 Nonuniform oil film in monochromatic light

At normal incidence, the propagation phase is $4\pi n_{\mathrm{oil}}d(x,y)/\lambda$. Different film thicknesses therefore produce bright and dark contours of equal thickness, with adjacent same-type fringes differing by $\boxed{\Delta d=\lambda /(2n_{\mathrm{oil}})}$. Under monochromatic illumination these are intensity variations of the incident colour, rather than a rainbow.

For the usual case $n_{\mathrm{oil}}>n_{\mathrm{water}}>1$, reflection at air-oil reverses phase while reflection at oil-water does not. The additional relative $\pi$ makes reflected minima occur at $2n_{\mathrm{oil}}d=m\lambda$ and maxima at $2n_{\mathrm{oil}}d=(m+1/2)\lambda$. If the oil index were below the water index, the phase-reversal pattern would change and these conditions would interchange. The thinnest regions need not be completely dark because the two reflected amplitudes need not match; at vanishing thickness the residual reflection approaches that of the air-water boundary. Surface motion changes the thickness contours and moves the fringes.