Physics · Units 3 & 4

Diffraction and Interference of Light

Understand diffraction and interference of light the easy way, with plain English intuition, an interactive simulation, Young's double slit experiment, the fringe spacing formula, worked examples and an auto marked practice test. VCE Physics Units 3 and 4.

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Shine light through two tiny slits and instead of two bright lines you get a whole row of bright and dark stripes marching across the screen. Light bends as it squeezes through each gap, the two spreading beams overlap, and where they meet they either add up into a bright band or cancel out into a dark one. That simple overlap is the fingerprint of a wave, and it is the strongest everyday clue that light is a wave.

Diffraction: waves spread through gaps

When a wave passes through a narrow gap it does not stay in a tidy beam. It fans out the other side, like ripples spreading after they slip through a gap in a harbour wall. This spreading is called diffraction.

The amount of spreading depends on one thing: how the wavelength compares with the size of the gap.

  • When the wavelength is large compared with the gap, the wave spreads out a lot. A big λ/w\lambda / w ratio means strong diffraction.
  • When the wavelength is small compared with the gap, the wave barely bends at all and carries straight on.

So a wave only really fans out when the gap is roughly its own wavelength or smaller. That is why light, with its tiny wavelength, needs incredibly narrow slits before the spreading becomes obvious.

Interference: two waves overlap

Now send light through two narrow slits side by side. Each slit diffracts the light, so two spreading beams sweep out across the screen and overlap. Where they meet, the waves combine. This is interference, and it is what builds the striped pattern.

There are only two outcomes when the waves meet:

  • They arrive in step, peak on peak, and reinforce. The light is bright. This is constructive interference.
  • They arrive out of step, peak on trough, and cancel. The screen is dark. This is destructive interference.

Whether a point is bright or dark comes down to the path difference, the extra distance one wave travels compared with the other to reach that point.

light sourcedslit 1slit 2Lpath differencebright fringescreen

The light source on the left sends wavefronts out to both slits at once, so the two slits stay perfectly in step. Each slit then acts as a fresh source, sending its own wavefronts forward (the grey arcs). The two blue lines are the paths from the two slits meeting at one point. Because that point is a whole number of wavelengths from both slits, the waves arrive in step and you get a bright fringe. The red bars show the full pattern of bright bands on the screen.

The bright and dark rule

Everything comes down to the path difference. Count how many wavelengths of extra distance one wave travels compared with the other.

  • Bright (constructive) when the path difference is a whole number of wavelengths: nλn\lambda for n=0,1,2,…n = 0, 1, 2, \dots The centre of the screen (n=0n = 0, zero path difference) is the brightest fringe.
  • Dark (destructive) when the path difference is a half number of wavelengths: (n+12)λ\left(n + \tfrac{1}{2}\right)\lambda for n=0,1,2,…n = 0, 1, 2, \dots

A whole number means the peaks line up and reinforce. A half number means a peak meets a trough and they wipe each other out.

How far apart are the fringes?

For Young’s double slit setup there is a neat formula for the gap between one bright fringe and the next:

Δy=λLd\Delta y = \frac{\lambda L}{d}

Here λ\lambda is the wavelength, LL is the distance from the slits to the screen, and dd is the separation between the two slits. The formula works when the screen is much further away than the slits are apart, that is when L≫dL \gg d, which is almost always the case in the lab.

Read it like a story. A longer wavelength or a further screen spreads the fringes out. Wider slit separation squeezes them back together.

See it for yourself

Drag the sources, change the wavelength and the slit spacing, and watch the bright and dark bands shift. Try widening the slit separation and notice the fringes crowd closer together, then stretch the wavelength and watch them spread back out.

Interactive simulation, Wave Interference Source: PhET Interactive Simulations, University of Colorado Boulder (CC BY 4.0)

How to actually solve one

The method is a short, repeatable recipe.

  1. Convert every length into metres so nothing trips you up later.
  2. To decide bright or dark, write the path difference as a multiple of λ\lambda. A whole number is bright, a half number is dark.
  3. For fringe spacing, use Δy=λLd\Delta y = \dfrac{\lambda L}{d} straight off.
  4. Check the answer is sensible. Fringe spacings are usually a few millimetres, not metres.

Keep the wavelength tiny. Visible light sits around 55 to 7×10−77 \times 10^{-7} m, so if your λ\lambda comes out near a metre, something has gone wrong with the unit conversion.

Try one: at a point on the screen, light from the two slits has a path difference of 2.5λ2.5\lambda. Is the point bright or dark, and why?

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

When is diffraction most noticeable?
What is the difference between diffraction and interference?
State the constructive (bright) condition in terms of path difference.
State the destructive (dark) condition in terms of path difference.
Write the fringe spacing formula and say what each symbol is.
If the slit separation dd is increased, what happens to the fringes?
Recall · The Wave Model of Light
Why do all colours of light travel at the same speed in a vacuum?
Recall · Matter Waves
What does an electron diffraction ring pattern prove?

See the recipe in action in the Worked Examples tab, then test yourself in Try It.

