AcademyDiffraction
Academy
Multiple-Slit Patterns
Level 1 - Physics topic page in Diffraction.
Principle
Many equally spaced slits produce sharp principal maxima. If each slit has finite width, the multiple-slit interference pattern is modulated by a single-slit diffraction envelope.
Notation
\(N\)
number of slits
1
\(d\)
slit spacing
\(\mathrm{m}\)
\(a\)
slit width
\(\mathrm{m}\)
\(\delta\)
phase difference between adjacent slits
\(\mathrm{rad}\)
\(\theta\)
observation angle
\(\mathrm{rad}\)
\(\lambda\)
wavelength
\(\mathrm{m}\)
Method
Derivation 1: Adjacent-slit phase difference
Adjacent slits have path difference \(d\sin\theta\), giving a phase difference \(\delta\).
Path difference
\[\Delta r=d\sin\theta\]
Phase difference
\[\delta=\frac{2\pi d\sin\theta}{\lambda}\]
Derivation 2: Principal maxima
Principal maxima occur when every slit is in phase with the next.
In-phase condition
\[\delta=2\pi m\]
Principal maxima
\[d\sin\theta=m\lambda\]
Derivation 3: Diffraction envelope
Finite slit width creates a single-slit envelope. A principal maximum can be missing if it lies at a single-slit minimum.
Envelope minima
\[a\sin\theta=p\lambda\]
Missing order condition
\[\frac{m}{d}=\frac{p}{a}\]
Rules
Principal maxima
\[d\sin\theta=m\lambda\]
Adjacent-slit phase
\[\delta=\frac{2\pi d\sin\theta}{\lambda}\]
Equal-slit principal intensity
\[I_{\max}\propto N^2\]
Single-slit envelope minima
\[a\sin\theta=p\lambda\]
Examples
Question
Five identical slits are exactly in phase at a principal maximum. How does the peak intensity compare with one slit alone?
Answer
Field amplitude is
\[5\]
times larger, so intensity is \[5^2=25\]
times larger.Checks
- More slits make principal maxima narrower and brighter.
- Slit spacing sets where the principal maxima occur.
- Slit width sets the diffraction envelope.
- Missing orders occur only when a grating maximum coincides with an envelope zero.