By Joseph W. Goodman
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Additional info for Introduction To Fourier Optics, Third Edition, Problem Solutions
6-13. Recall that fo = w w = , λzi λf where the last step holds because the object is at infinite distance from the lens, and the lens law implies that zi = f . The F-number of the lens is F# = f . 2w Solving for f in the equation above, and substituting that expression in the first equation yields fo = 1 . 2λF # 6-14. Let s(u) = |h(u, 0)|2 . Then the Sparrow resolution distance (in the image space) will be the δ that satisfies the equation δ δ d2 +s u+ s u− = 0. du2 2 2 u=0 2 d (a) By the symmetry of s(u), du 2 s(u) is also symmetric in u, as proved by the following argument.
6-41) and repeated below, H(fX , fY ) = × fY fX Λ 2fo 2fo 8Wm fX sinc λ 2fo Λ 1− |fX | 2fo 8Wm λ sinc fY 2fo 1− |fY | 2fo , with the product of the diffraction-limited OTF, given by H(fX , fY ) = Λ fX 2fo Λ fX 2fo , and the geometrical-optics OTF (from Eq. (6-42)) fX 2fo 8Wm λ H(fX , fY ) = sinc sinc fY 2fo 8Wm λ . It is clear that the first equation is not the product of the second and third equations, due to the presence of the terms 1 − |f2fXo| and 1 − |f2fYo| in the arguments of the sinc functions.
51 6-8. If the point-spread function is to be the convolution of the diffraction-limited spread function with the geometrical-optics spread function, the OTF must be the product of the two corresponding OTFs. We focus on the OTFs from this point on. We wish to compare the OTF of a misfocused system, given by Eq. (6-41) and repeated below, H(fX , fY ) = × fY fX Λ 2fo 2fo 8Wm fX sinc λ 2fo Λ 1− |fX | 2fo 8Wm λ sinc fY 2fo 1− |fY | 2fo , with the product of the diffraction-limited OTF, given by H(fX , fY ) = Λ fX 2fo Λ fX 2fo , and the geometrical-optics OTF (from Eq.
Introduction To Fourier Optics, Third Edition, Problem Solutions by Joseph W. Goodman