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Diffraction gratings and spectra: WACE Year 12 Physics Unit 4

Syllabus dot point

“Apply the diffraction grating equation to analyse the dispersion of light into spectra”

8 min read

Quick answer

A diffraction grating is many thousands of equally spaced slits. Light interferes constructively only where dsin⁡θ=mλd\sin\theta=m\lambda, the same condition as two slits, but with thousands of slits the bright maxima become very narrow and sharp. Because the angle depends on wavelength, a grating spreads white light into a spectrum and separates spectral lines far better than a double slit or prism.

What this dot point is asking

WACE wants you to use the grating equation, understand why many slits sharpen the maxima, and explain how a grating produces a spectrum. The grating is the precision instrument of wave optics.

The grating equation

For a grating with slit spacing dd (the distance between adjacent slits), bright maxima occur where

dsin⁡θ=mλ,m=0,1,2,…d\sin\theta=m\lambda,\qquad m=0,1,2,\dots

Here mm is the order of the maximum. The slit spacing is usually given as lines per millimetre NN, so d=1/Nd=1/N in millimetres, which you convert to metres. The zeroth order (m=0m=0) is straight through; higher orders appear at larger angles.

Why many slits give sharp lines

With only two slits the bright fringes are broad and gradually fade. With thousands of slits, light from all of them must be in phase to give a maximum, which happens only at very precise angles. A tiny deviation from the exact angle causes waves from different parts of the grating to cancel, so the maxima become extremely narrow and bright. This sharpness is what makes gratings able to resolve closely spaced wavelengths.

Dispersing light into a spectrum

Because the diffraction angle depends on wavelength, each colour in white light emerges at a slightly different angle in each order. The result is a spectrum spread out by wavelength, with violet (short wavelength) deviated least and red (long wavelength) most. Unlike a prism, a grating spreads the colours in a predictable, calculable way, which is why gratings are used in spectrometers to measure the wavelengths emitted by light sources.

Limits on the orders

Since sin⁡θ\sin\theta cannot exceed 11, only a finite number of orders exist for a given grating and wavelength. The highest visible order is the largest integer mm for which mλ/d≤1m\lambda/d\le 1. Long wavelengths and finely ruled gratings (small dd) produce fewer orders.

Diffraction angle for a spectral line

A grating dispersing green light

A grating has 600600 lines per millimetre. Green light of wavelength 546 nm546\ \text{nm} shines on it. Find the angle of the first-order maximum.

Step 1: Calculate the slit spacing.

The grating specification gives the number of lines per millimetre, not the spacing. The slit spacing dd is the reciprocal of the line density, converted to metres so the grating equation gives an angle directly:

d=1600 mm−1=1.667×10−3 mm=1.667×10−6 m.d=\frac{1}{600\ \text{mm}^{-1}}=1.667\times10^{-3}\ \text{mm}=1.667\times10^{-6}\ \text{m}.

Step 2: Apply the grating equation for the first-order maximum.

The grating equation dsin⁡θ=mλd\sin\theta = m\lambda links the slit spacing, the wavelength and the diffraction angle. For the first-order maximum, m=1m=1, so rearranging gives sin⁡θ=mλ/d\sin\theta = m\lambda/d:

sin⁡θ=mλd=(1)(546×10−9)1.667×10−6=0.3276.\sin\theta=\frac{m\lambda}{d}=\frac{(1)(546\times10^{-9})}{1.667\times10^{-6}}=0.3276.

Step 3: Solve for the diffraction angle.

Taking the inverse sine of the result from Step 2 gives the angle at which the first-order bright maximum appears on the screen:

θ=sin⁡−1(0.3276)=19.1∘.\theta=\sin^{-1}(0.3276)=19.1^\circ.

Finding an unknown wavelength

The grating is most often used in reverse: measure the angle of a known order and solve for the wavelength. Rearranging the grating equation gives λ=dsin⁡θm\lambda=\dfrac{d\sin\theta}{m}, so a measured first-order angle and a known slit spacing give the wavelength directly. Using a higher order improves precision, because the same wavelength is spread to a larger angle where an angular measurement error matters less. This is exactly how a spectrometer determines the wavelengths of the bright lines emitted by an excited gas, linking this topic to atomic spectra: the grating disperses the light, the angles are measured, and each line's wavelength (and hence the energy-level transition that produced it) is calculated.

