Year 12 Physics Module 7 ⏱ ~40 min 5 MC · 2 Short Answer Lesson 3 of 14

Interference, Young's Double Slit

In 1801, Thomas Young put the wave theory of light and the principle of interference to the Royal Society, and demonstrated it by splitting one beam of light into two coherent sources. Where the beams overlapped he saw evenly spaced bright and dark bands, a pattern no stream of particles can produce. In a modern version of the demonstration, light of wavelength 589 nm through slits 0.5 mm apart, with a screen 1 m behind, gives a fringe spacing of about 1.17 mm. Young's work opened the case for the wave model; Newton's particle theory was displaced only over the following two decades.

Today's hook: Send light of wavelength 589 nm through two slits just 0.5 mm apart onto a screen 1 m away. If light were a stream of particles you would expect two bright stripes. Instead you get dozens of alternating bright and dark bands, spaced about 1.17 mm apart. Why does splitting light into two beams create more stripes, not fewer?
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Orient and predict

Predict what two slits do to a laser beam, then set your goals and vocabulary for interference.

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 Apply Application & data response Master Mastery tasks Exam Exam-style questions
Before you read, predict

Imagine shining a laser through two very thin parallel slits cut in a piece of cardboard, with a screen placed behind them.

Before reading on, answer:

  1. If light behaved as a stream of particles, what pattern would you expect on the screen?
  2. If light behaves as a wave, what additional pattern might appear? Why?
  3. What would happen to the pattern if you used red light instead of blue light?

Warm-up: In Young's double slit experiment, bright fringes (maxima) occur where the path difference from the two slits is:

Learning Intentions
goals

Know, Double Slit Experiment

  • Light passes through two slits and produces an interference pattern
  • Bright fringes: constructive interference
  • Dark fringes: destructive interference

Understand, Path Difference and Coherence

  • Path difference $= n\lambda$ (constructive), $(n+\frac{1}{2})\lambda$ (destructive)
  • Coherent sources: same frequency and constant phase relationship
  • Fringe spacing $\Delta x = \lambda L/d$

Can Do, Solve Interference Problems

  • Calculate fringe spacing from experimental parameters
  • Determine wavelength from measured fringe separation
  • Predict how changes to setup affect the pattern
Scan these before reading
vocab
InterferenceThe superposition of waves from different sources producing regions of reinforcement and cancellation.
Constructive interferenceWaves arrive in phase; amplitudes add, producing a bright fringe.
Destructive interferenceWaves arrive out of phase by half a wavelength; amplitudes cancel, producing a dark fringe.
Coherent sourcesSources with the same frequency and a constant phase relationship.
Fringe spacingThe distance $\Delta x$ between adjacent bright (or dark) fringes: $\Delta x = \lambda L/d$.
Cross-lesson links: L02 established that EM waves are transverse and require no medium. L03 shows the experimental evidence: Young's double-slit demonstration, from 1801 onwards, produced interference fringes, a phenomenon only waves can produce. In a modern version, 589 nm light through slits 0.5 mm apart with a 1 m screen gives a fringe spacing of about 1.17 mm. In L04 you extend this to single-slit diffraction and diffraction gratings, which give even sharper evidence for the wave model.

Core Content

2

See how two slits build a fringe pattern

Trace the apparatus, then link path difference to bright and dark fringes and to $\Delta x = \lambda L/d$.

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Young's Double Slit Experiment
+5 XP

The experiment that proved light is a wave

In 1801, Thomas Young passed monochromatic light through a single slit to create a coherent source, then through two closely spaced parallel slits. On a screen placed behind the slits, he observed a pattern of evenly spaced bright and dark fringes, an interference pattern.

This result is impossible to explain if light is a stream of particles. Particles would produce two bright patches, one behind each slit. The observed pattern of multiple fringes is the signature of wave behaviour: light waves from the two slits overlap, and where they meet in phase they reinforce (bright fringe), where they meet out of phase they cancel (dark fringe).

Laser Single slit Double slit Screen Constructive Constructive Destructive L (slit to screen) d

Figure 1, Young's double slit: coherent light from two slits overlaps on a screen, producing interference fringes.

Constructive interference (bright fringe) occurs when the path difference from the two slits to the screen is a whole number of wavelengths:

$$\text{Path difference} = n\lambda \quad (n = 0, 1, 2, \ldots)$$

Destructive interference (dark fringe) occurs when the path difference is a half-integer number of wavelengths:

$$\text{Path difference} = \left(n + \frac{1}{2}\right)\lambda \quad (n = 0, 1, 2, \ldots)$$

For small angles, the distance from the central maximum to the $n$th bright fringe is:

$$x_n = \dfrac{n\lambda L}{d}$$

The fringe spacing (distance between adjacent bright fringes) is therefore:

$$\Delta x = \dfrac{\lambda L}{d}$$

where $\lambda$ is the wavelength, $L$ is the slit-to-screen distance, and $d$ is the slit separation.

