Year 12 Physics Module 7 ⏱ ~40 min 5 MC · 2 Short Answer Lesson 4 of 14

Diffraction and Diffraction Gratings

In 1821 Joseph von Fraunhofer in Munich ruled a grating with 3,000 lines per centimetre and used it to measure the sodium D-line at 588.7 nm, within 0.05% of the modern value of 589.0 nm. He also catalogued 574 dark absorption lines in the solar spectrum. Fraunhofer's gratings turned light's tendency to bend through narrow openings into the most precise wavelength-measuring tool of the nineteenth century.

Today's hook: In 1821 Joseph von Fraunhofer in Munich ruled a grating with 3,000 lines per centimetre and measured the sodium D-line to 588.7 nm, more precisely than any prism could manage. Why does passing light through thousands of tiny slits give a sharper, more precise result than passing it through just one?

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Orient and predict

Predict what a laser does at a narrow slit, then set your goals and vocabulary for diffraction.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice

Before you read, predict

A laser beam passes through a single narrow slit and strikes a screen.

Before reading on, answer:

Would you expect to see a single bright line on the screen, or a spread-out pattern?
What would happen to the pattern if the slit were made narrower?
A diffraction grating has thousands of slits instead of one. How might the pattern differ from a single slit?

Warm-up, diffraction is most pronounced when the aperture width is…

Learning Intentions

goals

Know, Diffraction Patterns

Single slit diffraction: central maximum, minima at $a\sin\theta = n\lambda$
Diffraction gratings: many slits produce sharp maxima
Grating equation: $d\sin\theta = n\lambda$

Understand, Multiple Slits

More slits → narrower, brighter maxima
Why diffraction gratings are superior to prisms for spectroscopy
Diffraction limits optical resolution

Can Do, Analyse Spectra

Calculate angles for maxima using grating equation
Determine unknown wavelength from diffraction angles
Compare single slit, double slit and grating patterns

Scan these before reading

vocab

DiffractionThe spreading of waves as they pass through an aperture or around an obstacle.

Diffraction gratingAn optical component with many closely spaced parallel slits that disperses light into its component wavelengths.

Grating element ($d$)The distance between adjacent slits in a diffraction grating; $d = 1/N$ where $N$ is lines per metre.

Order ($n$)An integer (0, 1, 2, …) labelling each maximum in a diffraction pattern. $n = 0$ is the central maximum.

Resolving powerThe ability of an optical instrument to distinguish two closely spaced wavelengths; increases with more illuminated slits.

Cross-lesson links: L03 showed double-slit interference. L04 extends diffraction to single slits and gratings, Joseph von Fraunhofer's 1821 Munich grating (3,000 lines/cm, sodium D-line 588.7 nm) demonstrated that thousands of slits produce sharper maxima and higher resolving power than two. In L05 the same wave model explains polarisation, the final wave-model evidence block before the module turns to quantum phenomena.

Core Content

Spread light through one narrow slit

See why a narrower slit widens the pattern, and locate the minima with $a\sin\theta = n\lambda$.

Single Slit Diffraction

+5 XP

When one opening is enough to spread light

Shine a laser at a slit barely wider than the beam. Instead of a single sharp dot on the screen, you see a wide central bright band flanked by dimmer side bands, and the narrower you make the slit, the wider the central band grows. This counterintuitive result is single slit diffraction: when monochromatic light of wavelength $\lambda$ passes through a narrow slit of width $a$, it spreads out and produces a pattern of bright and dark bands on a screen.

The pattern consists of:

A wide, bright central maximum
Dimmer secondary maxima on either side
Minima (dark bands) where light from different parts of the slit cancels

The positions of the minima are given by:

$$a\sin\theta = n\lambda \quad (n = 1, 2, 3, \ldots)$$

where $a$ is the slit width, $\theta$ is the angle to the minimum, and $n$ is the order number. The central maximum is twice as wide as the secondary maxima.

For small angles, $\sin\theta \approx \theta \approx x/L$, so the width of the central maximum is approximately $2\lambda L/a$.

Figure 1, Single slit diffraction: wide central maximum with progressively dimmer side maxima

Key properties of single slit diffraction:

Narrower slit → wider diffraction pattern ($\theta \propto 1/a$)
Longer wavelength → wider pattern ($\theta \propto \lambda$)
The intensity of secondary maxima drops off rapidly

Single Slit Minima

$a\sin\theta = n\lambda$

$a$ = slit width (m) · $\theta$ = angle to minimum · $n$ = order (1, 2, 3, …) · $\lambda$ = wavelength (m)

Stop & Check

A slit of width 0.10 mm is illuminated with red light ($\lambda = 650$ nm). The screen is 2.0 m away. Calculate the width of the central maximum on the screen.

