Intensity in Double-Slit Interference — Question 7

Question 7

The slit envelope suppresses selected nominal interference orders

Two identical slits of width a and center spacing $d=4a$ have far-field intensity $I=4I_s(\sin\beta/\beta)^2\cos^2\alpha$, with $\beta=\pi a\sin\theta/\lambda$ and $\alpha=\pi d\sin\theta/\lambda$. Here $I_s$ is one slit’s on-axis intensity.

1. Find which nominal interference orders $d\sin\theta=m\lambda$ coincide with zeros of the slit envelope.

2. Count the nominal bright orders strictly inside the central envelope and evaluate the intensity at nominal order $m=2$.

Question 7 - Solution

The slit envelope suppresses selected nominal interference orders

(a) Method and calculation

$$\beta_m=\pi am/d=m\pi/4;\quad\sin\beta_m=0\ \text{for}\ m=4q,\ q\ne0.$$$$\boxed{m=\pm4,\pm8,\ldots\ \text{are missing nominal orders}}.$$

(b) Interpretation and further calculation

$$|a\sin\theta|<\lambda\Longrightarrow|m|<4\Longrightarrow\boxed{m=-3,\ldots,3:\ 7\ \text{nominal bright orders}}.$$$$m=2:\quad\beta=\pi/2,\ \cos^2\alpha=1\Longrightarrow\boxed{I=16I_s/\pi^2=1.62114I_s}.$$

Physical check

The envelope shifts the true local maxima slightly away from nominal interference orders; the requested intensity is evaluated at the nominal direction.