Diffraction by a Double Slit — Question 5

Question 5

Unequal slit widths produce angle-dependent visibility

Two uniformly illuminated slits have widths a and 2a, with the same incident field amplitude $A_0$. At fixed q their individual amplitudes are $F_1=A_0a\operatorname{sinc}\beta$ and $F_2=2A_0a\operatorname{sinc}(2\beta)$, where $\beta=qa/2$. A controlled phase is scanned between them.

1. Find the on-axis visibility.

2. At $\beta=\pi/2$, find the visibility and intensity relative to the in-phase on-axis intensity of both slits.

Question 5 - Solution

Unequal slit widths produce angle-dependent visibility

(a) Method and calculation

$$q=0:\quad |F_1|:|F_2|=1:2\Longrightarrow\boxed{V_0=\frac{2(1)(2)}{1^2+2^2}=0.80}.$$$$I_{c,\mathrm{both}}\propto|A_0a+2A_0a|^2=9A_0^2a^2.$$

(b) Interpretation and further calculation

$$\beta=\pi/2:\quad F_1=2A_0a/\pi,\quad F_2=0\Longrightarrow\boxed{V=0}.$$$$\boxed{I/I_{c,\mathrm{both}}=4/(9\pi^2)=0.0450316}.$$

Physical check

At the wider slit’s first zero, only the narrower slit contributes, so changing relative phase cannot modulate the intensity.