Stop and check

In a double slit experiment, $d = 0.20$ mm, $L = 1.5$ m, and the fringe spacing is 4.5 mm. Calculate the wavelength of the light. Is this light red, green, or blue?

Young's double slit produces alternating bright and dark interference fringes. Constructive (bright) when path difference $= n\lambda$; destructive (dark) when path difference $= (n+\tfrac{1}{2})\lambda$. Fringe spacing: $\Delta x = \lambda L/d$, wider fringes for longer $\lambda$, larger $L$, or smaller $d$.

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

In a double slit experiment, $d = 0.30$ mm, $L = 2.0$ m, $\lambda = 600$ nm. The fringe spacing $\Delta x$ is:

3

Work out what interference needs, and what it predicts

See why sources must be coherent, then read off how $\lambda$, $L$ and $d$ change the pattern.

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Conditions for Interference
+5 XP

Why ordinary light bulbs do not produce interference patterns

We just saw that Young's double slit produces interference fringes described by $\Delta x = \lambda L/d$. That raises a question: why can't two ordinary light bulbs produce the same pattern? This card answers it → the requirement for coherent sources with the same frequency and a constant phase relationship.

For a stable interference pattern to be observed, the two sources must be coherent:

  • They must have the same frequency (same wavelength).
  • They must maintain a constant phase relationship over time.

Ordinary light sources (like light bulbs or the Sun) are incoherent: atoms emit light randomly and independently, so the phase relationship between waves from different parts of the source fluctuates rapidly. No stable interference pattern can form.

Young solved this by using a single slit first. Light passing through the single slit spreads by diffraction, and the portion that reaches each of the two double slits comes from the same wavefront, guaranteeing coherence. Modern experiments use lasers, which are naturally coherent.

Double Slit Equations

$\Delta x = \dfrac{\lambda L}{d}$, Fringe spacing

$x_n = \dfrac{n\lambda L}{d}$, Position of $n$th bright fringe

$d\sin\theta = n\lambda$, Condition for bright fringes (exact)

$d\sin\theta = \left(n + \dfrac{1}{2}\right)\lambda$, Condition for dark fringes (exact)

Key predictions from $\Delta x = \lambda L/d$:

  • Longer wavelength ($\lambda$) → larger fringe spacing. Red light produces wider fringes than blue light.
  • Greater slit-to-screen distance ($L$) → larger fringe spacing.
  • Smaller slit separation ($d$) → larger fringe spacing.
  • If white light is used, the central fringe is white but outer fringes show rainbow colours (red outermost).
n=0 n=±1 n=±1 n=±2 n=±2 Δx

Figure 2, Fringe pattern on the screen. Bright fringes (constructive) alternate with dark regions (destructive). The central fringe is the brightest.

Stop and check

A student uses red light ($\lambda = 650$ nm) and observes a fringe spacing of 3.2 mm. She then switches to blue light ($\lambda = 450$ nm) without changing the apparatus. Predict the new fringe spacing. Explain why the pattern becomes harder to see if the slits are moved very far apart.

HSC Tip, Small Angle Approximation

The equation $\Delta x = \lambda L/d$ uses the small angle approximation ($\sin\theta \approx \tan\theta \approx \theta$ in radians). This is valid when the fringe spacing is much smaller than the slit-to-screen distance. For precise calculations at large angles, use the exact equation $d\sin\theta = n\lambda$ and $x = L\tan\theta$. A common exam trap: students forget to convert mm to m. Always check your units: $\lambda$ in metres, $d$ in metres, $L$ in metres gives $\Delta x$ in metres.

Coherent sources require the same frequency and a constant phase relationship. Incoherent sources (bulbs, Sun) produce randomly fluctuating phases, no stable fringe pattern. Young solved this with a single slit first; lasers are naturally coherent. Red light gives wider fringes than blue; larger $d$ gives narrower fringes.

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

Increasing the slit separation $d$ increases the fringe spacing $\Delta x$.

Two separate light bulbs cannot produce a stable interference pattern because they are incoherent.

Using red light instead of blue light (same apparatus) gives a larger fringe spacing.

4

Solve a full double slit calculation

Work fringe spacing, fringe position, an unknown wavelength and the white-light case step by step.