Single slit diffraction produces a wide central maximum flanked by dimmer side maxima. Minima at $a\sin\theta = n\lambda$ (where $n = 1, 2, 3\ldots$). Narrower slit → wider pattern ($\theta \propto 1/a$); longer wavelength → wider pattern. Width of central max $\approx 2\lambda L/a$.

Pause, copy the highlighted definition and formula into your book before moving on.

A single slit is illuminated by monochromatic light. The slit width is halved while keeping everything else the same. What happens to the width of the central diffraction maximum?

Use thousands of slits: the grating equation

Apply $d\sin\theta = n\lambda$, convert lines per mm into a grating element, and see why gratings beat prisms.

Diffraction Gratings

+5 XP

Thousands of slits produce precise spectra

We just saw that a single slit spreads light into a diffraction pattern. That raises a question: what happens when you use thousands of slits instead of one, and how does the pattern change? This card answers it → the diffraction grating equation $d\sin\theta = n\lambda$ and the production of sharp, bright spectral maxima.

A diffraction grating consists of many thousands of closely spaced parallel slits. When light passes through, each slit acts as a new source, and the waves from all slits interfere. The result is a series of very sharp, bright maxima separated by wide dark regions.

The condition for a maximum is the grating equation:

$$d\sin\theta = n\lambda \quad (n = 0, 1, 2, 3, \ldots)$$

where $d$ is the distance between adjacent slits (the grating element), $\theta$ is the angle of the maximum, and $n$ is the order. The grating element relates to lines per unit length $N$ by $d = 1/N$.

Diffraction Grating Equations

$d\sin\theta = n\lambda$

$d = 1/N$ grating element from lines per metre · $n$ = order of maximum (0, 1, 2, …)

Figure 2, Diffraction grating dispersing white light into spectra: violet deviated least, red most

Why diffraction gratings are better than prisms for spectroscopy:

The maxima are very sharp and narrow, making wavelength measurement precise.
The angular dispersion is linear ($\sin\theta \propto \lambda$), unlike prisms where dispersion is nonlinear.
The resolving power increases with the number of illuminated slits.
Different orders ($n = 1, 2, 3$…) give multiple spectra, allowing cross-checking of wavelengths.

When white light is passed through a diffraction grating, each wavelength is diffracted at a different angle. The result is a spectrum: violet (shortest $\lambda$) deviated least, red (longest $\lambda$) deviated most. The central maximum ($n = 0$) remains white for all wavelengths.

Stop & Check

A diffraction grating has 500 lines per mm. Calculate the grating element $d$. At what angle does the first-order maximum occur for green light ($\lambda = 550$ nm)? What is the highest order visible?

Diffraction grating: $d\sin\theta = n\lambda$ where $d = 1/N$ (N in lines/m). More slits → sharper, brighter maxima. White light: $n = 0$ is white; all other orders show a full spectrum (violet least deviated, red most). Maximum order: $n_\text{max} \leq d/\lambda$.

Add the highlighted grating equation and key results to your notes before the check below.

A diffraction grating has 600 lines/mm. What is the grating element $d$?

Measure a wavelength with a grating

Work a four-part example: wavelength, second-order angle, highest order, and what a finer grating does.

Worked Example: Grating and Wavelength

+5 XP

Using diffraction gratings to measure light

We just saw the grating equation and why gratings give sharper spectra than prisms. That raises a question: how do you apply $d\sin\theta = n\lambda$ to find unknown wavelengths, second-order angles, and maximum orders? This card answers it → a multi-part worked example with the grating equation.

Problem

A diffraction grating has 300 lines per mm. It is illuminated with monochromatic light, and the first-order maximum is observed at an angle of 18.5° from the central beam.

Calculate the wavelength of the light.
Calculate the angle of the second-order maximum for this wavelength.
Determine the highest order maximum that can be observed.
The experiment is repeated with a second grating of 600 lines/mm. Predict how the first-order angle changes.

Solution

Step 1: Wavelength

$d = 1/N = 1/(300 \times 10^3) = 3.33 \times 10^{-6}$ m

$\lambda = d\sin\theta/n = (3.33 \times 10^{-6})(\sin 18.5°)/1 = 1.06 \times 10^{-6}$ m = 1060 nm (infrared)

Step 2: Second-order angle

$\sin\theta_2 = 2\lambda/d = 2(1.06 \times 10^{-6})/(3.33 \times 10^{-6}) = 0.636$

$\theta_2 = \sin^{-1}(0.636) =$ 39.5°

Step 3: Maximum order

$n_\text{max} \leq d/\lambda = (3.33 \times 10^{-6})/(1.06 \times 10^{-6}) = 3.14$

Maximum observable order: n = 3

Step 4: Finer grating

With 600 lines/mm, $d' = d/2 = 1.67 \times 10^{-6}$ m.

