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Understanding the Michelson Interferometer

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tony-francisTony Francis 03/09/2025 at 18:160 Comments

A Michelson interferometer splits a beam of light into two paths using a beamsplitter.  These paths are then reflected back and recombined, creating an interference pattern.  The intensity of this pattern depends on the path difference between the two beams

Superposition of Two Sine Waves: Resultant Intensity at the Michelson Interferometer Detector

Example: Two Sine Waves

Let's consider two sine waves representing the two beams:

Where:

The path difference (Δ) and phase difference (φ) are related by: φ = (2π/λ) * Δ, where λ is the wavelength.

Specific Cases:

  1. Path Difference Δ = 0 (φ = 0): Constructive Interference
    • y2 = A * sin(kx)
    • y_total = y1 + y2 = 2A * sin(kx)
    • The waves are in phase, and the resulting wave has double the amplitude (constructive interference). The intensity is maximum.
  2. Path Difference Δ = λ/2 (φ = π): Destructive Interference
    • y2 = A * sin(kx + π) = -A * sin(kx)
    • y_total = y1 + y2 = A * sin(kx) - A * sin(kx) = 0
    • The waves are completely out of phase, and they cancel each other out (destructive interference). The intensity is zero.
  3. Path Difference Δ = λ (φ = 2π): Constructive Interference
    • y2 = A * sin(kx + 2π) = A * sin(kx)
    • y_total = y1 + y2 = 2A * sin(kx)
    • The waves are in phase, and the resulting wave has double the amplitude (constructive interference). The intensity is maximum.

The detector output, representing the interference of the two waves, is visualized as an intensity variation. When you digitize the interference pattern of a monochromatic laser in a Michelson interferometer with a mirror moving at a constant velocity, here's what you can infer:

Key Relationship:

Detector Output:

Inference of the Sine Wave (recorded at the detector):

So far, we've focused on one frequency of light. In the next section, we'll see how the interferometer reacts to light with a mix of frequencies

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