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Worked Example: Double Slit Calculations
+5 XP

From fringe spacing to wavelength

We just saw the coherence requirement and the fringe spacing formula. That raises a question: how do you apply $\Delta x = \lambda L/d$ to find an unknown wavelength, predict fringe positions, or deal with white light? This card answers it → worked calculations for each scenario.

Problem

Red laser light ($\lambda = 632.8$ nm) is directed at two slits separated by $d = 0.25$ mm. A screen is placed $L = 2.4$ m from the slits.

  1. Calculate the fringe spacing on the screen.
  2. Calculate the distance from the central maximum to the 4th bright fringe.
  3. The experiment is repeated with an unknown wavelength, and the fringe spacing is measured as 3.8 mm. Calculate the unknown wavelength.
  4. Explain what would be observed if white light were used instead of monochromatic light.
Solution
  1. Fringe spacing. $$\Delta x = \frac{\lambda L}{d} = \frac{(632.8 \times 10^{-9})(2.4)}{0.25 \times 10^{-3}} = 6.07 \times 10^{-3}\ \text{m} \approx \mathbf{6.1\ \text{mm}}$$
  2. 4th bright fringe. $$x_4 = 4\,\Delta x = 4 \times 6.07\ \text{mm} \approx \mathbf{24.3\ \text{mm}} \approx 2.4\ \text{cm from centre}$$
  3. Unknown wavelength. $$\lambda = \frac{\Delta x \cdot d}{L} = \frac{(3.8 \times 10^{-3})(0.25 \times 10^{-3})}{2.4} = 3.96 \times 10^{-7}\ \text{m} \approx \mathbf{396\ \text{nm}}\ \text{(near UV/violet)}$$
  4. White light. The central fringe ($n = 0$) is white because all wavelengths constructively interfere at the centre. Higher-order fringes show spectral colours red outermost (largest $\Delta x$), violet innermost (smallest $\Delta x$). Eventually the fringes overlap and blur into white again.
Stop and check

A double slit experiment produces fringes with spacing 2.5 mm using light of wavelength 500 nm. If the slit separation is halved and the screen distance is doubled, what is the new fringe spacing?

To find wavelength: rearrange to $\lambda = \Delta x \cdot d / L$. For white light: central fringe ($n = 0$) is white; outer fringes show a spectrum with red outermost (largest $\lambda$). Halving $d$ and doubling $L$ multiplies fringe spacing by 4.

Pause, write the highlighted rearrangement and white-light result into your book before the check below.

In a double slit experiment, $\lambda = 500$ nm, $d = 0.25$ mm, $L = 2.0$ m. The fringe spacing $\Delta x$ in mm is _____ mm.

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Apply and consolidate

Run your own fringe-spacing calculations, then reason out loud about why coherence matters.

Activity 1, Double Slit Simulator
ApplyBand 4

Explore how changing parameters affects the interference pattern

  1. Set $\lambda = 550$ nm (green), $d = 0.20$ mm, $L = 1.5$ m. Record the fringe spacing. Verify $\Delta x = \lambda L/d$ by hand.
  2. Change the wavelength to 450 nm (blue) without changing $d$ or $L$. What happens to the fringe spacing? Calculate the new spacing.
  3. Return to 550 nm. Double the slit separation to 0.40 mm. What happens to the fringe spacing? Explain why.
  4. A student wants to measure an unknown wavelength. She uses $d = 0.25$ mm and $L = 2.0$ m, and measures $\Delta x = 5.0$ mm. Calculate the unknown wavelength and identify its colour.
Activity 2, Reasoning About Coherence
UnderstandBand 5

Explain why coherence is essential for interference

  1. Explain in your own words why two separate light bulbs cannot produce a stable interference pattern, but the same bulb passed through two slits can.
  2. A student replaces the laser in Young's experiment with a sodium lamp (monochromatic but incoherent). Predict what pattern, if any, would appear on the screen and why.
  3. Explain why Young placed a single slit before the double slit in his original 1801 experiment.
6

Clear the two classic misconceptions

Fix what smaller $d$ really does, and what is actually happening at a dark fringe.

Misconceptions, Final Check
Wrong: "Moving the slits closer together (smaller $d$) would compress the fringes closer together."
Right: $\Delta x = \lambda L/d$. Decreasing $d$ increases $\Delta x$, the fringes spread further apart. This is counterintuitive but follows directly from the equation.
Wrong: "Dark fringes occur where no light arrives, the two beams don't reach those spots."
Right: Light from both slits reaches every point on the screen. At dark fringes, light from both paths arrives with equal amplitude but opposite phase, so they cancel by destructive interference.

Three of these statements about Young's double slit experiment are correct. Pick the odd one out.