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3.3.2 Double Slit Diffraction

By the end of this lesson you can:

  • Describe the double-slit experiment and the pattern it produces
  • Calculate wavelength or slit spacing, including for diffraction gratings
  • Explain why the pattern supports the wave model of light
Find your course: NSW Year 11NSW Year 12: Module 7, IQ2QLD Unit 4: Quantum theoryVIC Unit 4: Light, matter and relativityAP Physics 2: Unit 14IB: C.3 Wave phenomena
🎥 Watch📝 Practise🕹️ Explore🎯 Exam ready🍎 Teachers
🎥Watch
Learn the idea: a short explanation, then a check.

As we learned in the previous lesson, light was widely considered to be composed of particles—or "corpuscles"—at the start of the 19th century. This view was strongly supported by the scientific community, largely due to the enduring authority of Isaac Newton.

However, a pivotal experiment in the early 1800s dealt a severe blow to the corpuscular model.

Thomas Young designed an experiment to test the wave nature of light. What is now known as Young’s double-slit experiment provided compelling, empirical evidence that light behaves as a wave.

This lesson reviews the phenomena of diffraction and interference, demonstrating how they explain double-slit interference patterns to support the wave model of light. We will also examine the underlying mathematical models and explore how diffraction gratings allow for precise quantitative analysis of light.


Young's Double Slit Experiment

This video examines the basis of this experiment including a mathematical analysis for the behaviour.

In a rush?
Need a review?
This video provide a quick summary of the double slit experiment.

Diffraction Gratings

Instead of using one, two or three slits  to produce to diffraction and thus interference patterns (thereby demonstrating the way of nature of light), diffraction gratings are in essence many slits, like, well, a grating.
The result is much more defined diffraction patterns which increases the precision of the measurements.
This video discusses the physics principles behind the diffraction with some examples.
📝Practise
Worked examples first, then problems to try, then worksheets.

Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.

Picture

Check Your Understanding: Double Slit Interference

1. In a double slit experiment, what produces a bright fringe on the screen?
2. The two slits are moved closer together, and nothing else is changed. What happens to the fringe pattern on the screen?
3. A double slit is lit with red light and a pattern forms on a screen. The red light is replaced by blue light, with nothing else changed. What happens to the fringes?

Practice Problems: Double Slit Calculations

1 nm = 10-9 m and 1 mm = 10-3 m. The screen is far from the slits compared with the slit separation.

1. Light of wavelength 600 nm passes through two slits that are 0.20 mm apart. A screen is placed 2.0 m from the slits. What is the spacing between neighbouring bright fringes?

Given: λ = 600 nm = 6.00 × 10-7 m, d = 0.20 mm = 2.0 × 10-4 m, L = 2.0 m

Formula: Δx = λL / d

Calculation: Δx = (6.00 × 10-7 × 2.0) / (2.0 × 10-4) = 6.0 × 10-3 m

Answer: 6.0 mm
2. In a double slit experiment the slits are 0.50 mm apart and the screen is 1.5 m away. The distance from the central bright fringe to the fourth bright fringe is 7.2 mm. Calculate the wavelength of the light.

Given: d = 0.50 mm = 5.0 × 10-4 m, L = 1.5 m, 4 fringe spacings = 7.2 mm

Step 1 (Fringe spacing): Δx = 7.2 / 4 = 1.8 mm = 1.8 × 10-3 m

Step 2 (Wavelength): λ = Δx × d / L = (1.8 × 10-3 × 5.0 × 10-4) / 1.5 = 6.0 × 10-7 m

Answer: 6.0 × 10-7 m (600 nm)
3. Light of wavelength 500 nm is shone on two slits 0.010 mm apart. At what angle from the central axis is the second-order bright fringe (n = 2)?

Given: λ = 500 nm = 5.00 × 10-7 m, d = 0.010 mm = 1.0 × 10-5 m, n = 2

Formula: Bright fringes occur when d sinθ = nλ

Calculation: sinθ = nλ / d = (2 × 5.00 × 10-7) / (1.0 × 10-5) = 0.10

θ = sin-1(0.10) ≈ 5.7°

Answer: 5.7°
🕹️Explore
Test your predictions in a simulation.

Its now time for you to see the effect through this interactive (by Tom Walsh)

This interactive allows you to see the conditions when constructive takes place.
The dots align exactly when this takes place
Alter any of the sliders and see the effect of the superposition of the two waves.
It is important to appreciate the mathematical formula of mλ = dsinθ , that only certain combinations of Y, d, L and λ will result in constructive interference.
​You can do this by adjusting the various variables to generate constructive interference at the screen.
Interactive
​We can now explore the formula  mλ = dsinθ further. The animation lets you explore the effect of grating, wavelength and distance to the maxima separation.  (by Tom Walsh)
A good way to use this is to
  1. Set two variables yourself
  2. Measure the distance between the maxima
  3. Calculate the value for the remaining variable and compare it to the animation.