Coherence — Question 7

Question 7

Orthogonal polarization hides interference until a common projection is selected

Two mutually coherent beams of equal intensity $I_0$ have perpendicular linear polarizations. A polarization-insensitive detector receives both. An ideal linear analyzer is then inserted at $45^\circ$ to each input polarization; the relative optical phase is $\delta$.

1. Calculate the intensity and visibility before the analyzer.

2. Calculate the transmitted intensity as a function of phase, its extrema, and its visibility.

Question 7 - Solution

Orthogonal polarization hides interference until a common projection is selected

(a) Method and calculation

$$I=I_1+I_2+2\sqrt{I_1I_2}(\hat{\mathbf e}_1\cdot\hat{\mathbf e}_2)\cos\delta.$$$$\hat{\mathbf e}_1\cdot\hat{\mathbf e}_2=0\Longrightarrow\boxed{I=2I_0,\quad V=0}.$$

(b) Interpretation and further calculation

$$I_1^\prime=I_2^\prime=I_0\cos^245^\circ=I_0/2\Longrightarrow\boxed{I^\prime=I_0(1+\cos\delta)}.$$$$\boxed{I^\prime_{\max}=2I_0,\quad I^\prime_{\min}=0,\quad V^\prime=1}.$$

Physical check

The analyzer makes the two transmitted fields share a polarization. It does not manufacture a stable phase relationship between independent sources.