Diffraction by a Double Slit — Question 1

Question 1

Two displaced finite slits multiply an envelope by an interference factor

Two equal uniform slits of width a have centers at $u=\pm d/2$. One centered slit has amplitude $F_s(q)=A_0a\,\operatorname{sinc}(qa/2)$, with $\operatorname{sinc}x=\sin x/x$. Let $\beta=qa/2$ and $\alpha=qd/2$.

1. Use the translation phase factors to derive the total amplitude.

2. Derive the intensity in units of one slit’s axial intensity $I_s$, and identify the two independent zero conditions.

Question 1 - Solution

Two displaced finite slits multiply an envelope by an interference factor

(a) Method and calculation

$$F(q)=F_s(q)e^{iqd/2}+F_s(q)e^{-iqd/2}=\boxed{2A_0a\,\operatorname{sinc}\beta\cos\alpha}.$$$$\boxed{I=4I_s\operatorname{sinc}^2\beta\cos^2\alpha}.$$

(b) Interpretation and further calculation

$$\boxed{\beta=p\pi,\ p\ne0:\ \text{envelope zeros}}.$$$$\boxed{\alpha=(m+1/2)\pi:\ \text{interference zeros}}.$$

Physical check

A dark direction may be caused by either factor; attributing every minimum to slit separation can give an incorrect aperture estimate.