Diffraction by a Single Slit: Locating the Minima — Question 4

Question 4

Finite angular acceptance truncates the set of observable minima

A slit has width $a=3.20\,\mu\mathrm m$ and wavelength $\lambda=500\,\mathrm{nm}$. A detector collects directions satisfying $|\sin\theta|\leq0.60$. Normal uniform illumination is assumed.

1. List all diffraction minimum orders inside the detector acceptance.

2. Find the smallest acceptance half-angle needed to include the fourth positive minimum.

Question 4 - Solution

Finite angular acceptance truncates the set of observable minima

(a) Method and calculation

$$|m|\lambda/a\leq0.60\Longrightarrow|m|\leq3.84,\quad m\ne0.$$$$\boxed{m=\pm1,\pm2,\pm3:\ 6\ \text{minima}}.$$

(b) Interpretation and further calculation

$$\sin\theta_4=4\lambda/a=0.625.$$$$\boxed{\theta_{\mathrm{accept,min}}=\sin^{-1}(0.625)=38.6822^\circ}.$$

Physical check

An order can be physically propagating while still lying outside the collection aperture.