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Extra info for Introduction To Fourier Optics, Third Edition, Problem Solutions
6-16. The amplitude point-spread function for a coherent system with a square aperture of width 2w is given(up to a constant multiplier) by h(u, v) = sinc 2wu λzi 2wv λzi sinc . The input to the system is a one-dimensional coherent step with the step taking place along the η axis. The response σ(u, v) will be ∞ σ(u, v) = h(u, v) ⊗ s(u, v) = −∞ h(α, β)s(u − α, β) dαdβ where s(u, v) = 0 u<0 1 u>0 is the unit step function. Since s(α, β) is independent of β, integration with respect to that variable yields a constant, which we drop.
FLC: θt = 45◦ : LFLC = − sin π4 cos π4 cos π4 sin π4 = 1 2 1−j 1+j 1+j 1−j 1 0 e sin π4 cos π4 cos π4 − sin π4 0 −j π 2 . • Single pass matrix: Lsingle = LFLC Lpolarizer = 1−j 1+j 1+j 1−j 0 0 0 1 = 0 0 1+j 1−j 0 1+j 0 1−j . For a double pass, we have Ldouble = R Ltsingle Lsingle = −1 0 0 1 0 1+j 0 1−j = 0 0 0 0 . Thus, we see that in the state where the molecular axis is oriented away from the polarizer direction, no light is transmitted by the cell. When the molecular long axis is parallel to the polarizer, we have θt = 0◦ so that: LFLC = Lsingle = 1 0 0 −j LFLC Lpolarizer = 1 0 0 −j 0 0 0 1 0 0 0 −j 0 0 = 0 0 0 −j .
The normally incident plane wave at L1 will be transformed into a spherical wave diverging from a point at distance f to the left of that lens, or equivalently at distance 2f to the left of lens L2 . That point source will be imaged at distance 2f to the right of lens L2 , so the Fourier plane will appear at distance 2f to the right of lens L2 . As for the location of the image of the object, we can replace the object by a point-source in the object plane. According to the lens law, the negative lens produces an image of that point source at distance zi1 satisfying 1 1 1 + =− zi1 2f f or 2 zi1 = − f.