$\sin\theta'_1 = \lambda/d' = 2\lambda/d = 0.636$, so $\theta'_1 =$ 39.5°, equal to the second-order angle from the original grating. Spectra are more spread out.

Figure 3, Orders produced by a diffraction grating: symmetric about the central ($n = 0$) maximum

HSC Tip, Grating vs Double Slit Confusion

Students often confuse $\Delta x = \lambda L/d$ (double slit fringe spacing on a screen) with $d\sin\theta = n\lambda$ (grating equation). The grating equation works for any number of slits and gives you angles. The fringe spacing formula only works for exactly two slits when you measure on a screen. A common trap: forgetting to convert lines/mm to lines/m before finding $d$.

Stop & Check

A student uses a diffraction grating with 400 lines/mm and observes a first-order maximum at 22.0°. Calculate the wavelength. If the grating is replaced with one of 800 lines/mm, at what angle is the first-order maximum now observed?

To find wavelength: $\lambda = d\sin\theta/n$. Maximum order: $n_\text{max} = \lfloor d/\lambda \rfloor$ (set $\sin\theta = 1$). Doubling lines/mm halves $d$, doubling the sine of each angle so spectra spread further apart. Always convert N from lines/mm to lines/m before finding $d$.

Add the highlighted procedure to your notes before the check below.

A diffraction grating with 400 lines/mm produces a first-order maximum at 15° for an unknown wavelength. What is the wavelength?

Compare one slit, two slits and a grating

Put the three intensity patterns side by side and say what each one reveals about the physics.

Comparing Diffraction Patterns

+5 XP

Single slit, double slit and diffraction grating side by side

We just saw the grating equation applied in practice. That raises a question: how do the intensity patterns of single slit, double slit, and diffraction gratings compare, and what are the key differences? This card answers it → a direct comparison of all three patterns and what each reveals about the physics.

Understanding how the pattern changes as you increase from one slit to two slits to thousands of slits is a key conceptual skill tested in the HSC.

Figure 4, Intensity patterns compared: single slit, double slit (with sinc envelope) and diffraction grating

Summary of pattern comparison:

Single slit: Broad central maximum, rapidly fading side maxima
Double slit: Interference fringes modulated by single-slit diffraction envelope; maxima are broader than grating
Diffraction grating: Extremely sharp, bright maxima with wide dark regions; excellent for wavelength measurement

Stop & Check

A student observes that when she uses a diffraction grating with more lines per mm, the spectral lines appear at larger angles but are still very sharp. Explain why (a) larger angles occur and (b) sharpness is maintained.

Single slit: broad central max with fading side maxima. Double slit: interference fringes modulated by single-slit envelope. Diffraction grating: extremely sharp, bright maxima, more slits means destructive interference fills wider regions between them, improving sharpness. Grating resolving power $R = nN$.

Pause, write the highlighted comparison into your book before the check below.

A diffraction grating with more lines per mm produces maxima at larger angles (for the same wavelength and order).

The central maximum ($n = 0$) of a diffraction grating shows a spectrum of colours when white light is used.

The maximum observable order is limited by the condition $\sin\theta \leq 1$, giving $n_\text{max} \leq d/\lambda$.

Revisit your thinking

Check your opening predictions against what you now know, then head into Practice.

Revisit Your Thinking

reflect

Look back at your Think First predictions and review them:

Did you predict a spread-out pattern with a central bright band? Single slit diffraction produces exactly this, with the central maximum twice as wide as the side maxima.
Did you predict that a narrower slit would produce a wider pattern? Since $a\sin\theta = \lambda$ (for $n = 1$), smaller $a$ means larger $\theta$, the light spreads more.
Did you predict that many slits would produce sharper, brighter maxima? With many slits, interference becomes much more directional, giving narrow bright lines separated by wide dark regions.

I've worked through all the Learn content

Independent practice

Show what you can do without prompts

Answer a shuffled question-bank set, then write two HSC-style responses on diffraction and gratings.

Short Answer

SAQ 1

Analyse and Explain

5 marks

(a) Explain why a diffraction grating produces sharper and more widely separated spectral lines than a double slit with the same slit separation.

