CourseNotes.59 - 14.5 Problems Problem 33.42 In figure 39...

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Unformatted text preview: 14.5 Problems Problem 33.42 In figure 39, unpolarized light is sent into a system of three polarizing sheets, which transmits 0.05 of the initial light intensity. The polarizing direction of the first and third sheets are as angles θ1 = 0◦ and θ3 = 90◦ . What are the (a) smaller and (b) larger possible values of angle θ2 (< 90◦ ) for the polarizing direction of sheet 2? This is actually a very interesting problem because if the middle polarizer were not there, then no light would be transmitted. The first polarizer simply serves to transmit half of the original intensity of the light and to polarize the light in the y direction. The second polarizer will decrease the intensity of the light and rotate the polarization. The third polarizer will have the same effect as the second. We must be careful when using the intensity formula for a polarizer because the angle is defined as the angle between the polarization of the wave and the direction of the polarizer. The third polarizer will therefore have to be relative to the polarization leaving the second. I = I0 1 2 cos2 θ2 cos2 (θ3 − θ2 ) I0 cos2 θ2 cos2 (90◦ − θ2 ) 2 I0 cos2 θ2 sin2 θ2 = 2 I0 = sin2 (2θ2 ) 8 = Hence, θ2 = 1 sin−1 2 8I I0 Plugging in for I/I0 we find that the two values of θ2 less than 90◦ are θ2 = 19.6◦ and θ2 = 70.4◦ . Figure 39: Figure for Problem 33.42. Problem 33.53 In figure 40, light is incident ant an angle θ1 = 40.1◦ on a boundary between two transparent materials. Some of the light travels down through the next three layers of transparent materials, while some of it reflects upward and then escapes into the air. If n1 = 1.30, n2 = 1.40, n3 = 1.32, and n4 = 1.45, what is the value of (a) θ5 and (b) θ4 ? Lets begin by calculating θ5 . Since a ray reflects with the same angle that it was incident with, the ray will hit the other surface at the same angle of incidence, θ1 . Hence, when it leaves the 59 ...
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