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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 centertocenter 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.
 Fall '11
 KirkW.Madison
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