From: Dave Cook 12/10/99 16:51:22 Subject: Polarised Sunglasses post id: 43723 A friend recently showed me a neat trick with polarised glass... we've all seen the trick that when you turn two polarised glass sheets 90 degrees to each other they go completely opaque (a la Hitchhikers guide's 'Peril-Sensitive sunglasses'), but the real trick happened when he introduced a third glass at 45 degrees in between the two other glass panels. You could see through again! What gives? From: Matt 12/10/99 18:53:22 Subject: re: Polarised Sunglasses post id: 43752 Light polarised at 45deg has equal vertical and horizontal components so after the 45 deg polariser one of these components is transmitted by either a vertical or horizontal polariser. From: Dr. Ed G (Avatar) 12/10/99 21:35:56 Subject: re: Polarised Sunglasses post id: 43782 Just to extend Matt's answer with a bit of mathematics... First of all, the first polariser will remove half the light, so if the initial incident light intensity is I0, the the transmitted intensity after the first polariser will be, I1 = 0.5 I0 Now the amount of light transmitted by a polariser whose polarisation direction is different to the incident light will be given by the cosine squared of the angle between the polarisation direction of the light and the polariser, A. Therefore after the second polariser at 45 degrees the intensity will be, I2 = I1 COS2A = I1COS245o = 0.25I0 and after the third polariser at 45 degrees to the second will again be, I3 = I2 COS2A = I2COS245o = 0.125I0 So, the light will be antenuated by 87.5%, but some will still get through. If however, any of the adjacent polarisers are at 90 degrees, COS2A will be zero, and no light will get through. Basically, it's all the result of vector mathematics. Soupie twist, Ed G. From: Gary 14/10/99 1:29:38 Subject: re: Polarised Sunglasses post id: 44280 Would you still achieve the same result of .125I if the first polarised lense filtered out all of either the vertical or horizontal component of the light rays? (keep in mind that this is only working off first year physics, but this was how they explained it. That light had a horizontal and a vertical component) Wouldn't the first lense leave only a horizontal or vertical component, which the last polarised lense would block completely, regardless of a lense being in the middle? From: Martin B 14/10/99 12:55:20 Subject: re: Polarised Sunglasses post id: 44477 Ahhh, but that is where the subtle wierdness of the quantum theory enter the picture. The way you have described things is the classical way of looking at the situation. The EM waves have two components, horizontally polarized and vertically polarised. An appropriately oriented polariser blocks alll of one component leaving the other component to go through. With this way of looking at things it makes no sense that you can get more light transmitted by adding a third polariser between two crossed polariser. To explain that behaviour we have to consider the quantum nature of light. Like other properties of quantum objects, polarisation is not some that light objectively has, but is something that light can be observed to have. The way the property behaves when it is not being observed is very different to the way it behaves when it is being observed. Each photon can be in one of two states, horizontally or vertically polarised for any given direction. Normally the photon exists in a 'quantum superposition' of these two states. (There is a wave equation which describes how the state of the photon evolves over time in terms of the two basis states of polarisation.) We cannot say that a photon is in one state or the other until we do a measurement on it ie pass it through a polariser. At this point the photon will be found to be in one state - and be passed by the polariser - or be found to be in the other state - and be blocked by the polariser. So after the photons have passed through the first, say vertical polariser, they are all in a state of vertical polarisation. If they now came across a horizontal polariser, they would all be blocked, as we expect. So far so good. But if a vertically polarised photon comes across a 45o polariser, what happens? We know from the classical experiment that half of the light will pass and half will be blocked. But the quantum thing about it is that all the photons that pass the 45o polariser will now be polarised at 45o, even though they were vertically polarised to start off with. (To determine whther or not a vertically polarised photon will pass this polariser, we have to use a new basis consisting of 45o polarisation and - 45o. In this basis a vertically polarised photon has equal components in each direction, so it has a 50% chance of 'falling' into either component when it hits the polariser. If it 'falls' into the 45o component it will pass, but if it 'falls' into the - 45o component it will be blocked.) So now all the photons hittng the third polariser are polarised at 45o, and the same thing happens. Half of them pass the horizontal polariser - and end up horizontally polarised - and half of them are blocked, because they are found to 'fall' into a vertically polarised state. Remember in comparison, without the middle 45o polariser, all the photons hitting the horizontal polariser were vertically polarised and so were blocked. But with the middle polariser the photons hitting the horizontal polariser were at 45o polarisation, so half of them pass. Like they say quantum mechanics is weird... This forum is un-moderated. The views and opinions expressed are those of the individual poster and not the ABC. The ABC reserves the right to remove offensive or inappropriate messages.