18.2.2 Applications of Diffraction Grating

A diffraction grating is a glass (or metal) plate ruled with a large number of close, parallel, equally spaced slits — typically thousands per centimetre. When light passes through it, each slit acts as a coherent secondary source, and the combined diffraction and interference produce sharp, bright spectral maxima at well-defined angles.

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The Grating Element

The grating element $d$ is the centre-to-centre distance between adjacent slits:

$d=\frac{L}{N}=\frac{1}{N^{\prime }}$

where $L$ is the total ruled length, $N$ is the total number of lines, and $N^{\prime }$ is the number of lines per unit length (e.g. lines per metre).

Example: A grating with 500 lines/mm has $d=\frac{1}{500\times 10^3{\text{ m}}^{-1}}=2\times 10^{-6}\text{ m}$

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The Grating Equation

For constructive interference (principal maxima), the path difference between adjacent slits must equal a whole number of wavelengths:

$\boxed{d\sin \theta =n\lambda }$

The zeroth order ($n=0$) is the straight-through beam. Higher orders appear symmetrically on either side.

Effect of increasing $N$ (number of slits)

Keeping $d$ constant but increasing the number of slits makes the principal maxima narrower and sharper (higher resolving power) while the dark regions between them become broader. The angular positions of the maxima are unchanged.

Maximum observable order

Since $\sin \theta \le 1$: $n_{\max }\le \frac{d}{\lambda }$ If this gives $\theta =90°$ exactly, that order lies along the grating surface and is not physically observable; the practical maximum is the largest integer $n$ for which $\theta <90°$.

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Determining Wavelength with a Diffraction Grating

The diffraction grating is one of the most precise instruments for measuring the wavelength of light:

1. Set up the grating normal to the incident beam.
2. Locate the $n$-th order maximum by rotating a spectrometer telescope until the bright fringe is centred.
3. Read the angle $\theta$ from the spectrometer scale.
4. Calculate the wavelength: $\lambda =\frac{d\sin \theta }{n}$

Because the maxima are very sharp (many slits), the angle can be measured with high precision, giving an accurate value of $\lambda$.

Dispersion of white light

For white light, each order (except $n=0$) is spread into a spectrum. Since $\sin \theta \propto \lambda$, red light (longest $\lambda \approx 700\text{ nm}$) is diffracted through the largest angle and violet (shortest $\lambda \approx 400\text{ nm}$) through the smallest.

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Everyday Applications

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Worked Example

Problem: Light of unknown wavelength falls normally on a grating with 600 lines/mm. The second-order maximum is observed at $\theta =45°$. Find $\lambda$.

Solution: $d=\frac{1}{600\times 10^3}=1.667\times 10^{-6}\text{ m}$ $\lambda =\frac{d\sin \theta }{n}=\frac{1.667\times 10^{-6}\times \sin 45°}{2}=\frac{1.667\times 10^{-6}\times 0.7071}{2}\approx 589\text{ nm}$

This corresponds to the yellow sodium doublet — a classic spectroscopy result.

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