(b) A grating with 600 lines/mm is illuminated with light of wavelength 520 nm. Calculate the angle of the second-order maximum.

SAQ 2

Calculate and Predict

4 marks

(a) A single slit of width 0.20 mm is illuminated with light of wavelength 600 nm. The screen is 1.5 m away. Calculate the distance from the centre of the pattern to the first minimum.

(b) If the slit width is halved to 0.10 mm, predict what happens to (i) the position of the first minimum and (ii) the intensity of the central maximum. Justify your answers.

Show model answers

SAQ 1 model answer

(a) A diffraction grating has thousands of slits (vs two for a double slit). With many more slits, the condition for constructive interference ($d\sin\theta = n\lambda$) must be satisfied simultaneously by all slits. The maxima therefore become extremely narrow and bright because destructive interference fills the wide angular regions between them. The angular spacing between orders increases as the grating element $d$ decreases (more lines → smaller $d$ → larger $\theta$ for each order).

(b) $d = 1/(600 \times 10^3) = 1.67 \times 10^{-6}$ m. For $n = 2$: $\sin\theta_2 = 2\lambda/d = 2(520 \times 10^{-9})/(1.67 \times 10^{-6}) = 0.623$. $\theta_2 = \sin^{-1}(0.623) \approx 38.5°$.

(c) $n_\text{max} \leq d/\lambda = (1.67 \times 10^{-6})/(520 \times 10^{-9}) = 3.21$. Highest observable order: $n = 3$.

SAQ 2 model answer

(a) Position of first minimum: $a\sin\theta = \lambda$ → $\sin\theta \approx \theta = \lambda/a = (600 \times 10^{-9})/(0.20 \times 10^{-3}) = 3.0 \times 10^{-3}$ rad. Distance $x = \theta L = 3.0 \times 10^{-3} \times 1.5 = 4.5 \times 10^{-3}$ m = 4.5 mm.

(b)(i) Halving $a$ doubles the angle to the first minimum: new position is $9.0$ mm. (ii) The central maximum is dimmer, half the slit width means half the area, so less light is transmitted through the slit and the intensity decreases.

Final step

Retrieve, reflect and finish

Check what stuck, revisit your opening predictions, stretch yourself, and record the rule you will reuse.

Check what actually stuck

How did your thinking change?

At the start you were asked whether a laser through a narrow slit produces a single bright dot or a spread-out pattern, and whether a narrower slit makes the pattern wider or narrower.

The answer: the pattern spreads wider as the slit narrows (the central maximum angle $\theta \approx \lambda/a$ increases as $a$ decreases). This is wave behaviour, particles would simply pass straight through.

This lesson's anchor in history is Fraunhofer's 1821 Munich grating: 3,000 lines per centimetre, measuring the sodium D-line at 588.7 nm, within 0.05% of the modern value. Using thousands of slits rather than one sharpens the maxima dramatically, giving the high resolving power needed to distinguish closely spaced spectral lines. Fraunhofer's grating work directly enabled the spectroscopic discoveries you will see in L09.

Beyond the syllabus. Resolving power $R = \lambda/\Delta\lambda$ and telescope-resolution treatment are extension, so this challenge card is optional practice. You are still examined on qualitative single-slit diffraction and on quantitative grating work with $d\sin\theta = n\lambda$ — exam marks come from the grating equation, not from resolving power.

Stretch, Going Deeper

+10 XP

Challenge problem: A sodium vapour lamp emits two closely spaced yellow lines at wavelengths 589.0 nm and 589.6 nm. A diffraction grating with 500 lines/mm is used in first order ($n = 1$).

Calculate the angle difference $\Delta\theta$ between the two lines in first order.
The resolving power needed to separate two lines is $R = \lambda/\Delta\lambda$. Calculate the minimum number of grating lines required to resolve this pair.
If the grating is 2.0 cm wide and has 500 lines/mm, does it have enough resolving power?

Lesson Summary

key ideas

Single slit diffraction: minima at $a\sin\theta = n\lambda$; narrower slit → wider central max ($\theta \propto 1/a$)
Diffraction grating: maxima at $d\sin\theta = n\lambda$ where $d = 1/N$; many slits → sharp, bright maxima
White light through grating: $n = 0$ is white; all other orders show a full spectrum (violet least deviated, red most)
Maximum order: $n_\text{max} \leq d/\lambda$ (from $\sin\theta \leq 1$)
Gratings vs prisms: linear dispersion, higher resolving power, precision spectroscopy

Lesson complete, I can apply the grating equation and explain diffraction patterns

← Interference and Young’s Double Slit Next: Polarisation of Light →