36  Light: The Wave Model

36.1 Syllabus inquiry question

  • What evidence supports the wave model of light?
Feynman Insight

From The Feynman Lectures on Physics, Vol I, Chapter 30:

Young’s double-slit experiment is one of the most beautiful demonstrations in physics. It shows decisively that light behaves as a wave—spreading out, overlapping, and creating a pattern that no particle model can explain.

36.2 Learning Objectives

  • Explain diffraction as evidence for wave behaviour
  • Apply the interference condition: \(d\sin\theta = m\lambda\)
  • Analyse diffraction grating patterns
  • Apply Malus’s law for polarisation: \(I = I_0\cos^2\theta\)
  • Compare Newton’s and Huygens’ models of light

36.3 Content

36.3.1 Historical Models of Light

Newton’s Corpuscular Theory (1670s): - Light consists of tiny particles (corpuscles) - Explains reflection and refraction (incorrectly assumed faster in denser media) - Cannot explain diffraction or interference

Huygens’ Wave Theory (1678): - Light is a wave in a medium (ether) - Each point on a wavefront is a source of secondary wavelets - Correctly explains reflection, refraction, and diffraction

The Critical Test

Wave theory predicts light is slower in denser media. Particle theory predicts it’s faster. Foucault (1850) proved light is slower in water—confirming wave theory.

36.3.2 Double-Slit Interference

When light passes through two narrow slits, it creates an interference pattern:

\[d\sin\theta = m\lambda\]

where: - \(d\) = slit separation (m) - \(\theta\) = angle to the maximum - \(m\) = order number (0, ±1, ±2, …) - \(\lambda\) = wavelength (m)

For bright fringes (constructive interference): \(m = 0, 1, 2, ...\) For dark fringes (destructive interference): path difference = \((m + \frac{1}{2})\lambda\)

36.3.3 Interactive: Double-Slit Interference

36.3.4 Path Difference

Constructive interference occurs when waves arrive in phase:

\[\Delta = m\lambda\]

Destructive interference occurs when waves arrive out of phase:

\[\Delta = (m + \frac{1}{2})\lambda\]

36.3.5 Diffraction Gratings

A diffraction grating has thousands of slits per millimetre, producing sharp, bright maxima:

\[d\sin\theta = m\lambda\]

The grating equation is identical to double-slit, but: - Maxima are much sharper (many slits reinforce) - Higher orders are clearly separated - Used for precise wavelength measurement

Grating Spacing

If a grating has N lines per mm, then: \[d = \frac{1}{N \times 1000}\ \text{metres}\]

36.3.6 Interactive: Diffraction Grating

36.3.7 Polarisation

Polarisation is the restriction of a wave’s oscillation to a single plane. Only transverse waves can be polarised—proving light is a transverse wave.

Malus’s Law: When polarised light passes through an analyser:

\[I = I_0\cos^2\theta\]

where: - \(I_0\) = intensity after first polariser - \(\theta\) = angle between polariser axes - \(I\) = transmitted intensity

Key Angles
  • \(\theta = 0°\): Maximum transmission (\(I = I_0\))
  • \(\theta = 45°\): Half transmission (\(I = 0.5I_0\))
  • \(\theta = 90°\): No transmission (\(I = 0\))—crossed polarisers

36.3.8 Interactive: Polarisation

36.4 Worked Examples

36.4.1 Example 1: Double-slit interference

Light of wavelength 600 nm passes through two slits 0.20 mm apart. Find the angle to the first-order maximum.

Solution:

  1. Use \(d\sin\theta = m\lambda\) with \(m = 1\)

  2. \(\sin\theta = (1 \times 6.0 \times 10^{-7})/(2.0 \times 10^{-4})\)

  3. \(\sin\theta = 0.003\), so \(\theta = 0.17°\)

36.4.2 Example 2: Diffraction grating

A grating with 600 lines/mm is used with 500 nm light. Find the angle for the second-order maximum.

Solution:

  1. \(d = 1/(600 \times 1000) = 1.67 \times 10^{-6}\ \text{m}\)

  2. \(\sin\theta = m\lambda/d = (2 \times 5.0 \times 10^{-7})/(1.67 \times 10^{-6})\)

  3. \(\sin\theta = 0.60\), so \(\theta = 36.9°\)

36.4.3 Example 3: Maximum order

For the grating in Example 2, find the highest order visible.

