The diffraction grating equation

8 minAS Level8.4.1

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can use $d\sin\theta = n\lambda$ , find the slit spacing from the grating's lines per millimetre, and work out the highest order that can be seen.

Key to the symbols

The main symbols used in this lesson:

Symbol	Meaning
$\lambda$	wavelength

Thousands of slits, not two

A diffraction grating is a transparent plate ruled with thousands of equally spaced parallel slits, typically several hundred per millimetre.

With two slits the fringes are broad and rather fuzzy. With thousands, the maxima become narrow, sharp bright lines on a dark background, because only at very precise angles do the waves from every slit arrive in phase. At every other angle they cancel.

That sharpness is why a grating measures wavelength so accurately.

Definition

Slit spacing

If a grating has $N$ lines per metre, the spacing between neighbouring slits is

$d = \frac{1}{N}.$

For a grating quoted in lines per millimetre, $d$ in metres is $\dfrac{1}{N \times 10^{3}}$ . For example, $500$ lines per millimetre gives $d = \dfrac{1}{500 \times 10^{3}} = 2.0 \times 10^{-6}\ \text{m}$ .

Where the equation comes from

Neighbouring slits are a distance $d$ apart. Rays leaving them at an angle $\theta$ to the straight-through direction differ in path length by $d\sin\theta$ .

For a maximum, every slit must be in step with its neighbour, so that path difference must be a whole number of wavelengths:

$d\sin\theta = n\lambda, \qquad n = 0, 1, 2, \ldots$

$n = 0$ is the central maximum, straight through at $\theta = 0$ .
$n = 1$ gives the first order, $n = 2$ the second order, and so on.

The angle $\theta$ is always measured from the straight-through direction, not from the grating surface.

Worked example: Angles for a standard grating

A grating has $500$ lines per millimetre. Light of wavelength $6.0 \times 10^{-7}\ \text{m}$ is incident normally.

Find (a) the slit spacing and (b) the angle of the first-order maximum.

Have a go first?

How many orders can you see?

Since $\sin\theta$ cannot exceed 1, the equation requires

$n \le \frac{d}{\lambda}.$

For the grating above, $\dfrac{d}{\lambda} = \dfrac{2.0 \times 10^{-6}}{5.0 \times 10^{-7}} = 4$ , so $n$ can be 0, 1, 2, 3 — but not 4.

At $n = 4$ the equation would need $\sin\theta = 1$ , which means $\theta = 90^\circ$ : the light would have to travel along the surface of the grating, where no maximum is observed.

So the highest visible order is 3, and with $n = 3$ the angle is $\sin^{-1}(0.9) \approx 64^\circ$ .

What a grating shows with white light

With monochromatic light every order is a sharp line of the same colour.

With white light, the central maximum ( $n = 0$ ) is white, because every wavelength has zero path difference there. But the other orders disperse: longer wavelengths diffract through larger angles, so each order becomes a spectrum with red on the outside and violet on the inside.

That is exactly how a grating separates light into colours — and why it can be used to measure wavelength.

Common mistakes

d is the width of the grating.

$d$ is the spacing between neighbouring slits, not the size of the whole grating.

Use d = N when the grating is quoted in lines per millimetre.

$d = 1/N$ , and the units must be converted. 500 lines per mm gives $d = 2.0 \times 10^{-6}\ \text{m}$ , not 500 or $5.0 \times 10^{5}$ .

In the exam

Write $d\sin\theta = n\lambda$ first, then substitute. Show the conversion from lines per mm to $d$ in metres explicitly; that is where most marks are lost.

For the highest order, set $\sin\theta = 1$ and take the largest integer strictly below $d/\lambda$ if $d/\lambda$ is itself a whole number.

Quote angles to a sensible precision (usually 3 significant figures) and remember that $\theta$ is measured from the centre of the pattern.

Check yourself

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