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Unformatted text preview: FNT #8 A look at polarisation
August 22, 2005 Question 1 the fencepost model
Consider transverse waves on a horizontal rope that is strung between vertical fence posts. a) Which polarisations will pass through the fence posts? Which will not?
Rope that tries to move to the sides will hit the posts. So only rope travelling up and down (or that component) will get through. b) Why is this a bad model for light polarisation?
Let us treat the metal as the fenceposts. Then if the light comes along we have
11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1 1111111111111 0 0000000000000 1111111111111 0000000000000 1111111111111 0000000000000 1111111111111 0000000000000 1111111111111 0000000000000 1111111111111 0000000000000 1111111111111 0000000000000 1111111111111 0000000000000 1111111111111 0000000000000 1111111111111 0000000000000
Field energy goes into kinetic energy of electrons in metal. No signal But this is the light that the "fencepost model" would suggest would get through. If I keep the light the same, but turn the polariser around instead I have 1 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 Electrons cannot move much. Wave keeps energy instead of transferring 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 2 Polariser 1 Polariser 2 Polariser 3 0 1 2 3 Figure 1: The setup of the polarisers. We divide the problem into four regions (0,1,2 and 3). Question 2 Combinations of polarisers
We have a setup of three polarisers as shown in figure 1. In each question we change the orientation of the polarisers slightly to see what gets through. I am going to work through parts (a) and (b) throughly, then do the general case and apply the formula for parts (c) and (d). Important! The orientation of the polariser refers to which field can get through, NOT the direction of the material in the polariser a) Polariser 1: 0, Polariser 2: 40, Polariser 3: 75
Note that all angles are given with respect to the vertical. To start with we have vertically polarised light going through a vertical polariser. Not suprisingly, all the light gets through. So in region 1 we have all the light, and it is still vertically polarised. The second polariser is at 40 to the vertical. We need to break the polarisation vector into components along this direction and its normal
Blocked Component Allowed component 40 0 The green (component of the) electric field goes through while the red (component of the) electric field is blocked. So in region 2 we have light polarised at 40 to the vertical with an amplitude E2 = E0 cos 40 3 Going from region 2 to region 3 we have light polarised at 40 to the vertical passing through a polariser at 75 to the vertical. Diagramwise this is
Blocked Component 40
0 0 75 35 0 Allowed component So we can relate the final magnitude to the initial by using trigonometry. We have E3 = E2 cos 35 = (E0 cos 40 ) cos 35 = E0 cos 40 cos 35 0.6275E0 But we are interested in the intensity at the other end, which is related to the square of the amplitude by some constant. We have
2 I3 = kE3 = (0.6275)2(k(E0 )2 ) 2 = 0.3938(kE0 ) = 0.3938I0 We were told that I0 = 1550 W/m2 in the beginning of the question. So the final intensity is I3 = 610 W/m2 A general rule
When going through a polariser, we are trying to find the component of the electric field along the direction of the polariser. Let us look at the light before at an angle , and a polariser at angle .
Blocked Component Allowed component Let us take that central triangle and use trig. to find the components. We know the inner angle is  from the figure. Redrawing the triangle we have 4 Ei  Ei cos(  ) Ei sin() Here we see the amplitude of the next part of the field Ei+1 is related to Ei by Ei+1 = Ei cos(  ). Let us relate this more directly to the question we have. We start with light polarised at 0 , passing through a polariser at an angle 1 . So we have E1 = E0 cos 1 . The light that exits the first polariser is polarised at an angle 1 . So when it goes through the second polariser (at an angle of 2 to the vertical) we have E2 = E1 cos(2  1 ). We can relate this back to the amplitude of the original electric field by using the relationship above. E2 = (E0 cos 1 ) cos(2  1 ). The light is now polarised along the direction 2 , and it should be easy to show that E3 = E0 cos 1 cos(2  1 ) cos(3  2 ) To relate all of this back to intensities we have I3 = I0 cos2 1 cos2 (2  1 ) cos2 (3  2 ) b) 1 = 30, 2 = 30, 3 = 70
Here we have I3 = I0 cos2 30 cos2 (30  30 ) cos2 (70  30 ) = I0 (0.75)(1)(0.5868) = 0.44I0 which gives I3 = 0.44 1550 W/m2 = 682 W/m2 5 Sometimes some intuition gets lost when plugging into long formulas. Note that the second polariser did nothing. The light was polarised at 30 from the first polariser, so all the light passed through the second polariser. c) 1 = 0, 2 = 45, 3 = 90
This is pretty simple: I3 = I0 cos2 (0 ) cos2 (45  0 ) cos2 (90  45 ) = I0 (1)(0.5)(0.5) = 0.25I0 = 0.25 1550 W/m2 = 387.5 W/m2 d) 1 = 0, 2 = 90, 3 = 45
This should be zero after all we go through one polariser that makes the light vertical and then another that takes the horizontal component. Surely enough I3 = I0 cos2 0 cos2 (90  0 ) cos2 (45  90 ) = I0 (1)(0)(0.5) =0 6 Question 3 A lens question
A near sighted person wears contacts to correct for a far point that is 3.62 metres from his eyes. The near point of his unaided eyes us 25.0 cm from his eyes. If he does not remove his lenses when reading, how close can he hold the book and still read clearly? Answer: Let us recall what we know. The near point is fine at 25 cm, and so we are not getting the lens to correct for that. Instead the lens should make an image of an object a long way away appear to be approximately 3.62 metres from the eye. Then the eye is capable of focusing on the image created by the lens. As this is a contact lens, we know that the distance from the eye is the same as the distance from the lens. We are trying to correct for objects near infinity: ocontact = oeye = 3.62 m, so that the eye can focus. icontact = oeye = 3.62 m. Using the thin lens equation for the contact lenses we find 1 ocontact icontact 1 1 = 0 3.62 m fcontact So we conclude fcontact = 3.62 m We now want to know how close we can bring an object to the eye. The near point is 25 cm, so the object for the eye cannot be any closer than this. Remember that the object for the eye is the image created by the contact lens. See figure 2 for an idea of the layout. So we want icontact = 25 cm so that the image is 25 cm behind the contact lens (and hence 25 cm in front of the eye, at its near point). 7 + 1 = 1 fcontact i contact Object Image ocontact oeye i eye Focal Point This is the near point Figure 2: In the upper diagram we see the contact lens focusing an object to a (virtual) image. To find what happens when the eye looks at it we go to the second picture, where we treat the image as the object for the eye and ignore the contact lens. The eye can still focus on the green arrow as it is located at the near point. We apply the thin lens equation to the contact lens again 1 ocontact icontact fcontact 1 1 1 =  ocontact 0.25 m 3.62 m 1 1 1 + = 3.72376 m1 = ocontact 3.62 m 0.25 m This gives us an object distance of ocontact = 26.85 cm As expected from figure 2, this is further away than if the person had taken off their glasses to start with. + 1 = 1 8 ...
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 Spring '09
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