First minimum of single-slit diffraction
For a single slit of width b, the first diffraction minimum lies at the angle given by sin θ = λ/b. The mth minimum is at sin θ = mλ/b. The result comes from splitting the wavefront in the slit into two halves so that every wavelet in the top half is cancelled by a partner b/2 below it with a path difference of λ/2.
sin θ = λ/b (first minimum) · sin θₘ = mλ/b, m = 1, 2, 3, …What this actually means
Notice this equation gives a minimum, while d sin θ = nλ for a grating gives a maximum. Same-looking algebra, opposite meaning. Label which one you are using every time.
The pairing argument is the derivation the syllabus expects if asked. Divide the slit into two halves of width b/2. Pair each point in the upper half with the point b/2 below it. If those pairs differ by λ/2 they cancel, and since every point is in some pair, the whole slit cancels. That gives (b/2) sin θ = λ/2, hence sin θ = λ/b.
Narrower slit, wider pattern. Since sin θ = λ/b, halving b doubles sin θ, so the central maximum spreads out. This inverse relationship is the point of nearly every single-slit exam question.
The secondary maxima between the minima are much dimmer, under 5% of the central peak for the first one, and they get narrower as well as fainter.
Small angles are usual for light, so sin θ ≈ tan θ = y/D can be used. For a very narrow slit, or for sound and microwaves, angles can be large and you must keep the sine.
Treating sin θ = λ/b as the position of a maximum, when it locates the first minimum.
Prove it — watch it be true
- Switch to the single-slit sinc² pattern and read off the angle of the first zero of intensity.
- Compare it with the predicted sin θ = λ/b value shown alongside.
- Halve the slit width b and confirm the first minimum moves out to roughly double the angle.