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Electromagnetic Waves E y ( t ) = E y.max sin( kx – w t )

Electromagnetic Waves E y ( t ) = E y.max sin( kx – w t ) ( k = 2 p / l and w = 2 p f ) ( v = f l = w / k ) B z ( t ) = B z.max sin( kx – w t ) v = c = 1/√( e 0 m 0 ) = 3.00 x 10 8 m/s E ( t ) = cB ( t ). Electromagnetic Energy Intensity

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Electromagnetic Waves E y ( t ) = E y.max sin( kx – w t )

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  1. Oregon State University PH 213, Class #28

  2. Electromagnetic Waves Ey(t) = Ey.maxsin(kx – wt) (k = 2p/l and w = 2pf) (v = fl = w/k) Bz(t) = Bz.maxsin(kx – wt) v = c = 1/√(e0m0) = 3.00 x 108 m/s E(t) = cB(t) Oregon State University PH 213, Class #28

  3. Electromagnetic Energy Intensity Average intensity delivered: Iavg. = Pavg./A = (Average power delivered) ÷ (Area receiving delivery) For a spherically symmetric (“point”) source of EM waves, we can also write Iavg.like this: Iavg = Psource/(4pr2) where r is the distance from the source to the delivery area. But in general, in terms of the wave’s E-field strength: Energy intensity (inst.) delivered: Sx(t) = (1/m0)[EyxBz] = ce0Ey2 (W/m2) Energy intensity (avg.) delivered: Sx.avg = (1/2)ce0Ey.max2 Oregon State University PH 213, Class #28

  4. Electromagnetic Waves Example (prob. 34-19): Average intensity delivered: Iavg. = Pavg./A In this case: Pavg. = .001 A = p(.001/2)2 So: Iavg. = 1273 W/m2 But we also know: Iavg.= (1/2)ce0Emax2 Therefore: (1/2)ce0Emax2= 1273 Thus: Emax= √[2(1273)/(ce0)] = 979 V/m And: Bmax= Emax/c = 3.26 mT Oregon State University PH 213, Class #28

  5. Oregon State University PH 213, Class #28

  6. Oregon State University PH 213, Class #28

  7. Polarized EM Waves I = I0cos2q where q is the angle between the planes of polarization in the first and second filters. What if there is no “first filter”? What if we’re just sending un-polarized light through a single filter? Then, on average, all planes of polarization are represented equally (random probability), and the average of cos2q over all angles q is 0.5. So for un-polarized light: I = (1/2)I0 Oregon State University PH 213, Class #28

  8. When a ray of light is incident on two polarizers with their polari-zation axes perpendicular, no light is transmitted. If a third polar-izer is inserted between these two with its polarization axis at 45° to that of the other two, does any light get through to point P? Yes, the effective transmitted intensity is equal to ¼ of I1 (where I1 is the intensity transmitted by the first polarizer): If = I1[cos2(45°)][cos2(45°)] Oregon State University PH 213, Class #28

  9. When a third polarizer is inserted at 45° between two orthogonal polarizers, some light is transmitted. If, instead of a single polarizer at 45°, we insert a larger number Nof polarizers, each time rotating the axis of polarization over an angle 90°/(N + 1), then compared to the previous case, how much light gets through? 1. no light 2. less light 3. the same amount of light 4. more light The question is really asking for the limit of [cos2(90°/N)]N as N gets large. That is, you’re effectively reducing the light intensity, I1 (transmitted by the first polarizer alone) as follows: If = I1[cos2(q12)][cos2(q23)][cos2(q34)]…, etc. Try N = 2: If = I1[cos2(30°)][cos2(30°)][cos2(30°)] = 0.422I1 Oregon State University PH 213, Class #28

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