In one arm of a Mach-Zehnder interferometer a quarter-wave plate turns the linearly…

PHYSICS · Multi-part · Fermi University Physics · Optics & Photonics · Polarisation and anisotropic media · Interference and diffraction

Optics and Photonics, Ch Two-beam interference and interferometry

Problem

In one arm of a Mach-Zehnder interferometer a quarter-wave plate turns the linearly xx-polarised input beam into right-circularly polarised light; in the other arm an identical plate, rotated by 90 degrees, turns the same input into left-circularly polarised light. Under CV3 the Jones vectors reaching the output are E1=E02(1,−i)\mathbf{E}_1 = \dfrac{E_0}{\sqrt2}(1,-i) and E2=E02eiδ(1,i)\mathbf{E}_2 = \dfrac{E_0}{\sqrt2}e^{i\delta}(1,i), both of real amplitude E0E_0, where δ\delta is the extra phase set by a mirror position in one arm. The two arms recombine at the second beam splitter. ((a)) Show that the recombined irradiance I∝∣E1+E2∣2I \propto |\mathbf{E}_1+\mathbf{E}_2|^2 does not depend on δ\delta: the interferometer shows no fringes as the mirror is moved. ((b)) A linear polariser with transmission axis at angle θ\theta to xx is then placed after the recombination point, projecting each beam's field onto that axis before they add. Show that the transmitted irradiance is I(θ,δ)=E02[1+cos⁡(δ+2θ)]I(\theta,\delta) = E_0^2\big[1+\cos(\delta+2\theta)\big] (the constant 12cϵ0n\tfrac12c\epsilon_0n dropped), so fringes reappear with full visibility as δ\delta is scanned, shifted by an amount that depends on θ\theta.

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