Diffraction by a Double Slit — Question 7

Question 7

Polarization mismatch fills interference minima but preserves envelope zeros

Equal finite slits have individual intensity $I_s\operatorname{sinc}^2\beta$. Their transmitted fields are linearly polarized at angle $\chi=60^\circ$ to each other and have phase difference $2\alpha$. A polarization-insensitive detector measures their sum.

1. Derive the intensity and local visibility as phase is scanned at a fixed nonzero envelope value.

2. Find the maximum and minimum levels in units of $I_s\operatorname{sinc}^2\beta$, and evaluate the intensity at an envelope zero.

Question 7 - Solution

Polarization mismatch fills interference minima but preserves envelope zeros

(a) Method and calculation

$$\boxed{I=2I_s\operatorname{sinc}^2\beta[1+\cos\chi\cos(2\alpha)]}.$$$$\boxed{V=|\cos\chi|=1/2}.$$

(b) Interpretation and further calculation

$$\boxed{I_{\max}=3I_s\operatorname{sinc}^2\beta,\quad I_{\min}=I_s\operatorname{sinc}^2\beta}.$$$$\operatorname{sinc}\beta=0\Longrightarrow\boxed{I=0\ \text{for every relative phase}}.$$

Physical check

Polarization overlap controls the cross term, whereas an individual-slit zero removes both fields before they can interfere.