Diffraction gratings

Key ideas

A diffraction grating is thousands of slits per centimetre. The same condition as two slits gives the bright directions:

$d\sin\theta = n\lambda$

With many slits, a direction is only bright if every slit's wave arrives in phase — the slightest angle error leaves thousands of waves to cancel. So a grating produces:
- maxima at the same angles as a double slit with the same $d$
- but far sharper and brighter lines, with darkness in between.
Find the slit spacing from the ruling: a grating with $N$ lines per metre has

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

e.g. $500$ lines per mm $= 5.00\times10^{5}$ lines per m, so $d = 2.00\times10^{-6}$ m.
Maximum order: $\sin\theta$ cannot exceed 1, so $n \le \dfrac{d}{\lambda}$ — round down to the nearest whole number.
White light through a grating splits into spectra: each order fans out with violet (shortest $\lambda$ ) closest to the centre and red furthest, because larger $\lambda$ needs a larger angle.

Worked ExampleAngles and orders from a grating

Light of wavelength $600$ nm falls on a grating with $500$ lines per mm. Find the angle of the first-order maximum, and the highest order visible.

Step 1 — Slit spacing from the ruling

$d = \frac{1}{5.00\times10^{5}} = 2.00\times10^{-6} \text{ m}$

Step 2 — First-order angle from $d\sin\theta = n\lambda$

$\sin\theta = \frac{(1)(6.00\times10^{-7})}{2.00\times10^{-6}} = 0.300 \Rightarrow \theta = 17.5°$

Step 3 — Highest order ( $\sin\theta \le 1$ )

$n \le \frac{d}{\lambda} = \frac{2.00\times10^{-6}}{6.00\times10^{-7}} = 3.33 \Rightarrow n_{max} = 3$

Tips

"Lines per mm" needs two conversions: to lines per metre ( $\times 10^3$ ), then invert for $d$ . Write both steps down; doing it in your head breeds errors of $10^3$ .
If asked why grating lines are sharper than double-slit fringes, the answer is the many-slit cancellation: away from the exact angle, the thousands of waves are spread evenly around the phase circle and sum to zero.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A grating has $250$ lines per mm.

Show that the slit spacing is $4.0\times10^{-6}$ m.

Merit

Light of wavelength $450$ nm passes through a grating of spacing $d = 1.8\times10^{-6}$ m.

Calculate the angle of the second-order maximum.

Excellence

White light passes through a diffraction grating.

Explain why each order becomes a spectrum, why violet appears closer to the centre than red, and why the central maximum stays white.

Diffraction gratings

Covers AS91523

Key ideas

A diffraction grating is thousands of slits per centimetre. The same condition as two slits gives the bright directions:

$d\sin\theta = n\lambda$

With many slits, a direction is only bright if every slit's wave arrives in phase — the slightest angle error leaves thousands of waves to cancel. So a grating produces:
- maxima at the same angles as a double slit with the same $d$
- but far sharper and brighter lines, with darkness in between.
Find the slit spacing from the ruling: a grating with $N$ lines per metre has

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

e.g. $500$ lines per mm $= 5.00\times10^{5}$ lines per m, so $d = 2.00\times10^{-6}$ m.
Maximum order: $\sin\theta$ cannot exceed 1, so $n \le \dfrac{d}{\lambda}$ — round down to the nearest whole number.
White light through a grating splits into spectra: each order fans out with violet (shortest $\lambda$ ) closest to the centre and red furthest, because larger $\lambda$ needs a larger angle.

Worked ExampleAngles and orders from a grating

Light of wavelength $600$ nm falls on a grating with $500$ lines per mm. Find the angle of the first-order maximum, and the highest order visible.

Step 1 — Slit spacing from the ruling

$d = \frac{1}{5.00\times10^{5}} = 2.00\times10^{-6} \text{ m}$

Step 2 — First-order angle from $d\sin\theta = n\lambda$

$\sin\theta = \frac{(1)(6.00\times10^{-7})}{2.00\times10^{-6}} = 0.300 \Rightarrow \theta = 17.5°$

Step 3 — Highest order ( $\sin\theta \le 1$ )

$n \le \frac{d}{\lambda} = \frac{2.00\times10^{-6}}{6.00\times10^{-7}} = 3.33 \Rightarrow n_{max} = 3$

Tips

"Lines per mm" needs two conversions: to lines per metre ( $\times 10^3$ ), then invert for $d$ . Write both steps down; doing it in your head breeds errors of $10^3$ .
If asked why grating lines are sharper than double-slit fringes, the answer is the many-slit cancellation: away from the exact angle, the thousands of waves are spread evenly around the phase circle and sum to zero.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A grating has $250$ lines per mm.

Show that the slit spacing is $4.0\times10^{-6}$ m.

Merit

Light of wavelength $450$ nm passes through a grating of spacing $d = 1.8\times10^{-6}$ m.

Calculate the angle of the second-order maximum.

Excellence

White light passes through a diffraction grating.

Explain why each order becomes a spectrum, why violet appears closer to the centre than red, and why the central maximum stays white.

Diffraction gratings

Key ideas

Step 1 — Slit spacing from the ruling

Step 2 — First-order angle from dsin⁡θ=nλd\sin\theta = n\lambdadsinθ=nλ

Step 3 — Highest order (sin⁡θ≤1\sin\theta \le 1sinθ≤1)

Tips

Test yourself

Practice by grade

Diffraction gratings

Key ideas

Step 1 — Slit spacing from the ruling

Step 2 — First-order angle from dsin⁡θ=nλd\sin\theta = n\lambdadsinθ=nλ

Step 3 — Highest order (sin⁡θ≤1\sin\theta \le 1sinθ≤1)

Tips

Test yourself

Practice by grade

Step 2 — First-order angle from $d\sin\theta = n\lambda$

Step 3 — Highest order ( $\sin\theta \le 1$ )

Step 2 — First-order angle from $d\sin\theta = n\lambda$

Step 3 — Highest order ( $\sin\theta \le 1$ )