Diffraction
Diffraction appears when a wave meets an opening or obstacle of finite size. The wave does not simply continue as a perfectly sharp ray: contributions from different parts of the aperture superpose and produce angular spreading and an intensity pattern.
When diffraction becomes visible
If an aperture is many wavelengths wide, the central beam is narrow and ray optics works well. As the ratio λ/a grows, diffraction spreads the wave more strongly.
Single-slit minima
a sin θ = mλ, m = 1, 2, 3, ...
Here a is slit width. These angles locate the dark minima, not the secondary maxima. The central maximum lies between the first minima on either side.
First-minimum angle
For a single slit, a sin θ = λ for the first minimum.
Why the central maximum is special
The central maximum is the brightest and, measured between its first minima, about twice as wide as the neighboring maxima in the ideal single-slit pattern. Side maxima become progressively weaker.
The full intensity curve follows a sinc-squared shape; for most introductory problems, the minima relation already captures the most useful geometry.
Diffraction is not the same as two-source interference
Both phenomena come from superposition. In double-slit interference, the dominant phase comparison is between two separated coherent openings. In single-slit diffraction, contributions from different positions across one finite aperture interfere with one another.
Real double slits have finite width, so an interference fringe pattern can be modulated by a broader diffraction envelope.
How aperture size controls the pattern
The key dimensionless comparison is λ/a. Increasing wavelength or decreasing slit width makes the pattern spread farther from the forward direction. In the small-angle limit, the first-minimum position on a screen is y₁ ≈ Lλ/a, so the full central width is approximately 2Lλ/a.
Do not call the side peaks “interference from two edges” as a complete explanation. Every point across the aperture contributes a wavelet; the dark directions arise from cancellation across the whole slit. Pairing points is a useful derivation device, not a claim that only two locations matter.
Worked examples
1. First minimum
Light of wavelength 500 nm passes through a 5.0 μm slit. Find the first-minimum angle.
Solution
sin θ₁ = λ/a = 0.100, so θ₁ ≈ 5.74°.
2. Effect of narrowing a slit
If the slit width is halved while wavelength stays fixed, what happens to the central maximum?
Solution
The first minima satisfy sin θ₁ = λ/a. Halving a roughly doubles the small-angle θ₁, so the central maximum becomes wider.
3. Central width on a distant screen
For small angles, λ = 600 nm, a = 0.20 mm and screen distance L = 2.0 m. Estimate the central maximum width.
Solution
First-minimum position y₁ ≈ Lλ/a = 2.0(600 × 10−9)/(2.0 × 10−4) = 0.0060 m. The central width is 2y₁ = 12 mm.