Diffraction by a Single Slit: Locating the Minima — Question 1

Question 1

Pairing narrow strips proves every nonzero single-slit minimum

A uniformly illuminated slit has width a. At a trial direction $\theta_m$, assume $a\sin\theta_m=m\lambda$ for positive integer m. Partition the opening into $2m$ equal strips and pair neighboring strips point by point.

1. Find the path and phase difference within each pair and establish cancellation.

2. Explain why m=0 is excluded and count the strip pairs needed for the third minimum.

Question 1 - Solution

Pairing narrow strips proves every nonzero single-slit minimum

(a) Method and calculation

$$\Delta u=a/(2m)\Longrightarrow\Delta\ell=\Delta u\sin\theta_m=\lambda/2.$$$$\boxed{\Delta\phi=2\pi\Delta\ell/\lambda=\pi,\quad E_1+E_2=0}.$$

(b) Interpretation and further calculation

$$m=0:\quad\Delta\phi(u)=0\ \text{throughout the slit}\Longrightarrow\boxed{\text{central maximum}}.$$$$m=3:\quad2m=6\ \text{strips}\Longrightarrow\boxed{3\ \text{canceling pairs}}.$$

Physical check

Complete pairwise cancellation requires uniform coherent illumination; unequal strip weights would leave a residual field.