An otherwise clear

PHYSICS · Multi-part · Fermi University Physics · Optics & Photonics · Fourier optics and image formation · Fourier optics and image formation

What an Instrument Can See, Ch The transfer function of an imaging system

Problem

An otherwise clear, aberration-free exit pupil, P0(xp,yp)P_0(x_p,y_p), produces an amplitude point-spread function h0(x,y)=F{P0}h_0(x,y) = \mathscr{F}\{P_0\} evaluated at νx=x/λzi\nu_x = x/\lambda z_i, νy=y/λzi\nu_y = y/\lambda z_i. Now suppose the same pupil instead carries a pure wavefront tilt, W(xp,yp)=αxpW(x_p,y_p) = \alpha x_p, with α\alpha a dimensionless constant and no other aberration, so the pupil function becomes P(xp,yp)=P0(xp,yp) eikαxpP(x_p,y_p) = P_0(x_p,y_p)\,e^{ik\alpha x_p}. ((a)) Using the Fourier shift theorem, show that the resulting point-spread function h(x,y)h(x,y) is exactly h0h_0 displaced along xx by some amount Δx\Delta x, with the same shape and width as h0h_0 itself, and find Δx\Delta x in terms of α\alpha and the image distance ziz_i. ((b)) An imaging system with zi=50 mmz_i = 50\ \mathrm{mm} carries a tilt of α=2.0×10−4\alpha = 2.0\times10^{-4}. Compute Δx\Delta x, in micrometres.

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