Michelson Interferometer

What you are seeing: the detector intensity of a Michelson interferometer as one mirror moves, producing path difference $L = 2d$ . The intensity oscillates as $I(L) = \tfrac12 (1 + V(L) \cos(2 \pi L / \lambda))$ , where the visibility $V(L) = e^{-(L/L_c)^2}$ falls off over a coherence length $L_c$ . Fringes are sharp near $L = 0$ and wash out as $L \gg L_c$ .

A laser ( $L_c \sim$ kilometers) maintains visibility over enormous paths; sunlight ( $L_c \sim 0.4$ um) decoheres after a few fringes. The readout reports the implied spectral linewidth $\Delta \nu \approx 0.44 c / L_c$ .

Figure 1. Michelson fringes and Gaussian coherence envelope.

lambda (nm)550

log10 L_c (nm)3.00

WHAT TO TRY

Slide the mirror across zero path difference: the fringes blaze brightest where the two arms are matched, then fade as you move away, the visibility envelope set by the source coherence length L_c.
Shorten L_c for a broadband source and the envelope narrows to a few fringes around zero; lengthen it toward a laser and fringes persist over millimetres of mirror travel.
Change the wavelength lambda: the fringe spacing scales with it, the optical ruler Michelson used to measure the metre in wavelengths of light.