Worked examples

Worked Example 1Fringe spacing in a double slit experiment

In a Young double slit experiment, light of wavelength 600600 nm passes through slits 0.200.20 mm apart onto a screen 2.02.0 m away. Find the spacing between adjacent bright fringes.

  1. 1

    First write every quantity in metres so the units cancel cleanly. The wavelength is 600600 nm, the slit separation is 0.200.20 mm and the screen distance is 2.02.0 m.

    λ=6.0×10−7 m,d=2.0×10−4 m,L=2.0 m\lambda = 6.0 \times 10^{-7} \text{ m}, \qquad d = 2.0 \times 10^{-4} \text{ m}, \qquad L = 2.0 \text{ m}
  2. 2

    The spacing between adjacent bright fringes is the wavelength times the screen distance, divided by the slit separation.

    Δy=λLd\Delta y = \frac{\lambda L}{d}
  3. 3

    Substitute the values and work it through.

    Δy=(6.0×10−7)(2.0)2.0×10−4=6.0×10−3 m\Delta y = \frac{(6.0 \times 10^{-7})(2.0)}{2.0 \times 10^{-4}} = 6.0 \times 10^{-3} \text{ m}
  4. 4

    Convert to millimetres so the answer is easy to picture on the screen.

    Δy=6.0×10−3 m=6.0 mm\Delta y = 6.0 \times 10^{-3} \text{ m} = 6.0 \text{ mm}
Answer
Δy=λLd=6.0×10−3 m=6.0 mm\Delta y = \dfrac{\lambda L}{d} = 6.0 \times 10^{-3} \text{ m} = 6.0 \text{ mm}
Worked Example 2Bright or dark from the path difference

At a point on the screen, the path difference between light from the two slits is 1.51.5 wavelengths. State whether the point is bright or dark, and explain.

  1. 1

    Write the path difference as a multiple of the wavelength so you can test it against the two conditions.

    path difference=1.5 λ\text{path difference} = 1.5\,\lambda
  2. 2

    A bright fringe needs a whole number of wavelengths, nλn\lambda. A dark fringe needs a half number, (n+12)λ\left(n + \tfrac{1}{2}\right)\lambda. Check which one 1.5 λ1.5\,\lambda matches.

    1.5 λ=(1+12)λ1.5\,\lambda = \left(1 + \tfrac{1}{2}\right)\lambda
  3. 3

    That fits the destructive condition with n=1n = 1, so the two waves arrive exactly out of step and cancel.

    (n+12)λ with n=1  ⟹  destructive\left(n + \tfrac{1}{2}\right)\lambda \text{ with } n = 1 \implies \text{destructive}
Answer
The point is dark (destructive interference).\text{The point is } \textbf{dark} \text{ (destructive interference).}

Practice questions

Practice test

Try it yourself

6 questions, 11 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.Diffraction, the spreading of a wave as it passes through a gap, becomes more noticeable when:

1mark
Need a hint?
Think about the ratio of wavelength to gap size. A large lambda over w means more spreading.

Q2.The pattern of bright and dark fringes in Young's double slit experiment is strong evidence that light:

1mark
Need a hint?
Only waves can add and cancel. Particles cannot produce dark bands by overlapping.

Q3.A point on the screen has a path difference of exactly 2λ2\lambda from the two slits. This point is:

1mark
Need a hint?
A whole number of wavelengths, n times lambda, gives constructive interference.

Q4.In a double slit experiment the fringe spacing is Δy=λLd\Delta y = \dfrac{\lambda L}{d}. If the slit separation dd is increased while everything else stays the same, the fringes will:

1mark
Need a hint?
d is on the bottom of the fraction. A bigger bottom makes the spacing smaller.

Q5.In a Young double slit experiment, light of wavelength 5.0×10−75.0 \times 10^{-7} m passes through slits 1.0×10−41.0 \times 10^{-4} m apart onto a screen 1.01.0 m away. Find the spacing between adjacent bright fringes. Show your working.

3marks

Work this on paper. The worked solution appears once you submit.

Q6.In a Young's double-slit experiment, green light of wavelength 550550 nm shines on two slits and the bright bands on a screen 2.002.00 m away are 2.752.75 mm apart. (a) Calculate the slit separation dd. (b) The green source is then replaced by one of lower frequency. Explain the effect on the spacing of the bands.

4marks

Work this on paper. The worked solution appears once you submit.

VCAA 2025 Physics Exam, Section B Q15

Frequently asked questions

What is the difference between diffraction and interference?
Diffraction is one wave spreading out as it passes through a gap or around an obstacle. Interference is what happens when two or more waves overlap and either add together or cancel out. In the double slit experiment the light diffracts at each slit first, and then the two spread out waves interfere on the screen.
Why are there dark fringes at all?
A dark fringe appears where the two waves arrive exactly out of step, with one peak landing on the other's trough. This happens when the path difference is a half number of wavelengths. The waves cancel, leaving no light, which is why interference is such strong evidence that light is a wave.
What does the fringe spacing formula tell me?
The formula delta y equals lambda L over d gives the distance between adjacent bright fringes. Longer wavelengths and a bigger screen distance spread the fringes out, while pushing the slits further apart squeezes them closer together. It only works when the screen is much further away than the slit separation.