Diffraction by a Double Slit — Question 10

Question 10

Opposite illumination tilts can separate the two diffraction envelopes

Two equal slits have normalized field factors $f_1(\beta)=\operatorname{sinc}(\beta-b)$ and $f_2(\beta)=\operatorname{sinc}(\beta+b)$ because their incident wavefront tilts differ. Set $b=\pi/2$. A controllable relative phase is scanned; $V=2|f_1f_2|/(|f_1|^2+|f_2|^2)$ where the denominator is nonzero.

1. Calculate visibility at $\beta=0$ and the maximum normalized intensity there.

2. Calculate visibility and normalized intensity at $\beta=\pi/2$. Use unit intensity for a single field factor of magnitude one.

Question 10 - Solution

Opposite illumination tilts can separate the two diffraction envelopes

(a) Method and calculation

$$\beta=0:\quad f_1=f_2=2/\pi\Longrightarrow\boxed{V=1}.$$$$\boxed{I_{\max}(0)=(4/\pi)^2=16/\pi^2}.$$

(b) Interpretation and further calculation

$$\beta=\pi/2:\quad f_1=1,\ f_2=0\Longrightarrow\boxed{V=0}.$$$$\boxed{I(\pi/2)=1\quad\text{independent of relative phase}}.$$

Physical check

Equal slit widths do not guarantee equal amplitudes at every observation angle when the incident phase slopes differ.