Interference

12PHY - Wave Systems

Finn Le Sueur

2024

Ngā Whāinga Ako

  • Explain the interference of two wave patterns in terms of nodes, anti-nodes and path difference

Diffraction

  • When waves pass through an opening or bump against a barrier, they will diffract
  • Openings: The most diffraction occurs when the opening width and wavelength are similar

  • Barriers: Longer wavelengths diffract best around barriers e.g. AM radio has a longer wavelength than FM and can be better heard in the mountains
  • Notice that the wavelength does not change when diffraction occurs.

Constructive vs Destructive Interference

  • Recall the following:
    • In phase waves constructively interfere
    • Out of phase (180deg) waves destructively interfere
    • Waves not exactly (180deg) out of phase partially interfere

Two Point Source Interference

  • Two point sources with the same frequency and amplitude are placed near each other
  • Their waves will interfere to create points of zero amplitude, and points of 2x amplitude of the original wave

Interference of Light

  • For light to interfere we need:
    • Waves of the same frequency
    • Waves with the same amplitude
    • Waves with a stable phase relationship (coherent)
  • This is usually achieved by passing light from a single source through two very narrow and close slits.

Path Difference

  • Moving away from the centre point, the distance each travels becomes unequal
  • This means the waves will not keep a constant phase relationship
  • We call the difference in the distances that they travel the path difference

Worksheet

  • Collect a worksheet
  • Fill out the table by counting the number of wavelengths to each point
  • Use the table to determine the path difference for each of the anti-nodal lines
  • Highlight the statements in the two boxes on the second page
  • Finish the rest of the questions on the sheet

Calculating Path Difference

  • Path difference can be written as \(pd\)
  • Antinodal Lines: \(pd = n\lambda\)
  • Nodal Lines: \(pd = (n - \frac{1}{2})\lambda\)

Mahi Tuatahi

  • If you used the equations, verify by counting wavelengths on the diagram.
  • If you used the diagram, verify your answer by doing calculations.

Interference Formula

Assumption: The angle \(\theta\) is small, therefore \(\theta \approx \frac{x}{L}\)

This assumption lets us formulate this equation which we can use to calculate a variety of things:

\[ \begin{aligned} pd &= \frac{dx}{L} \end{aligned} \]

pd = path difference
d = distance between the slits
x = distance of the fringe (bright spot) from the center
L = distance between slits and screen

What does it tell us?

\[ \begin{aligned} pd &= \frac{dx}{L} \newline \text{Rearranging for x and substiting $pd = n\lambda$:} \newline x &= \frac{n\lambda L}{d} \end{aligned} \]

  • as wavelength (\(\lambda\)) increases, so does the fringe spacing (x)
  • as screen distance (L) increases, so does the fringe spacing (x)
  • as slit separation (d) increases, the fringe spacing decreases

Interference Formula

All this futzing leaves us with a series of formula:

\[ \begin{aligned} pd &= n\lambda \text{ antinodes} \newline pd &= (n- \frac{1}{2})\lambda \text{ nodes} \newline pd &= \frac{dx}{L} \newline \text{OR} \newline pd &= dsin(\theta) \newline &\text{ \^ assumes } \theta \approx \theta ' \end{aligned} \]

Practice

  • Take some time to try the questions from Textbook Activity 4B
  • Check your answers or find some help from the answers on Classroom