Diffraction Grating Equation

Equivalent forms

\sin\theta = m \lambda / d

d (\sin\theta_i + \sin\theta_m) = m \lambda

Every spectrometer in chemistry, astronomy, and physics rests on this single integer relation.

Symbol	Name	SI unit	Dimension	Range
$d$	Slit spacing Distance between adjacent grating slits	m	L	1e-7 – 0.0001
$theta$	Diffraction angleoutput Angle of the m-th order maximum from the normal	rad	1	0 – 1.5708
$m$	Order number Integer order of the diffraction maximum	dimensionless	1	1 – 5
$lambda$	Wavelength Wavelength of the incident light	m	L	4e-7 – 7e-7

Unit systems

Symbol	SI	CGS	Imperial
$d$	m	cm	m
$lambda$	m	cm	m
$theta$	rad	rad	deg

Where it holds

Valid for plane-wave illumination at normal incidence. For oblique incidence or blazed gratings, the general form

d(\sin \theta _i + \sin \theta _m) = m*\lambda

applies.

Discovery

Joseph von Fraunhofer · 1821

Fraunhofer built the first practical diffraction gratings using fine wires, then ruled gratings on glass. He used them to measure solar absorption lines (Fraunhofer lines), founding spectroscopy.

Try this

Why does a CD show a rainbow when held under light?

A grating with 600 lines/mm is illuminated by light of wavelength 650 nm. Find the angle of the first-order maximum.

Research status: stable

Real-world applications

Spectroscopy: chemical identification by atomic emission/absorption lines.
CD/DVD storage: pit spacing acts as a reflective grating, producing the rainbow.
Wavelength division multiplexing in fiber-optic communication.
Astronomical spectrographs for stellar composition and redshift measurement.

Common misconceptions

Gratings split white light into rainbows like prisms — gratings disperse the opposite way (red bends more than blue).
Higher N means new orders appear — N controls peak sharpness, not which orders exist.
Reflection gratings need a different formula — same condition applies with sign convention.

Experimental verification

Verified by Fraunhofer with mechanical gratings; modern verification uses laser beams through commercial gratings with angles measured to arcsecond precision.

Derivation

Each of N slits at spacing d acts as a coherent source.

Path difference to a far-field point at angle theta is d*sin(theta) between adjacent slits.

Constructive interference requires path difference

= m*\lambda

, giving

d*\sin (\theta ) = m*\lambda

.

Sharpness of maxima scales as 1/N.

Limiting cases

\lambda \to 0

⟶

\theta \to 0

X-rays barely diffract through optical gratings.

m*lambda > d⟶ No solutionOrder does not exist — sin(theta) cannot exceed 1.

m = 0

⟶

\theta = 0

Zero-order beam passes straight through, regardless of wavelength.

What if…

What if the wavelength were doubled?

sin(theta) doubles. For first-order red at $22.95^{\circ}$ , doubling lambda may push $m=1$ past $90^{\circ}$ , eliminating that order entirely.

What if the grating is illuminated at oblique angle theta_i?

Generalized condition: $d(\sin \theta _i + \sin \theta _m) = m*\lambda$ . Blazed gratings exploit this to maximize a chosen order.

1

First-order diffraction angle for red light

Given ·

d:: 0.000001667
m:: 1
lambda:: 6.5e-7

Find · theta

Steps

Compute slit spacing: $d = 1/600000\,\mathrm{m} = 1.667e-6\,\mathrm{m}$
Apply grating equation: $\sin (\theta ) = m*\lambda /d$
Substitute: $\sin (\theta ) = 1 * 6.5e-7 / 1.667e-6 = 0.39$
$\theta = \arcsin (0.39) = 0.4006\,\mathrm{rad} \approx 22.95^{\circ}$

Result ·

\sin (\theta ) = m*\lambda /d = 1 * 6.5e-7 / 1.667e-6 = 0.39

, so

\theta = \arcsin (0.39) = 22.95^{\circ} = 0.4006\,\mathrm{rad}

Young's Double Slit Interference→Single Slit Diffraction→Rayleigh Criterion→