Interference and Diffraction

By the end of this lesson, you will be able to describe constructive and destructive interference, explain the double-slit experiment, and apply diffraction concepts.

The Principle of Superposition

When two waves overlap, the total displacement is the sum of the individual displacements.

Constructive interference: waves in phase → amplitudes add → larger wave.

Destructive interference: waves out of phase (by 180°) → amplitudes subtract → cancellation.

Conditions for Interference

For two sources with the same wavelength, at a point P:

Constructive: path difference = $m\lambda$ ($m = 0, 1, 2, ...$)

Destructive: path difference = $(m + \frac{1}{2})\lambda$

Young's Double-Slit Experiment

Two slits separated by distance $d$, at distance $L$ from a screen.

Bright fringes (maxima) at: $y_m = \frac{m\lambda L}{d}$

Dark fringes (minima) at: $y_m = \frac{(m + \frac{1}{2})\lambda L}{d}$

This experiment definitively proved the wave nature of light (1801, Thomas Young).

Example — Double Slit

Two slits separated by 0.50 mm with 550 nm green light, 2.0 m from screen. Find fringe spacing.

$y_1 = \frac{1 \times 550 \times 10^{-9} \times 2.0}{0.50 \times 10^{-3}} = 2.2 \times 10^{-3}$ m = 2.2 mm

Diffraction

Diffraction: bending of waves around obstacles or through openings.

Significant when obstacle/opening size ≈ wavelength.

Single-slit diffraction: produces a central bright fringe (wider) flanked by narrower side fringes.

Diffraction grating: many slits; sharp, bright maxima at $d\sin\theta = m\lambda$.

Thin Film Interference

Light reflecting from the two surfaces of a thin film (like a soap bubble) interferes.

The condition for constructive/destructive interference depends on the film thickness and phase changes at the boundaries.

Why soap bubbles show colors: different wavelengths interfere constructively at different thicknesses → color patterns.

Summary

Superposition: waves add at every point.

Constructive: path difference = whole number of wavelengths.

Destructive: path difference = half-integer number of wavelengths.

Double-slit: $d\sin\theta = m\lambda$ for maxima; fringes equally spaced.