AcademyDiffraction
Academy
Diffraction Gratings
Level 1 - Physics topic page in Diffraction.
Principle
A diffraction grating has many equally spaced lines or slits. It produces sharp angular maxima that separate wavelengths, making gratings useful for spectroscopy.
Notation
\(d\)
grating spacing
\(\mathrm{m}\)
\(n_L\)
line density
\(\mathrm{m^{-1}}\)
\(m\)
diffraction order
1
\(\lambda\)
wavelength
\(\mathrm{m}\)
\(\theta_m\)
angle of order m
\(\mathrm{rad}\)
\(N\)
number of illuminated slits
1
\(R\)
resolving power
1
Method
Derivation 1: Convert line density to spacing
Line density is the reciprocal of grating spacing.
Spacing
\[d=\frac{1}{n_L}\]
Lines per millimetre
\[n_L=(\mathrm{lines\,mm^{-1}})\times10^3\]
Derivation 2: Find bright orders
Constructive interference from adjacent grating lines requires a whole-wavelength path difference.
Grating equation
\[d\sin\theta_m=m\lambda\]
Allowed orders
\[|m|\lambda\le d\]
Derivation 3: Resolve close wavelengths
A grating resolves two nearby wavelengths when their maxima are separated enough compared with peak width.
Resolving power
\[R=\frac{\lambda}{\Delta\lambda}=mN\]
Smallest resolvable difference
\[\Delta\lambda=\frac{\lambda}{mN}\]
Rules
Grating equation
\[d\sin\theta_m=m\lambda\]
Spacing from line density
\[d=\frac{1}{n_L}\]
Maximum possible order
\[m_{\max}=\left\lfloor\frac{d}{\lambda}\right\rfloor\]
Resolving power
\[R=mN\]
Examples
Question
A grating has
\[600\,\mathrm{lines\,mm^{-1}}\]
Find the spacing.Answer
\[d=\frac{1}{600\times10^3}=1.67\times10^{-6}\,\mathrm{m}\]
Checks
- Higher line density means smaller \(d\), so orders spread farther apart.
- The zero order is undeviated for normal incidence.
- A proposed order is impossible if \(|m|\lambda>d\).
- Resolving power improves with order and with the number of illuminated lines.