Diffraction by a Double Slit — Question 2

Question 2

Envelope width and fringe spacing independently determine the aperture geometry

A paraxial double-slit pattern at wavelength $600\,\mathrm{nm}$ is recorded $L=1.00\,\mathrm m$ away. The central envelope extends $24.0\,\mathrm{mm}$ between its first zeros; nominal interference maxima are separated by $4.80\,\mathrm{mm}$.

1. Infer the width a of each slit and their center spacing d.

2. Count the nominal interference maxima strictly inside the central envelope and find the first positive missing order.

Question 2 - Solution

Envelope width and fringe spacing independently determine the aperture geometry

(a) Method and calculation

$$W=2\lambda L/a\Longrightarrow\boxed{a=50.0\,\mu\mathrm m};\quad p=\lambda L/d\Longrightarrow\boxed{d=125\,\mu\mathrm m}.$$$$\boxed{d/a=2.50};\quad |m|<d/a\Longrightarrow\boxed{m=-2,-1,0,1,2}.$$

(b) Interpretation and further calculation

$$\text{Missing:}\quad m\lambda/d=p_e\lambda/a\Longrightarrow m=(5/2)p_e.$$$$\boxed{m=5\ \text{first missing, at envelope order }p_e=2}.$$

Physical check

Nominal interference orders are not exactly the local maxima of the envelope-weighted pattern; they remain useful geometric markers.