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# Scan10002 - Physics 7C Polarization Using Reﬂection...

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Unformatted text preview: Physics 7C Polarization Using Reﬂection: Brewster's Angle A second way to generate linearly polarized light is by using reﬂection. If you shine light of any polarization at a boundary between two media, at a particular angle of incidence the reﬂected beam is linearly polarized. The particular angle of incidence is called Brewster's angle, 913- We learned in a previous lab how the percentage of reﬂected and transmitted intensity of light hitting a boundary (R and T) depends on n,, I12 and 91 (Reﬂection & Refraction lab, Equation 5). The formula we used is actually only true for the component of the incident light linearly polarized perpendicular to the plane of incidence, which is the plane deﬁned by the incident and reﬂected beams (see Figure 3). The other component, which is linearly polarized in the plane of incidence, has a different formula for R and T (which also depends on n1, 112 and 91). In the Reﬂection & Refraction lab our lasers generated linearly polarized light perpendicular to the plane of incidence. At the angle of incidence GB, R for the component polarized in the plane of incidence is zero. Therefore none of the light polarized in the plane of incidence gets reﬂected, so the reﬂected beam comes entirely from light polarized perpendicular to the plane of incidence and has this polarization. Fig. 3: A light beam striking a boundary at Brewster’s angle. The plane of the page is called the plane of incidence. The reﬂected and transmitted beams are at right angles. The reﬂected beam is linearly polarized perpendicular to the plane of incidence. It can be shown from Maxwell’s equations that when the angle of incidence is GB, the reﬂected and transmitted beams are at right angles. Starting with Snell’s law and using this fact, we can derive an expression for GB: 62 = 90° —63 = n1 sine,5 = 722 sin(90° —05) = 722 0030,, => tanGB =ﬂ Eq.2 n. Brewster's angle depends on the two indices of refraction of the boundary, and so is different for each boundary. Notice, however, that Equation 2 has a solution for any combination of 111 and n2, and so all POLARIZATION 7 4 58 ...
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