PHYS 408 HOMEWORK 9 SOLUTIONS

22 b the wave propagates towards increasing x so a

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Unformatted text preview: hion, which is to say the wave is left elliptically polarized. 2.2 (b) The wave propagates towards increasing −x, so a possible ‘new’ set of ‘x’ and ‘y’ polarization axes are ‘z’ and ‘y’ axes, respectively (as ex × ey = ez , we now have ez × ey = −ex = e−x ). The Jones vector and normalized Jones vector are ˆ ˆ ˆ ˆ ˆ ˆ ˆ ￿ J = Ax￿ Ay ￿ = Az Ay = 2(i + 1) 3(i + 1)eiπ The phase difference is φ = π . Thus the wave is linearly polarized. 2.3 1 2 ˆ ￿ J=√ 13 3e+iπ (6) (c) The wave propagates towards increasing −z , so a possible ‘new’ set of ‘x’ and ‘y’ polarization axes are ‘y’ and ‘x’ axes, respectively (as ex × ey = ez , we now have ey × ex = e−z ). The Jones vector and normalized Jones vector are ˆ ˆ ˆ ˆ ˆ ˆ ￿ J = Ax￿ Ay ￿ = Ay Ax √ −1 2ei tan (1/8) = = √ i+8 65e−iπ/4 i−1 √ 2 1 ˆ ￿ J=√ √ −i(π/4+tan−1 (1/8)) 67 65e (7) The phase difference is φ = −π /4 − tan−1 (1/8) ≈ −0.9 rad. Thus the wave is left elliptically polarized. 2.4 (d) The Jones vector and normalized Jones vector are ￿ J = Ax Ay = 3i 3 1 1 ˆ ￿ J=√ 2 e−iπ/2 The phase difference is φ = φy − φx = 0 − π /2 and |J1 | = |J2 | so the wave is left circularly polarized. 3 (8) 2.5 (e) The wave propagates along towards increasing −x, so a possible ‘new’ set of ‘x’ and ‘y’ polarization axes are ‘z’ and ‘y’ axes, respectively (as ex × ey = ez , we now have ez × ey = −ex = e−x ). The Jones vector and normalized Jones vector are ˆ ˆ ˆ ˆ ˆ ˆ ˆ ￿ J = Ax￿ Ay ￿ = Az Ay = 2(i + 1) 3(i − 1)eiπ √ 22 1 ˆ ￿ J=√ √ 26 3 2e−iπ/2 The phase difference is φ = φy − φz = −π /2 but |J1 | = |J2 | so the wave is left elliptically polarized. ￿ 4 (9) 3 Fraunhofer diffraction In the Fraunhofer approximation, the output image is proportional to the fourier transformation of the input image, evaluated g (x, y ) ∝ F{f (x, y )}￿ ⊥ =(k/d)￿ ⊥ k x (10) The input image is of constant amplitude over the opening, so the fourier transform is an integral of a pure complex exponential over the boundary of the opening. Below is the extent of the opening, expressed as two larger squared from which the two small opaque squares are subtracted, assuming D is the center-to-center distance of the two apertures (although some students assumed D differently, although no points were deducted): mark (0,0) at point of highest symmetry...
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This document was uploaded on 03/11/2014 for the course PHYS 408 at University of British Columbia.

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