Interference — Young's Double Slit
In 1801, Thomas Young at the Royal Institution London shone a sodium lamp (589 nm) through two slits 0.5 mm apart and placed a screen 1 m behind them. He measured fringe spacing of 1.17 mm and calculated a wavelength of 585 nm — within 1% of the modern value. This was the death blow to Newton's particle theory and the decisive experimental proof that light behaves as a wave.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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:
- If light behaved as a stream of particles, what pattern would you expect on the screen?
- If light behaves as a wave, what additional pattern might appear? Why?
- 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:
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
Core Content
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).
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.
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:
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.
$\Delta x = \dfrac{\lambda L}{d}$ — 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).
Figure 2 — Fringe pattern on the screen. Bright fringes (constructive) alternate with dark regions (destructive). The central fringe is the brightest.
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.
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.
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.
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.
- Calculate the fringe spacing on the screen.
- Calculate the distance from the central maximum to the 4th bright fringe.
- The experiment is repeated with an unknown wavelength, and the fringe spacing is measured as 3.8 mm. Calculate the unknown wavelength.
- Explain what would be observed if white light were used instead of monochromatic light.
- Fringe spacing. $$\Delta x = \frac{\lambda L}{d} = \frac{(632.8 \times 10^{-9})(2.4)}{0.25 \times 10^{-3}} = 6.07 \times 10^{-3}\ \text{m} \approx \mathbf{6.1\ \text{mm}}$$
- 4th bright fringe. $$x_4 = 4\,\Delta x = 4 \times 6.07\ \text{mm} \approx \mathbf{24.3\ \text{mm}} \approx 2.4\ \text{cm from centre}$$
- Unknown wavelength. $$\lambda = \frac{\Delta x \cdot d}{L} = \frac{(3.8 \times 10^{-3})(0.25 \times 10^{-3})}{2.4} = 3.96 \times 10^{-7}\ \text{m} \approx \mathbf{396\ \text{nm}}\ \text{(near UV/violet)}$$
- 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.
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.
Explore how changing parameters affects the interference pattern
- 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.
- Change the wavelength to 450 nm (blue) without changing $d$ or $L$. What happens to the fringe spacing? Calculate the new spacing.
- Return to 550 nm. Double the slit separation to 0.40 mm. What happens to the fringe spacing? Explain why.
- 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.
Explain why coherence is essential for interference
- 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.
- 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.
- Explain why Young placed a single slit before the double slit in his original 1801 experiment.
Three of these statements about Young's double slit experiment are correct. Pick the odd one out.
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ApplyBand 4(3 marks) 1. In a double slit experiment, $d = 0.20$ mm, $L = 1.6$ m, and the 3rd bright fringe is observed 12 mm from the central maximum. (a) Calculate the fringe spacing. (b) Calculate the wavelength of the light used. (c) State whether this light is more likely red, green or violet.
1 mark: correct fringe spacing · 1 mark: correct wavelength · 1 mark: correct colour identification
AnalyseBand 5(5 marks) 2. (a) Explain why two separate light bulbs cannot produce a stable interference pattern but a single coherent source passing through two slits can. (b) In a double slit experiment the 3rd bright fringe is 12 mm from the central maximum, with $d = 0.20$ mm and $L = 1.6$ m. Calculate the wavelength. (c) The experiment is repeated in water ($n = 1.33$). Explain what happens to the fringe spacing.
1 mark: explain incoherence · 1 mark: explain why single source works · 1 mark: correct wavelength · 1 mark: fringe spacing decreases · 1 mark: correct reasoning (wavelength shortens in water)
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Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Short Answer — Model Answers
Q1 (3 marks): (a) $\Delta x = x_3 / 3 = 12/3 = 4.0$ mm (1 mark). (b) $\lambda = \Delta x \cdot d / L = (4.0 \times 10^{-3})(0.20 \times 10^{-3})/1.6 = 5.0 \times 10^{-7}$ m $= 500$ nm (1 mark). (c) 500 nm is green light (1 mark).
Q2 (5 marks): (a) Two separate bulbs are incoherent — atoms emit light randomly so the phase difference between the two sources fluctuates constantly, preventing a stable fringe pattern (1 mark). A single source through two slits creates two secondary sources from the same coherent wavefront, guaranteeing a fixed phase relationship (1 mark). (b) From Q1: $\lambda = 500$ nm (1 mark). (c) In water, the speed of light decreases by factor $n = 1.33$, so the wavelength decreases: $\lambda_{\text{water}} = \lambda / n = 500/1.33 \approx 376$ nm. Since $\Delta x = \lambda L/d$, the fringe spacing decreases (fringes move closer together) (1 mark for correct direction + 1 mark for correct reasoning).
At the start you were asked: if light behaved as particles, what pattern would appear? Two bright patches — one per slit — is the particle prediction.
Instead, in 1801 Thomas Young at the Royal Institution London observed many alternating bright and dark fringes through slits 0.5 mm apart on a screen 1 m away — measuring a fringe spacing of 1.17 mm and calculating a wavelength of 585 nm, within 1% of the modern sodium value of 589 nm. Young's 1801 Royal Institution experiment was the definitive refutation of Newton's particle theory. Since $\Delta x = \lambda L/d$, red light (larger $\lambda$) produces wider fringes than blue light (smaller $\lambda$).
Review your Think First predictions:
- Did you predict that particles would produce two bright patches (one per slit)? This is what Newton's particle theory predicted.
- Did you predict alternating bright and dark fringes for wave behaviour? This is the key signature of interference.
- Did you predict that red light would produce wider-spaced fringes than blue light? Since $\Delta x = \lambda L/d$, longer wavelength means larger spacing.
Extend: A double slit experiment uses monochromatic light of wavelength 480 nm with $d = 0.15$ mm and $L = 2.5$ m. Calculate the fringe spacing. If the experiment is then submerged in water ($n = 1.33$), what is the new fringe spacing? Explain the change using the wave model.
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