analyser, at an angle theta, the Greek letter on screen, to the field. The amplitude, small a, becomes a times cos theta, and the intensity becomes I times cos squared theta. This is Malus's law. Okay? So two polaroids at angle theta give I nought over two, times cos squared theta, the label on the right of the figure. Now, the small circle on screen is the degree sign. What if the two axes are at ninety degrees? Nothing gets through, correct, because cos of ninety degrees is zero. Such polaroids are called crossed, and that is the dark screen from our puzzle. Right?

2:31A third polaroid between two crossed ones

Alright, now our puzzle. Slide a third polaroid between the crossed pair, at angle theta to the first. After this middle one, we have I nought over two, times cos squared theta. Now, what angle does the last polaroid make with the middle one? Ninety degrees minus theta, yes, exactly! The last one is at ninety degrees to the first, and the middle one has already turned by theta. So multiply by cos squared of ninety degrees minus theta, which is sine squared theta. Okay? So out comes I nought over two, times cos squared theta, times sine squared theta. Since sine two theta is two sine theta cos theta, that is I nought over eight, times sine squared two theta, and it is not zero any more! At what angle theta is it brightest? At forty five degrees, very good! Then sine two theta is one, and we get the greatest value, I nought over eight. The middle polaroid sets the field along its own axis, no longer perpendicular to the last one, and that solves our puzzle. Clear?

3:38Polarisation by reflection: Brewster's law

Now, a second way to polarise light is reflection. In the figure, unpolarised light from air falls on a medium of refractive index n; the dashed line is the normal. Dots show the field perpendicular to the page, double arrows the field in the page. At the polarising angle, or Brewster angle, written small i with a capital B below, the reflected light is completely plane polarised. Okay? Its electric field is perpendicular to the plane of incidence, the plane of the incident ray and the normal, so the reflected ray carries only dots. Now, what angle does the small square mark between the reflected and refracted rays show? Ninety degrees, correct! So i B plus r equals ninety degrees, where r is the angle of refraction. And Brewster's law says tan i B equals n, for light coming from air. Clear? Now, the reflected light is fully polarised, so the refracted light is too, right? No, no, that is the trap, the refracted light is only partially polarised. In the figure, it still carries both dots and arrows. With me? And from medium one into medium two, tan i B equals n two over n one, the small one and two naming the media. For air, n one is one, and we are back to tan i B equals n. Got it?

5:02Scattering, and where polaroids are used

Here is one more nice fact. Sunlight scattered by the sky through ninety degrees is plane polarised too. Okay? And polaroids are used in sunglasses, camera filters, LCD screens, the liquid crystal displays, and three D cinema. Right?

5:18Example 1: three polaroids, middle one at 30 degrees

Now let us put numbers on our puzzle. Example one: unpolarised light of intensity I nought passes through three polaroids. The first and third are crossed, and the middle one makes thirty degrees with the first. Find the emergent intensity. Pause the video and try it first. Done? Okay, let's check. The first polaroid passes half, so we have I nought over two. With me? The middle one is at thirty degrees, so multiply by cos squared thirty, which is three over four. Now, the last polaroid is at thirty degrees again, right? No, careful, that is the trap: the last one is at ninety degrees to the first, so it is at ninety minus thirty, which is sixty degrees, to the middle one. And cos squared sixty is one quarter. Okay? So the intensity is I nought over two, times three over four, times one over four. That is three I nought over thirty two, the light that came back in our puzzle. Clear?

6:22Example 2: Ip over Iu from a 5:1 max-min intensity ratio

Now a harder one. Example two: a beam is a mixture of unpolarised light of intensity I u, and plane polarised light of intensity I p. The small u and p below each I stand for unpolarised and polarised. When a polaroid is rotated in the beam, the transmitted intensity varies between a maximum and a minimum in the ratio five to one. Find I p over I u. Pause the video and try it, and treat the two parts separately. Done? Alright, let's go. The unpolarised part always gives half, I u over two, at any angle. Okay? The polarised part follows Malus's law, I p times cos squared theta, so it swings from all of I p down to zero. So when is the light brightest, and when is it dimmest? Brightest when the polarised part passes fully: I max, with max written small, equals I u over two, plus I p. Dimmest when it is blocked: I min is just I u over two. Careful, the minimum is not zero: the unpolarised half always gets through. With me? The ratio is five, so I u over two, plus I p, equals five times I u over two. The double arrow means it follows that I p equals two times I u, and I p over I u is two. Got it?

7:45Quick revision: I0/2, Malus cos squared, Brewster tan iB = n

So, quick revision. Unpolarised light through one polaroid gives I nought over two, and then Malus's law: I times cos squared theta. Crossed polaroids pass nothing, but a middle one at theta lets through I nought over eight, times sine squared two theta. And Brewster's law: tan i B equals n, i B plus r is ninety degrees, and the reflected light is fully polarised. Now go and try the practice problems on this page. All the best!