Grating versus prism

Both a grating and a prism disperse white light into a spectrum, but they do so by different physics and in opposite orders. A prism relies on refraction, where the refractive index varies with wavelength, so violet (short wavelength) is bent most and red least. A grating relies on interference through dsin⁡θ=mλd\sin\theta=m\lambda, so the long-wavelength red light is diffracted to the largest angle and violet to the smallest, the reverse of a prism. The grating also produces several orders and gives a wavelength that can be calculated exactly from the geometry, whereas a prism's dispersion must be calibrated. These differences are a frequent comparison question.

Working with line densities

Convert lines per millimetre to a spacing dd in metres before substituting, and remember mm must be a whole number. If a calculation gives sin⁡θ>1\sin\theta>1 for some order, that order does not exist for that wavelength.

Confusing lines per millimetre with the slit spacing

Students often substitute the number of lines per millimetre directly as dd. The slit spacing is the reciprocal: d=1/Nd=1/N. For 600600 lines per millimetre, d=1/600 mm≈1.67×10−6 md=1/600\ \text{mm}\approx1.67\times10^{-6}\ \text{m}, not 600600. Getting this inverse wrong throws every angle off completely.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style7 marks
A diffraction grating with 500500 lines per millimetre is illuminated normally by light of wavelength 589 nm589\ \text{nm}. (a) Calculate the slit spacing. (b) Calculate the angle of the second-order maximum. (c) Determine the highest order that can be observed for this wavelength.
Show worked answer →

A 7 mark calculation rewards the spacing, a grating-equation angle and the maximum order.

(a) Slit spacing
d=1500 mm−1=2.0×10−3 mm=2.0×10−6 md=\dfrac{1}{500\ \text{mm}^{-1}}=2.0\times10^{-3}\ \text{mm}=2.0\times10^{-6}\ \text{m}.
(b) Second-order angle
sin⁡θ=mλd=(2)(589×10−9)2.0×10−6=0.589\sin\theta=\dfrac{m\lambda}{d}=\dfrac{(2)(589\times10^{-9})}{2.0\times10^{-6}}=0.589, so θ=sin⁡−1(0.589)=36.1∘\theta=\sin^{-1}(0.589)=36.1^\circ.
(c) Highest order
Set sin⁡θ=1\sin\theta=1: mmax⁡=dλ=2.0×10−6589×10−9=3.4m_{\max}=\dfrac{d}{\lambda}=\dfrac{2.0\times10^{-6}}{589\times10^{-9}}=3.4. The highest whole-number order is m=3m=3.

Markers reward d=1/Nd=1/N in metres, sin⁡θ=mλ/d\sin\theta=m\lambda/d giving 36.1∘36.1^\circ and the maximum order m=3m=3 from m≤d/λm\le d/\lambda.

WACE 2021-style5 marks
Explain why a diffraction grating with many slits produces sharper, brighter spectral lines than a double slit, and explain why this makes a grating better for measuring wavelengths.
Show worked answer →

A 5 mark explanation needs the multi-slit interference argument and its practical consequence.

Sharper maxima
A maximum occurs only where light from every slit arrives in phase. With thousands of slits, even a tiny deviation from the exact angle dsin⁡θ=mλd\sin\theta=m\lambda causes waves from across the grating to fall out of phase and cancel, so the bright maxima are extremely narrow. A double slit, with only two contributions, gives broad fringes that fade gradually.
Brighter
Many slits also let through more light and concentrate it into those narrow maxima, so each line is bright as well as sharp.
Better measurement
Narrow, well-separated maxima mean closely spaced wavelengths land at clearly distinct angles, so a grating can resolve and measure wavelengths far more precisely than a double slit.

Markers reward the all-slits-in-phase narrowing, the increased brightness and the improved resolution for wavelength measurement.

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