Solution:

  1. Maximum when \(\sin\theta = 1\) (θ = 90°)

  2. \(m_{max} = d/\lambda = (1.67 \times 10^{-6})/(5.0 \times 10^{-7}) = 3.33\)

  3. Since m must be an integer: \(m_{max} = 3\)

36.4.4 Example 4: Malus’s law

Polarised light of intensity 8.0 mW passes through an analyser at 60° to the polarisation direction. Find the transmitted intensity.

Solution:

  1. Use \(I = I_0\cos^2\theta\)

  2. \(I = 8.0 \times \cos^2(60°) = 8.0 \times (0.5)^2\)

  3. \(I = 8.0 \times 0.25 = 2.0\ \text{mW}\)

36.4.5 Example 5: Two polarisers

Unpolarised light passes through a polariser, reducing intensity to \(I_0\). It then passes through an analyser at 45°. Find the final intensity as a fraction of the original unpolarised intensity.

Solution:

  1. First polariser: intensity becomes \(I_0\) (half of unpolarised)

  2. Second polariser at 45°: \(I = I_0\cos^2(45°) = I_0 \times 0.5\)

  3. Final intensity = 0.5 × 0.5 = 0.25 of original unpolarised intensity

36.5 Common Misconceptions

Common Misconceptions

  • Misconception: Diffraction only occurs at edges. Correction: Diffraction occurs whenever a wave encounters an obstacle or aperture, but it’s most noticeable when the obstacle size is comparable to the wavelength.

  • Misconception: In double-slit, m = 0 is the first bright fringe. Correction: m = 0 is the central maximum. m = 1 is the first-order maximum (first bright fringe from centre).

  • Misconception: Gratings with more lines produce more light. Correction: More lines produce sharper maxima, not brighter ones overall. Total energy is conserved.

  • Misconception: Polarisation removes half the light. Correction: A single polariser removes half the intensity from unpolarised light, but Malus’s law applies to already polarised light.

  • Misconception: Light must be slowed by passing through a polariser. Correction: Speed is unaffected; only the electric field component perpendicular to the axis is blocked.

36.6 Practice Questions

36.6.1 Easy (2 marks)

Light of wavelength 550 nm passes through two slits 0.25 mm apart. Calculate the angle to the first-order maximum.

  • Use \(\sin\theta = m\lambda/d\) with m = 1 (1)
  • \(\sin\theta = (5.5 \times 10^{-7})/(2.5 \times 10^{-4}) = 0.0022\), θ = 0.13° (1)

Answer: θ = 0.13°

36.6.2 Medium (4 marks)

A diffraction grating with 500 lines/mm is illuminated with white light. Calculate the angular width of the first-order visible spectrum (400–700 nm).

  • \(d = 1/(500 \times 1000) = 2.0 \times 10^{-6}\) m (1)
  • For 400 nm: \(\sin\theta = (1 \times 4.0 \times 10^{-7})/(2.0 \times 10^{-6}) = 0.20\), θ = 11.5° (1)
  • For 700 nm: \(\sin\theta = (1 \times 7.0 \times 10^{-7})/(2.0 \times 10^{-6}) = 0.35\), θ = 20.5° (1)
  • Angular width = 20.5° - 11.5° = 9.0° (1)

Answer: Angular width = 9.0°

36.6.3 Hard (5 marks)

Unpolarised light passes through three polarisers. The first is vertical, the second is at 30° to vertical, and the third is horizontal. Calculate the fraction of the original intensity transmitted.

  • After first polariser: \(I_1 = I_0/2\) (1)
  • Second at 30°: \(I_2 = I_1 \cos^2(30°) = I_1 \times 0.75\) (1)
  • Third at 60° to second (90° - 30°): \(I_3 = I_2 \cos^2(60°) = I_2 \times 0.25\) (1)
  • Total: \(I_3 = 0.5 \times 0.75 \times 0.25 = 0.094\) (1)
  • Without middle polariser: crossed polarisers give zero—middle polariser allows some transmission (1)

Answer: 9.4% (or 0.094) of original intensity

36.7 Multiple Choice Questions

Test your understanding with these interactive questions:

36.8 Summary

Key Takeaways
  • Double-slit/grating: \(d\sin\theta = m\lambda\)
  • Constructive interference: path difference = mλ
  • Destructive interference: path difference = (m + ½)λ
  • Gratings produce sharp maxima; more lines = sharper peaks
  • Polarisation proves light is a transverse wave
  • Malus’s law: \(I = I_0\cos^2\theta\)
  • Foucault proved light slows in denser media → wave model confirmed

36.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: